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Transparencies
Provides transparencies with answers for each lesson in the Student Edition

ISBN 0-07-828001-X

90000

9 780078 280016

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Chapter 1 Solving Equations and Inequalities
Lesson 1-1 Expressions and Formulas
Pages 8–10
1. First, find the sum of c and
d. Divide this sum by e.
Multiply the quotient by b.

3. b; The sum of the cost of adult and children tickets should be subtracted from
50. Therefore, parentheses need to be inserted around this sum to insure that this addition is done before subtraction. 4. 72

5. 6

6. 23

7. 1

14 Ϫ 4
5

8. Ϫ2

9. 119

10. 0

11. Ϫ23

12. 18

13. \$432

14. \$1875

15. \$1162.50

16. 20

17. 3

18. 29

19. 25

20. 54

21. Ϫ34

22. 19

23. 5

24. 11

25. Ϫ31

26. 7

27. 14

28. Ϫ15

29. Ϫ3

30. Ϫ52

31. 162

32. 15.3

33. 2.56

34. Ϫ7

35. 25

1
3

36. about 1.8 lb

37. 31.25 drops per min

38. 3.4

39. 2

40. 45

1

Algebra 2

Chapter 1

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41. Ϫ4.2

42. 5.3

43. Ϫ4

44. 75

45. 1.4

46. Ϫ4

47. Ϫ8

48. 36.01

49. 2

yϩ5 2 b 2

1
6

50. ␲ a

51. Ϫ16

52. 30

53. \$8266.03

54. 400 ft

4Ϫ4ϩ4Ϭ4ϭ1
4Ϭ4ϩ4Ϭ4ϭ2
14 ϩ 4 ϩ 42 Ϭ 4 ϭ 3
4 ϫ 14 Ϫ 42 ϩ 4 ϭ 4
14 ϫ 4 ϩ 42 Ϭ 4 ϭ 5
14 ϩ 42 Ϭ 4 ϩ 4 ϭ 6
44 Ϭ 4 Ϫ 4 ϭ 7
14 ϩ 42 ϫ 14 Ϭ 42 ϭ 8
4ϩ4ϩ4Ϭ4ϭ9
144 Ϫ 42 Ϭ 4 ϭ 10

56. Nurses use formulas to calculate a drug dosage given a supply dosage and a doctor’s drug order. They also use formulas to calculate IV flow rates.
Answers should include the following. • A table of IV flow rates is limited to those situations listed, while a formula can be used to find any IV flow rate. • If a formula used in a nursing setting is applied incorrectly, a patient could die. 57. C

58. D

59. 3

60. 4

61. 10

62. 13

63. Ϫ2

64. Ϫ5

65.

2
3

66.

2

6
7

Algebra 2

Chapter 1

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Lesson 1-2
1a.
1b.
1c.
1d.
1e.
1f.

Sample
Sample
Sample
Sample
Sample
Sample

Page 3 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:

2. A rational number is the ratio of two integers. Since 13 is
13
not an integer, is not a
2
rational number.

Properties of Real Numbers
Pages 14–18

3. 0; Zero does not have a multiplicative inverse since is undefined.

4. Z, Q, R

1
0

5. N, W, Z, Q, R

6. Q, R

7. Multiplicative Identity

8. Associative Property (ϩ)
1
8

10. 8, Ϫ

1
3

2
3

11. Ϫ , 3

12. Ϫ1.5,

13. Ϫ2x ϩ 4y

14. 13p

15. 3c ϩ 18d

16. Ϫ17a Ϫ 1

17. 1.5(10 ϩ 15 ϩ 12 ϩ 8 ϩ
19 ϩ 22 ϩ 31) or 1.5(10) ϩ
1.5(15) ϩ 1.5(12) ϩ 1.5(8) ϩ
1.5(19) ϩ 1.5(22) ϩ 1.5(31)

18. \$175.50

19. W, Z, Q, R

20. Q, R

21. N, W, Z, Q, R

22. Q, R

23. I, R

24. Z, Q, R

25. N, W, Z, Q, R

26. I, R

27. Q, R; 2.4, 2.49, 2.49, 2.49, 2.9

29. Associative Property (ϫ)

31. Associative Property (ϩ)

32. Commutative Property (ϩ)

33. Multiplicative Inverse

34. Distributive

35. Multiplicative Identity

36. 0

37. Ϫm; Additive Inverse

38.

3

1
;
m

Multiplicative Inverse
Algebra 2

Chapter 1

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41. 12 units
39. 1

40. natural numbers
42. The square root of 2 is irrational and therefore cannot be described by a natural number.

1
10

43. 10; Ϫ

44. Ϫ2.5; 0.4

45. 0.125; Ϫ8

46.

5
;
8

Ϫ

3
5

4 3
3 4

8
5
5
23

47. Ϫ ,

48. 4 , Ϫ

49. 3a Ϫ 2b

50. 10x ϩ 2y

51. 40x Ϫ 7y

52. 11m ϩ 10a

53. Ϫ12r ϩ 4t

54. 32c Ϫ 46d

55. Ϫ3.4m ϩ 1.8n

56. 4.4p Ϫ 2.9q

57. Ϫ8 ϩ 9y

58.

59. true

60. false; Ϫ3

61. false; 6

62. true

63. 6.5(4.5 ϩ 4.25 ϩ 5.25 ϩ
6.5 ϩ 5) or 6.5(4.5) ϩ
6.5(4.25) ϩ (6.5)5.25 ϩ
6.5(6.5) ϩ 6.5(5)

64. 3.6; \$327.60

4

9 x 10

Ϫ

19 y 6

Algebra 2

Chapter 1

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4

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1
8

65. 3 a2 b ϩ 2 a1 b

66. 50(47 ϩ 47); 50(47) ϩ 50(47)

1
4

1
8
Def. of a mixed number
1
1
3 122 ϩ 3 a b ϩ 2 112 ϩ 2 a b
4
8
Distributive
3
1
6ϩ ϩ2ϩ
Multiply.
4
4

ϭ 3 a2 ϩ b ϩ 2 a1 ϩ b ϭ ϭ

3
1
ϩ
4
4
3
1 ϭ8ϩ ϩ
4
4
1
3 ϭ8ϩa ϩ b
4
4

Comm. (ϩ)

ϭ 8 ϩ 1 or 9

ϭ6ϩ2ϩ

Assoc. (ϩ)

67. 4700 ft2

68. \$113(0.36 ϩ 0.19);
\$113(0.36) ϩ \$113(0.19)

69. \$62.15

70. Yes; ϭ ϩ ϭ 7;dividing
2
2
2
by a number is the same as multiplying by its reciprocal.

71. Answers should include the following. • Instead of doubling each coupon value and then adding these values together, the Distributive
Property could be applied allowing you to add the coupon values first and then double the sum.

72. B

6ϩ8

5

6

8

Algebra 2

Chapter 1

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• If a store had a 25% off sale on all merchandise, the Distributive Property could be used to calculate these savings. For example, the savings on a
\$15 shirt, \$40 pair of jeans, and \$25 pair of slacks could be calculated as 0.25(15) ϩ 0.25(40) ϩ
0.25(25) or as 0.25(15 ϩ
40 ϩ 25) using the
Distributive Property.
73. C

74. true

75. False; 0 Ϫ 1 ϭ Ϫ1, which is not a whole number.

76. true

2
3

77. False; 2 Ϭ 3 ϭ , which is not

78. 9

a whole number.
79. 6

80. Ϫ5

81. Ϫ2.75

82. 358 in2

83. Ϫ11

84.

85. Ϫ4.3

86. 36

7
10

Chapter 1
Practice Quiz 1
Page 18
1. 14

2. Ϫ9

3. 6

4. Ϫ1

5. 2 amperes

6. Q, R

7. N, W, Z, Q, R

6 7
7 6

9. Ϫ ,

10. 50x Ϫ 64y

6

Algebra 2

Chapter 1

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Lesson 1-3 Solving Equations
Pages 24–27
1. Sample answer: 2x ϭ Ϫ14

2. Sometimes true; only when the expression you are dividing by does not equal zero. 3. Jamal; his method can be confirmed by solving the equation using an alternative method. 4. 5 ϩ 4n

9
cC
5

ϩ

5
1322
9
5
1322 d
9

9
C
5

ϭ

5
1F Ϫ 322
9
5
5
F Ϫ 1322
9
9
5
F
9

ϭF

ϩ 32 ϭ F

5. 2n Ϫ n3

6. Sample answer: 9 times a number decreased by 3 is 6.

7. Sample answer: 5 plus
3 times the square of a number is twice that number.

8. Reflexive Property of
Equality
10. Ϫ21

9. Addition Property of Equality
11. 14

12. Ϫ4

13. Ϫ4.8

14. 1.5

15. 16

16. y ϭ

17. p ϭ

I rt 9 ϩ 2n
4

18. D

19. 5 ϩ 3n

20. 10n ϩ 7

21. n 2 Ϫ 4

22. Ϫ6n3

23. 519 ϩ n2

24. 21n ϩ 82

25.

a

26. 1n Ϫ 72 3

n 2 b 4

7

Algebra 2

Chapter 1

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28. 2␲r 1h ϩ r 2

27. 2␲rh ϩ 2␲r 2
29. Sample answer: 5 less than a number is 12.

30. Sample answer: Twice a number plus 3 is Ϫ1.

31. Sample answer: A number squared is equal to 4 times the number.

32. Sample answer: Three times the cube of a number is equal to the number plus 4.

33. Sample answer: A number divided by 4 is equal to twice the sum of that number and 1.

34. Sample answer: 7 minus half a number is equal to 3 divided by the square of x.

35. Substitution Property (ϭ)

36. Subtraction Property (ϭ)

37. Transitive Property (ϭ)

38. Addition Property (ϭ)

39. Symmetric Property (ϭ)

40. Multiplication Property (ϭ)

41. 7

42. 8

43. 3.2

44. 2.5

45.

3
4

1
12

46. Ϫ

47. Ϫ8

48. Ϫ11

49. Ϫ7

50.

51. 1

52. Ϫ12

53.

1
4

55. Ϫ

2
3

54. 19
10
17

55
2

56.

ϭr

58. a ϭ

57.

d t 59.

3V
␲r 2

60.

x 1c Ϫ 32 a ϩ2

2A h 62.

ϭh

61. b ϭ

Ϫb
2x

4x
1Ϫx

Ϫaϭb ϭy 63. n ϭ number of games;
2(1.50) ϩ n(2.50) ϭ 16.75; 5

64. s ϭ length of a side; 8s ϭ
124, 15.5 in.

65. x ϭ cost of gasoline per mile;
972 ϩ 114 ϩ 105 ϩ 7600x ϭ
1837; 8.5¢/mi

66. n ϭ number of students that can attend each meeting;
2n ϩ 3 ϭ 83; 40 students

8

Algebra 2

Chapter 1

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67. a ϭ Chun-Wei’s age; a ϩ
(2a ϩ 8) ϩ (2a ϩ 8 ϩ 3) ϭ 94;
Chun-Wei: 15 yrs old, mother:
38 yrs old, father: 41 yrs old

68. c ϭ cost per student;
50
50130 Ϫ c2 ϩ 1452 ϭ
5
1800; \$3

69. n ϭ number of lamps broken;
12(125) Ϫ 45n ϭ 1365;
3 lamps

70. h ϭ height of can A;

71. 15.1 mi/month

72. Central: 690 mi.; Union:
1085 mi

73. The Central Pacific had to lay their track through the
Rocky Mountains, while the
Union Pacific mainly built track over flat prairie.

74. \$295

75. the product of 3 and the difference of a number and 5 added to the product of four times the number and the sum of the number and 1

76. To find the most effective level of intensity for your workout, you need to use your age and 10-second pulse count. You must also be able to solve the formula given for A. Answers should include the following.
• Substitute 0.80 for I and
27 for P in the formula
I ϭ 6 ϫ P Ϭ 1220 Ϫ A2 and solve for A. To solve this equation, divide the product of 6 and 28 by 0.8.
Then subtract 220 and divide by Ϫ1. The result is
17.5. This means that this
1
person is 17 years old.

␲11.22 2h ϭ ␲122 23; 8 units
1
3

2

9

Algebra 2

Chapter 1

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• To find the intensity level for different values of A and P would require solving a new equation but using the same steps as described above. Solving for A would mean that for future calculations of A you would only need to simplify an expression, 220 Ϫ

6P
,
I

rather

than solve an equation.
77. B

78. D

79. Ϫ6x ϩ 8y ϩ 4z

80. 11a ϩ 8b

81. 6.6

82. 7.44

83. 105 cm2

84. Ϫ5

85. 3

86. Ϫ2.5
1
4

87. Ϫ

88. 3x

89. Ϫ5 ϩ 6y

Lesson 1-4

Solving Absolute Value Equations
Pages 30–32

1. 0 a 0 ϭ Ϫa when a is a negative number and the negative of a negative number is positive.

2a. 0 x 0 ϭ 4
2b. 0 x Ϫ 6 0 ϭ 2
4. Sample answer: 0 4 Ϫ 6 0 ; 2

3. Always; since the opposite of
0 is still 0, this equation has only one case, ax ϩ b ϭ 0.
The solution is

Ϫb
.
a

6. 9

5. 8

8. 5Ϫ21, 136

7. Ϫ17

10. 5Ϫ11, 296

9. 5Ϫ18, Ϫ126

11. 5Ϫ32, 366

12. л
10

Algebra 2

Chapter 1

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13. 586

14. 0 x Ϫ 160 0 ϭ 2

17. 15

18. 24

19. 0

20. 4

21. 3

22. 13

23. Ϫ4

24. Ϫ7.8

25. Ϫ9.4

26. 5

27. 55

28. Ϫ22

16. 162ЊF; This would ensure a minimum internal temperature of 160ЊF.

15. least: 158ЊF; greatest: 162ЊF

29. {8, 42}

30. 512, Ϫ306

33. 5Ϫ2, 166

34.

32. 5Ϫ28, 206

31. 5Ϫ45, 216

35.

3 e f
2

37.

e 2,

e 2,

16 f 3

Ϫ

36. л
38. 5Ϫ4, Ϫ16

9 f 2

39. л

40. л

41. {Ϫ5, 11}

42. {3, 15}

43.

11 eϪ ,
3

Ϫ3 f

44.

5 f 3

46. 5Ϫ46

45. {8}

48. 0 x Ϫ 16 0 ϭ 0.3; heaviest:
16.3 oz, lightest: 15.7 oz

47. 0 x Ϫ 200 0 ϭ 5; maximum:
205ЊF; minimum: 195ЊF
49. 0 x Ϫ 13 0 ϭ 5; maximum:
18 km, minimum: 8 km

50. sometimes; true only if a Ն 0 and b Ն 0 or if a Յ 0 and bՅ0 52. Answers should include the following. • This equation needs to show that the difference of the estimate E from the originally stated magnitude of 6.1 could be plus 0.3 or

51. sometimes; true only if c Ն 0

e 3,

11

Algebra 2

Chapter 1

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minus 0.3, as shown in the graph below. Instead of writing two equations,
E Ϫ 6.1 ϭ 0.3 and
E Ϫ 6.1 ϭ Ϫ0.3, absolute value symbols can be used to account for both possibilities, 0 E Ϫ 6.1 0 ϭ 0.3.
0.3 units

5.6

5.7

5.8

5.9

6.0

0.3 units

6.1

6.2

6.3

6.4

6.5

6.6

6.7

• Using an original magnitude of 5.9, the equation to represent the estimated extremes would be 0 E Ϫ 5.9 0 ϭ 0.3.

54. A

53. B

55. 0 x ϩ 1 0 ϩ 2 ϭ x ϩ 4;
0 x ϩ 1 0 ϩ 2 ϭ Ϫ1x ϩ 42

56. x ϩ 1 ϩ 2 ϭ x ϩ 4;
Ϫx Ϫ 1 ϩ 2 ϭ x ϩ 4; x ϩ 1 ϩ 2 ϭ Ϫx Ϫ 4;
Ϫx Ϫ 1 ϩ 2 ϭ Ϫx Ϫ 4

57. 5Ϫ1.56

58. 8
60. 5n 2

59. 21n Ϫ 112
16
3

68. false; 23

69. true

70. true

71. false; 1.2

72.

73. 364 ft2

74. 2

75. 8

76. Ϫ2

61.

62. Ϫ2

63. 14

64. Commutative Property (ϩ)

65. Distributive Property

66. Multiplicative Inverse

77.

2
3

1
1x
2

ϩ 321x ϩ 52

78. 6
3
4

79. Ϫ

12

Algebra 2

Chapter 1

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Lesson 1-5 Solving Inequalities
Pages 37–39
1. Dividing by a number is the same as multiplying by its inverse. 2. Sample answer: Ϫ2n Ͼ Ϫ6
4. 5a 0 a Ͻ 1.56 or 1Ϫϱ, 1.52

3. Sample answer: xϩ2Ͻxϩ1 Ϫ2

5
3

1

2

Ϫ10

9. 5p 0 p Ͼ 156 or 115, ϩϱ2

0

2

Ϫ6

26

Ϫ6

Ϫ5

Ϫ4

Ϫ3

Ϫ4

0

Ϫ2

18. 5d 0 d Ͼ Ϫ86 or 1Ϫ8, ϩϱ2
Ϫ10

28

Ϫ8

Ϫ6

Ϫ4

Ϫ2

20. 5p 0 p Յ Ϫ36 or 1Ϫϱ, Ϫ34

30

Ϫ6

21. 5k 0 k Ն Ϫ3.56 or 3Ϫ3.5, ϩϱ2
Ϫ7

Ϫ6

10 11 12 13 14 15 16 17 18 19 20 21

Ϫ4

19. 5g 0 g Յ 276 or 1Ϫϱ, 274
24

Ϫ8

16. 5b 0 b Յ 186 or 1Ϫϱ, 184

Ϫ1 0 1 2 3 4 5 6 7 8 9 10

22

8

14. at least 92

17. 5x 0 x Ͻ 76 or 1Ϫϱ, 72

20

6

4

15. 5n 0 n Ն Ϫ116 or 3Ϫ11, ϩ ϱ2
Ϫ8

4

12. 12n Ͼ 36; n Ͼ 3

13. 2n Ϫ 3 Յ 5; n Յ 4

Ϫ14 Ϫ12 Ϫ10

2

Ϫ30 Ϫ28 Ϫ26 Ϫ24 Ϫ22 Ϫ20

11. all real numbers or 1Ϫϱ, ϩϱ2
Ϫ2

3

10. 5n 0 n Յ Ϫ246 or 1Ϫϱ, Ϫ244

8 9 10 11 12 13 14 15 16 17 18 19

Ϫ4

2

8. 5w 0 w Ͻ Ϫ76 or 1Ϫϱ, Ϫ72

Ϫ1 0 1 2 3 4 5 6 7 8 9 10

Ϫ6

0

Ϫ2

3

7. 5y 0 y Ͼ 66 or 16, ϩϱ2

1

6. 5c 0 c Ն 36 or 33, ϩϱ2

5
3

5. e x ` x Յ f or aϪϱ, d f
0

0

Ϫ1

Ϫ4

0

Ϫ2

22. 5y 0 y Ͻ 56 or 1Ϫϱ, 52

Ϫ2

Ϫ4

13

Ϫ2

0

2

2

4

Algebra 2

0

4

6

Chapter 1

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23. 5m 0 m Ͼ Ϫ46 or 1Ϫ4, ϩϱ2
Ϫ6

Ϫ4

0

Ϫ2

2

2
3

4

0

Ϫ2

2

0.5

1

1.5

4

6

Ϫ2

2

2.5

Ϫ7

20

0

Ϫ2

Ϫ4

0

Ϫ2

4

6

1
20

or aϪ , ϩϱb

Ϫ1 Ϫ 3 Ϫ 1
4

20

1
20

20

2.0

2.2

2.4

5
7

2.6

2.8

4

5
7

0 1 2 3 4 5 6 1 8 9 10 11

1
5

3
5

Ϫ6

1

7 7

Ϫ4

Ϫ2

0

2

4

3
2

6

3
2

38. e n ` n Յ Ϫ f or aϪϱ, Ϫ d

37. л
Ϫ6

7 7

36. 5p 0 p Ͼ 06 or 10, ϩϱ2

1
5

35. e y ` y Ͻ f or aϪϱ, b
1
5

3.0

34. e a ` a Ն f or c , ϩϱb
7 7 7 7 7 7

3
5

3
20

32. 5z 0 z Ͼ 2.66 or 12.6, ϩ ϱ2

2

2

1
5

8

Ϫ20 Ϫ18 Ϫ16 Ϫ14 Ϫ12 Ϫ10

33. 5g 0 g Ͻ 26 or 1Ϫϱ, 22
Ϫ6

2

30. 5c 0 c Ͼ Ϫ186 or 1Ϫ18, ϩϱ2

31. 5d 0 d Ն Ϫ56 or 3Ϫ5, ϩϱ2
Ϫ4

2

1 f 20

Ϫ286 Ϫ284 Ϫ282 Ϫ280 Ϫ278 Ϫ276

Ϫ6

0

28. ew ` w Ͼ Ϫ

29. 5x 0 x Ͻ Ϫ2796 or 1Ϫϱ, Ϫ2792

Ϫ8

1

26. 5r 0 r Յ 66 or 1Ϫϱ, 64

27. 5n 0 n Ն 1.756 or 31.75, ϩϱ2
0

0

Ϫ1

25. 5t 0 t Յ 06 or 1Ϫϱ, 0 4
Ϫ4

2
3

24. e b ` b Ն f or c ϩϱb

Ϫ4

Ϫ2

0

2

4

Ϫ4

Ϫ3

Ϫ2

0

Ϫ1

39. at least 25 h

40. no more than 14 rides

41. n ϩ 8 Ͼ 2; n Ͼ Ϫ6

1

42. Ϫ4n Ն 35; n Յ Ϫ8.75

43.

1 n 2

44. Ϫ3n ϩ 1 Ͻ 16; n Ͼ Ϫ5

Ϫ 7 Ն 5; n Ն 24

n
2

46. n Ϫ 9 Յ ; n Յ 18

45. 21n ϩ 52 Յ 3n ϩ 11; n Ն Ϫ1
17

47. 217m2 Ն 17; m Ն , at least
14
2 child-care staff members

48. \$24,000 ϩ 0.015130,500n2 Ն
40,000
14

Algebra 2

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85 ϩ 91 ϩ 89 ϩ 94 ϩ s
5

49. n Ն 34.97; She must sell at least 35 cars.

50.

51. s Ն 91; Ahmik must score at least 91 on her next test to have an A test average.

52a. It holds only for Յ or Ն; 2 Х 2.
52b. 1 Ͻ 2 but 2 1
52c. For all real numbers a, b, and c, if a Ͻ b and b Ͻ c then a Ͻ c.

53. Answers should include the following. • 150 Ͻ 400
• Let n equal the number of minutes used. Write an expression representing the cost of Plan 1 and for Plan
2 for n minutes. The cost for Plan 1 would include a
\$35 monthly access fee plus 40¢ for each minute over 150 minutes or
35 ϩ 0.41n Ϫ 1502. The cost for Plan 2 for 400 minutes or less would be
\$55. To find where Plan 2 would cost less than Plan
1 solve 55 Ͻ 35 ϩ
0.41n Ϫ 1502 for n. The solution set is 5n 0 n Ͼ 2006, which means that for more than 200 minutes of calls,
Plan 2 is cheaper.

54. D

55. D

56. x Ͼ Ϫ3

57. x Ն Ϫ2

58. x Ն Ϫ1

59. 5Ϫ14, 206

60.

Ն 90

5 11 eϪ , f 4 4

61. л

62. b ϭ online browsers each year; 6b ϩ 19.2 ϭ 106.6; about 14.6 million online browsers each year

63. N, W, Z, Q, R

64. Q, R

15

Algebra 2

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65. I, R

66. 4.25(5.5 ϩ 8); 4.25(5.5) ϩ
4.25(8)

67. 5Ϫ7, 76
69.

e 4,

68. 513, Ϫ236
70. 511, 256

4 f 5

Ϫ

71. 5Ϫ11, Ϫ16

72. 5Ϫ18, 106

Chapter 1
Practice Quiz 2
Page 39
2s
t2

1. 0.5

2.

3. 14

4. eϪ , 5 f

ϭg
19
3

4
9

4
9

5. e m ` m Ͼ f or a , ϩ ϱb
Ϫ2
9

0

2
9

4
9

2
3

8
1
9

Lesson 1-6 Solving Compound and
Absolute Value Inequalities
Pages 43–46
1. 5 Յ c Յ 15

2. Sample answer: x Ͻ Ϫ3 and xϾ2 4. 0 n 0 Ͻ 8

3. Sabrina; an absolute value inequality of the form 0 a 0 Ͼ b should be rewritten as an or compound inequality, a Ͼ b or a Ͻ b.

Ϫ12

5. 0 n 0 Ͼ 3
Ϫ6

Ϫ4

7. 0 n 0 Ͻ 2

Ϫ2

0

2

Ϫ4

0

4

8

6. 0 n 0 Ն 4

4

8. 5y 0 y Ͼ 4 or y Ͻ Ϫ16
Ϫ4

Ϫ8

16

Ϫ2

0

2

4
Algebra 2

6
Chapter 1

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9. 5d 0 Ϫ2 Ͻ d Ͻ 36
Ϫ4

0

Ϫ2

10. 5a 0 a Ն 5 or a Յ Ϫ56
2

4

11. 5g 0 Ϫ13 Յ g Յ 56
Ϫ16 Ϫ12

Ϫ8

0

Ϫ4

6

Ϫ8

4

2

4

2

4

8

6

12

8

14. 55 Յ
Յ 60; 343.75 Յ c Յ
6.25
375; between \$343.75 and
\$375

6

16. 0 n 0 Ͻ 7

Ϫ8

Ϫ4

0

4

8

12

Ϫ4

Ϫ2

0

2

4

6

Ϫ8

Ϫ4

0

4

8

12

17. 0 n 0 Ͻ 4
19. 0 n 0 Ͼ 8
21. 0 n 0 Ͼ 1

Ϫ8

8

12

Ϫ4

0

4

8

12

Ϫ1.4 Ϫ1.2

0

1.2

1.4

1.6

4

6

8

2

4

6

2

4

6

22. 0 n 0 Յ 5
24. 0 n 0 Ͻ 6

0

4

29. 5x 0 Ϫ2 Ͻ x Ͻ 46
0

2

31. 5f 0 Ϫ7 Ͻ f Ͻ Ϫ56

26. 0 n Ϫ 1 0 Յ 3

28. 5t 0 1 Ͻ t Ͻ 36
8

4

12

Ϫ2

Ϫ4

Ϫ2

2

0

32. all real numbers

Ϫ8

Ϫ6

Ϫ4

Ϫ2

Ϫ8

Ϫ4

0

4

8

12

Ϫ4

Ϫ2

0

2

4

Ϫ4

0

6

33. 5g 0 Ϫ9 Յ g Յ 96

0

30. 5c 0 c Ͻ Ϫ2 or c Ն 16

6

Ϫ10

Ϫ2

0

34. 5m 0 m Ն 4 or m Յ Ϫ46
Ϫ4

Ϫ2

0

2

4

6

Ϫ8

Ϫ4

0

4

8

12

36. 5y 0 Ϫ7 Ͻ y Ͻ 76

35. л

4

20. 0 n 0 Յ 1.2

27. 5p 0 p Յ 2 or p Ն 86

Ϫ2

0

Ϫ8

25. 0 n ϩ 1 0 Ͼ 1

Ϫ4

Ϫ4

18. 0 n 0 Ն 6

23. 0 n 0 Ն 1.5

Ϫ4

0

Ϫ2

15. 0 n 0 Ն 5

Ϫ8

4

c

0

Ϫ2

0

12. 5k 0 Ϫ3 Ͻ k Ͻ 76

13. all real numbers
Ϫ4

Ϫ4

17

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38. 5r 0 Ϫ3 Ͻ r Ͻ 46

37. 5b 0 b Ͼ 10 or b Ͻ Ϫ26
0

Ϫ4

4

8

12

Ϫ4

7
3

39. e w ` Ϫ Յ w Յ 1 f
Ϫ2

1

en ` n

ϭ

0

43.

0

Ϫ2

2

4

6

4

6

Ϫ2

0

2

4

6

0

2

4

6

Ϫ4

Ϫ2

44. 5n 0 n Ͼ 1.56

7 f 2

1

2

42. 5n 0 n Ն 06

41. all real numbers
Ϫ4

0

40. л

0

Ϫ1

Ϫ2

Ϫ4

16

2

3

4

Ϫ2

5

45. 6.8 Ͻ x Ͻ 7.4

Ϫ1

0

1

2

3

46. 45 Յ s Յ 65

47. 45 Յ s Յ 55

48. 0 t Ϫ 98.6 0 Ն 8; 5b 0 b Ͼ 106.6 or b Ͻ 90.66

49. 108 in. Ͻ L ϩ D Յ 130 in.

50. 84 in. Ͻ L Յ 106 in.

51. a ϩ b Ͼ c, a ϩ c Ͼ b, bϩcϾa 52. a Ϫ b Ͻ c Ͻ a ϩ b

53a.
Ϫ4

Ϫ2

0

2

4

Ϫ4

Ϫ2

0

2

4

6

Ϫ4

Ϫ2

0

2

4

54. Compound inequalities can be used to describe the acceptable time frame for the fasting state before a glucose tolerance test is administered to a patient suspected of having diabetes. Answers should include the following.
• Use the word and when both inequalities must be satisfied. Use the word or when only one or the other of the inequalities must be satisfied.
• 10 Յ h Յ 16

6

6

53b.
53c.

53d. 3 Ͻ 0 x ϩ 2 0 Յ 8 can be rewritten as 0 x ϩ 2 0 Ͼ 3 and
0 x ϩ 2 0 Յ 8. The solution of
0 x ϩ 2 0 Ͼ 3 is x Ͼ 1 or x Ͻ Ϫ5. The solution of
0 x ϩ 2 0 Ն 8 is Ϫ10 Յ x Յ 6.
Therefore, the union of these two sets is 1x Ͼ 1 or x Ͻ Ϫ52

18

Algebra 2

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and 1Ϫ10 Յ x Յ 6). The union of the graph of x Ͼ 1 or x Ͻ Ϫ5 and the graph of
Ϫ10 Յ x Յ 6 is shown below. From this we can see that solution can be rewritten as 1Ϫ10 Յ x Ͻ Ϫ52 or
11 Ͻ x Յ 62.
Ϫ12

Ϫ8

0

Ϫ4

4

• 12 hours would be an acceptable fasting state for this test since it is part of the solution set of 10 Յ h
Յ 16, as indicated on the graph below.
8 9 10 11 12 13 14 15 16 17 18 19

8

55. x Ͼ Ϫ5 or x Ͻ Ϫ6

56. D

57.

58. 2 Ͻ x Ͻ 3

59. 15x ϩ 2 Ն 32 or
15x ϩ 2 Յ Ϫ32; 5x 0 x Ն 0.2 or x Յ Ϫ16

60. abs12x Ϫ 62 Ͼ 10;
5x Ϳ x Ͻ Ϫ2 or x Ͼ 86

61. d Ն Ϫ6 or 3Ϫ6, Ϫϱ2
Ϫ8

Ϫ6

Ϫ4

0

Ϫ2

2

63. n Ͻ Ϫ1 or 1Ϫϱ, Ϫ12
Ϫ4

Ϫ2

0

2

62. x Ͻ 4 or 1Ϫϱ, 42

4

Ϫ4

Ϫ2

0

2

4

6

64. 0 x Ϫ 587 0 ϭ 5; highest: 592 keys, lowest: 582 keys

6

66. 5Ϫ11, 46

65. {Ϫ10, 16}

68. Addition Property of Equality

67. л
69. Symmetric Property (ϭ)

70. Transitive Property of Equality

71. 3a ϩ 7b

72. Ϫ2m Ϫ 7n Ϫ 18

73. 2

74. 92

75. Ϫ7

19

Algebra 2

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Chapter 2 Linear Relations and Functions
Lesson 2-1 Relations and Functions
Pages 60–62

1. Sample answer: {(Ϫ4, 3),
(Ϫ2, 3), (1, 5), (Ϫ2, 1)}

y

x

O

3. Molly; to find g(2a), replace x with 2a. Teisha found
2g(a), not g(2a).

4. yes

5. yes

6. no

7. D ϭ {7}, R ϭ {Ϫ1, 2, 5, 8}, no

8. D ϭ {3, 4, 6}, R ϭ {2.5}, yes y (7, 8)

y

(7, 5)

(4, 2.5)
(3, 2.5)

(7, 2)

x

O
O

(7, Ϫ1)

(6, 2.5)

x

10. D ϭ 5x 0 x Ն 06,
R ϭ all reals, no

9. D ϭ all reals,
R ϭ all reals, yes

y

y

x ϭ y2
O

x

O

x

y ϭ Ϫ2x ϩ 1

12. Ϫ7

11. 10

20

Algebra 2

Chapter 2

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13. D ϭ {70, 72, 88},
R ϭ {95, 97, 105, 114}

14. {(88, 97), (70, 114), (88, 95),
(72, 105)}

15. Record High Temperatures

16. No; the domain value 88 is paired with two range values.

115

July

110
105
100
95
0

70
80
January

90

17. yes

18. no

19. no

20. yes

21. yes

22. no

23. D ϭ {Ϫ3, 1, 2},
R ϭ {0, 1, 5}; yes

24. D ϭ {3, 4, 6}, R ϭ {5}; yes y y

(3, 5) (4, 5)

(1, 5)

(6, 5)

(2, 1)

(Ϫ3, 0)

x

O

x

O

26. D ϭ {3, 4, 5, 6},
R ϭ {3, 4, 5, 6}; yes

25. D ϭ {Ϫ2, 3}, R ϭ {5, 7, 8}; no
(Ϫ2, 8)

y

(3, 7)

y
(5, 6)
(3, 4)

(Ϫ2, 5)

(6, 5)

(4, 3)
O

x

O

21

x

Algebra 2

Chapter 2

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27. D ϭ {Ϫ3.6, 0, 1.4, 2},
R ϭ {Ϫ3, Ϫ1.1, 2, 8}; yes

28. D ϭ {Ϫ2.5, Ϫ1, 0},
R ϭ {Ϫ1, 1}; no y y
(Ϫ3.6, 8)

(Ϫ2.5, 1) (Ϫ1, 1)
(0, 1)

x

O
(Ϫ1, Ϫ1)

(1.4, 2)

x
O
(0, Ϫ1.1)
(2, Ϫ3)

30. D ϭ all reals,
R ϭ all reals; yes

29. D ϭ all reals,
R ϭ all reals; yes

y

y

y ϭ 3x x O

x

O

y ϭ Ϫ5x

31. D ϭ all reals,
R ϭ all reals; yes

32. D ϭ all reals,
R ϭ all reals; yes y y

O
O

x

x

y ϭ 7x Ϫ 6

y ϭ 3x Ϫ 4

22

Algebra 2

Chapter 2

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34. D ϭ 5x 0 x Ն Ϫ36,
R ϭ all reals; no

33. D ϭ all reals,
R ϭ 5y 0 y Ն 06, yes y y

O

yϭx

x

2

x ϭ 2y 2 Ϫ 3

x

O

36. D ϭ {47, 48, 52, 56},
R ϭ {145, 147, 148, 157, 165}

35.
170
165
RBI

160
155
150
145
140
0

48

50

52

54

56

HR

37. No; the domain value 56 is paired with two different range values.

38. {(1997, 39), (1998, 43),
(1999, 48), (2000, 55),
(2001, 61), (2002, 52)}

39.

40. D ϭ {1997, 1998, 1999,
2000, 2001, 2002},
R ϭ {39, 43, 48, 52, 55, 61}

70

Stock Price

60
Price (\$)

50
40
30
20
10
0
1996

1998

2000 2002
Year

2004

42. {(1987, 12), (1989, 13),
(1991, 11), (1993, 12),
(1995, 9), (1997, 6), (1999, 3)}

41. Yes; each domain value is paired with only one range value. ©Glencoe/McGraw-Hill

23

Algebra 2

Chapter 2

PQ245-6457F-P02[020-055].qxd

43.
Representatives

14

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Page 24 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:

44. D ϭ {1987, 1989, 1991,
1993, 1995, 1997, 1999},
R ϭ {3, 6, 9, 11, 12, 13}

30+ Years of Service

12
10
8
6
4
2
0

’87

’91
’95
Year

’99

45. Yes; no; each domain value is paired with only one range value so the relation is a function, but the range value
12 is paired with two domain values so the function is not one-to-one. 46. Ϫ14

47. 6

48. Ϫ

49. Ϫ3

50. 3a Ϫ 5

51. 25n 2 Ϫ 5n

52. Ϫ4

53. 11

54. 39

55. f(x) ϭ 4x Ϫ 3

56. Relations and functions can be used to represent biological data. Answers should include the following.
• If the data are written as ordered pairs, then those ordered pairs are a relation. • The maximum lifetime of an animal is not a function of its average lifetime.

57. B

58. C

59. discrete

60. continuous

61. discrete

62. continuous

65. 5x 0 x Ͻ 5.16

66. \$2.85

2
9

63. 5y 0 Ϫ8 Ͻ y Ͻ 66

64. 5m 0 4 Ͻ m Ͻ 66
24

Algebra 2

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67. \$29.82

68. 43

69. 31a ϩ 10b

70. Ϫ1

71. 2

72. 6

73. 15

Lesson 2-2 Linear Equations
Pages 65–67
2. 5, Ϫ2

1. The function can be written as
1
2

f(x) ϭ x ϩ 1, so it is of the form f(x) ϭ mx ϩ b, where mϭ 1
2

and b ϭ 1.

3. Sample answer: x ϩ y ϭ 2

4. No, the variables have an exponent other than 1.

5. yes

6. 3x Ϫ y ϭ 5; 3, Ϫ1, 5

7. 2x Ϫ 5y ϭ 3; 2, Ϫ5, 3

8. 2x Ϫ 3y ϭ Ϫ3; 2, Ϫ3, Ϫ3

5
3

9. Ϫ , Ϫ5

10. 2, Ϫ2 y y

x

O

x

O

y ϭ Ϫ3x Ϫ 5

xϪyϪ2ϭ0

11. 2, 3

12. 3, y 3
2
y

3x ϩ 2y ϭ 6
O

x

x

O
4x ϩ 8y ϭ 12

25

Algebra 2

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13. \$177.62

14. 563.00 euros

15. yes

16. No; x appears in a denominator. 17. No; y is inside a square root.

18. No; x has exponents other than 1.

19. No; x appears in a denominator. 20. yes

21. No; x has an exponent other than 1.

22. No; x is inside a square root.

23. x 2 ϩ 5y ϭ 0

24. h(x ) ϭ x 3 Ϫ x 2 ϩ 3x

25. 7200 m

26. Sound travels only 1715 m in
5 seconds in air, so it travels faster underwater.

27. 3x ϩ y ϭ 4; 3, 1, 4

28. 12x Ϫ y ϭ 0; 12, Ϫ1, 0

29. x Ϫ 4y ϭ Ϫ5; 1, Ϫ4, Ϫ5

30. x Ϫ 7y ϭ 2; 1, Ϫ7, 2

31. 2x Ϫ y ϭ 5; 2, Ϫ1, 5

32. x Ϫ 2y ϭ Ϫ3; 1, Ϫ2, Ϫ3

33. x ϩ y ϭ 12; 1, 1, 12

34. x Ϫ y ϭ Ϫ6; 1, Ϫ1, Ϫ6

35. x ϭ 6; 1, 0, 6

36. y ϭ 40; 0, 1, 40

37. 25x ϩ 2y ϭ 9; 25, 2, 9

38. 5x Ϫ 4y ϭ 2; 5, Ϫ4, 2

39. 3, 5

40. 6, Ϫ2 y y

2x Ϫ 6y ϭ 12

5x ϩ 3y ϭ 15

x
O

O

26

x

Algebra 2

Chapter 2

PQ245-6457F-P02[020-055].qxd

41.

10
,
3

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Page 27 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:

5
2

42. 5, 2

Ϫ

y

y

3x Ϫ 4y Ϫ 10 ϭ 0

x

O

x

O

2x ϩ 5y Ϫ 10 ϭ 0

43. 0, 0

44. y 1
,
2

Ϫ2 y yϭx
O

x

x

O

y ϭ 4x Ϫ 2

45. none, Ϫ2

46. none, 4

y

y yϭ4 x

O

x

O

y ϭ Ϫ2

48. 1, none

47. 8, none
8
6
4
2
Ϫ8 Ϫ6Ϫ4 Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8

y

y

xϭ1

xϭ8
O
2 4 6

O

x

x

27

Algebra 2

Chapter 2

PQ245-6457F-P02[020-055].qxd

49.

1
,
4

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Page 28 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:

50. 6, Ϫ3

Ϫ1

g (x )

f (x ) f (x ) ϭ 4x Ϫ 1

g (x ) ϭ 0.5x Ϫ 3
O

x

O

51.

y

x

52. Sample answer: x ϩ y ϭ 2

x ϩy ϭ 5

x
O

xϩyϭ0 x ϩ y ϭ Ϫ5

The lines are parallel but have different y-intercepts.
53. 90ЊC

54. 4 km

55.

160
120
80
40

56. 1.75b ϩ 1.5c ϭ 525

T (d )

O1 2 3 4 d
Ϫ40
Ϫ80
Ϫ120 T (d ) ϭ 35d ϩ 20
Ϫ160

Ϫ4 Ϫ3Ϫ2

57.

58. Yes; the graph passes the vertical line test.

c
350
300
250
200
150
100
50
0

1.75b ϩ 1.5c ϭ 525

100

200

400b

28

Algebra 2

Chapter 2

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Page 29 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:

21
2

59. no

60.

61. A linear equation can be used to relate the amounts of time that a student spends on each of two subjects if the total amount of time is fixed. Answers should include the following.
• x and y must be nonnegative because
Lolita cannot spend a negative amount of time studying a subject.
• The intercepts represent
Lolita spending all of her time on one subject. The x-intercept represents her spending all of her time on math, and the y-intercept represents her spending all of her time on chemistry.
63. B

units2

62. B

64. D ϭ {Ϫ1, 1, 2, 4},
R ϭ {Ϫ4, 3, 5}; yes y (Ϫ1, 5) (
1, 3)

(4, 3)

x

O
(2, Ϫ4)

66. 5x 0 Ϫ1 Ͻ x Ͻ 26

65. D ϭ {0, 1, 2},
R ϭ {Ϫ1, 0, 2, 3}; no y (0, 2)

(1, 3)
(1, 0)

O

x
(2, Ϫ1)

29

Algebra 2

Chapter 2

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67. 5x 0 x Ͻ Ϫ6 or x Ͼ Ϫ26

68. \$7.95

69. 3s ϩ 14

70. 4
1
4

1
3

72. Ϫ

73. 2

74. Ϫ

75. Ϫ5

76.

77. 0.4

78. Ϫ0.8

71.

3
2

4
15

Lesson 2-3 Slope
Pages 71–74
1. Sample answer: y ϭ 1

2. Sometimes; the slope of a vertical line is undefined.

3. Luisa; Mark did not subtract in a consistent manner when using the slope formula. If y2 ϭ 5 and y1 ϭ 4, then x2 must be Ϫ1 and x1 must be
2, not vice versa.

4. 0

1
2

6. 1

5. Ϫ
7.

O

8.

y

x

y

O

30

x

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10.

y

y

x

O

O

11.

x

12. 5.5Њ/hr

y

O

x

13. 1.25Њ/hr

14. 2:00 P.M.–4:00 P.M.

5
2

16. 13

15. Ϫ
17.

3
5

18. 4

19. 0

20. Ϫ1

21. 8

22. undefined

23. Ϫ4

24. Ϫ

25. undefined

26. 0

27. 1

28. 9

31.

32.

5
4

y

y

O

O

x

x

31

Algebra 2

Chapter 2

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34.

y

y

x

O

O

35.

36.

y

x

O

x

y

O

x

37. about 68 million per year

38. about Ϫ32 million per year

39. The number of cassette tapes shipped has been decreasing. 40. 55 mph

41. 45 mph

42. speed or velocity

43.

O

45.

x

y

O

46.

y

O

44.

y

x

y

O

32

x

x

Algebra 2

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48.

y

O

49.

x

y

O

50.

y

x

y

x
O
O

51. Yes; slopes show that adjacent sides are perpendicular. 52. Ϫ1

53. The grade or steepness of a road can be interpreted mathematically as a slope.
Answers should include the following. • Think of the diagram at the beginning of the lesson as being in a coordinate plane. Then the rise is a change in y-coordinates and the horizontal distance is a change in x-coordinates. Thus, the grade is a slope expressed as a percent.

x

54. D

33

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y

x y ϭ 0.08x
O

55. D

56. The graphs have the same y-intercept. As the slopes increase, the lines get steeper. 57. The graphs have the same y-intercept. As the slopes become more negative, the lines get steeper.

58. Ϫ10, 4
8
Ϫ2x ϩ 5y ϭ 20 6
4
2
Ϫ10Ϫ8 Ϫ6Ϫ4 Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8

59. Ϫ2,

8
3

y

O
2 4x

60. 0, 0 y y

O

y ϭ 7x

x
O

4x Ϫ 3y ϩ 8 ϭ 0

61. Ϫ7

62. 5

5
2

63. Ϫ

64. 3a Ϫ4

65. 5x 0 Ϫ1 Ͻ x Ͻ 36

x

66. 5z 0 z Ն 7356

34

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67. at least 8

68. 17a Ϫb

69. 9

70. y ϭ 9 Ϫ x

71. y ϭ Ϫ4x ϩ 2

72. y ϭ Ϫ3x ϩ 7

5
2

73. y ϭ x Ϫ

1
2

2
3

75. y ϭ Ϫ x ϩ

3
5

74. y ϭ x ϩ

4
5

11
3

Chapter 2
Practice Quiz 1
Page 74
1. D ϭ {Ϫ7, Ϫ3, 0, 2},
R ϭ {Ϫ2, 1, 2, 4, 5}

2. 375

3. 6x ϩ y ϭ 4

4. 10, 6 y 3x ϩ 5y ϭ 30

x
O

5.

y
O

x

35

Algebra 2

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Lesson 2-4

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Writing Linear Equations
Pages 78–80

1. Sample answer: y ϭ 3x ϩ 2

2. 6, 0

3. Solve the equation for y to get
2
3 y ϭ x Ϫ . The slope of this

4. 2, Ϫ5

5

3
.
5

5

line is The slope of a parallel line is the same.
3
2

6. y ϭ 0.5x ϩ 1

5. Ϫ , 5

5
2

3
4

8. y ϭ Ϫ x ϩ 16

7. y ϭ Ϫ x ϩ 2
3
5

9. y ϭ Ϫ x ϩ

16
5

10. y ϭ Ϫx Ϫ 2

5
4

12. B

11. y ϭ x ϩ 7
2
3

14.

13. Ϫ , Ϫ4
15.

1
,
2

3
,
4

0
3
5

5
2

16. Ϫ , 6

Ϫ

17. undefined, none

18. Ϫc, d

19. y ϭ 0.8x

20. y ϭ Ϫ x ϩ

21. y ϭ Ϫ4

22. y ϭ 2

23. y ϭ 3x Ϫ 6

24. y ϭ 0.25x ϩ 4

1
2

25. y ϭ Ϫ x ϩ

5
3

3
2

7
2

26. y ϭ x ϩ

4
5

17
2

28. y ϭ 4x

27. y ϭ Ϫ0.5x Ϫ 2
29. y ϭ Ϫ x ϩ

29
3

17
5

30. no slope-intercept form for xϭ7 3
2

31. y ϭ 0

32. y ϭ x

33. y ϭ x ϩ 4

34. y ϭ x Ϫ

3
4

36

1
4

Algebra 2

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3

35. y ϭ x ϩ

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10
3

1
15

37. y ϭ Ϫ x Ϫ

36. y ϭ Ϫ4x ϩ 3
23
5

38. y ϭ Ϫx Ϫ 4

39. y ϭ 3x Ϫ 2

40. y ϭ Ϫ2x ϩ 6

41. d ϭ 180c Ϫ 360

42. 180, Ϫ360

43. 540Њ

44. y ϭ 75x ϩ 6000

45. 10 mi

46. y ϭ x ϩ 32

9
5

80
60
40
Ϫ30

Ϫ10
Ϫ20
Ϫ40

y y ϭ 9 x ϩ 32
5

x
O
10 20 30

47. 68ЊF

48. Ϫ40Њ

49. y ϭ 0.35x ϩ 1.25

50. \$11.75

51. y ϭ 2x ϩ 4

52. A linear equation can sometimes be used to relate a company’s cost to the number they produce of a product. Answers should include the following.
• The y-intercept, 5400, is the cost the company must pay if they produce
0 units, so it is the fixed cost. The slope, 1.37, means that it costs \$1.37 to produce each unit. The variable cost is 1.37x.
• \$6770

53. C x y
55.
5 Ϫ ϭ1

54. A

2

56.

5

57. Ϫ2

5
,
2

Ϫ5

58. 3
37

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59. 0

60. 0.55 s

62. 5x 0 x Ն Ϫ66

61. л

63. 5r 0 r Ն 66

64. 3

65. 6.5

66. 323.5

67. 5.85

Lesson 2-5 Modeling Real-World Data:
Using Scatter Plots
Pages 83–86
1. d

2. D ϭ {Ϫ1, 1, 2, 4},
R ϭ {0, 2, 3}; Sample answer using (Ϫ1, 0) and
(2, 2): 4

3. Sample answer using
(4, 130.0) and (6, 140.0): y ϭ 5x ϩ 110

4a.

Atmospheric Temperature
Temperature (˚C)

16
14
12
10
8
6
4
2
0

1000

2000 3000
Altitude (ft)

4000

5000

4b. Sample answer using
(2000, 11.0) and (3000, 9.1): y ϭ Ϫ0.0019x ϩ 14.8
4c. Sample answer: 5.3ЊC
5a.

6a.

60
50
40
30
20
10
0
’88 ’90 ’92 ’94 ’96 ’98 ’00
Year

Lives Saved by
Minimum Drinking Age
Lives (thousands)

Households (millions)

Cable Television
80
70

38

25
20
15
10
5
0
’94 ’95 ’96 ’97 ’98 ’99 ’00
Year

Algebra 2

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5b. Sample answer using
(1992, 57) and (1998, 67): y ϭ 1.67x Ϫ 3269.64
87 million

6b. Sample answer using
(1996, 16.5) and (1998, 18.2): y ϭ 0.85x Ϫ 1680.1
6c. Sample answer: 28,400

7a.

8a.

2000–2001
Detroit Red Wings

Bottled Water Consumption
14
12
Gallons

Assists

60
50
40
30
20
10
0

10
8
6
4
2
0

10 20 30 40
Goals

’91

’93

’95
Year

’97

’99

7b. Sample answer using (4, 5) and (32, 37): y ϭ 1.14x ϩ 0.44

8b. Sample answer using (1993,
9.4) and (1996, 12.5): y ϭ 1.03x Ϫ 2043.39

9a.

10. Sample answer using
(1990, 563) and (1995, 739): y ϭ 35.2x Ϫ 69,485

Revenue (\$ millions)

Play Revenue
700
600
500
400
300
200
100
0

1 2 3 4
Seasons Since ’95–’96

9b. Sample answer using
(1, 499) and (3, 588): y ϭ 44.5x ϩ 454.5, where x is the number of seasons since 1995–1996
\$1078 million or \$1.1 billion
11. Sample answer: \$1091

12. The value predicted by the equation is somewhat lower than the one given in the graph. 39

Algebra 2

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13. Sample answer: Using the data for August and
November, a prediction equation for Company 1 is y ϭ Ϫ0.86x ϩ 25.13, where x is the number of months since August. The negative slope suggests that the value of Company 1’s stock is going down. Using the data for October and November, a prediction equation for
Company 2 is y ϭ 0.38x ϩ
31.3, where x is the number of months since August. The positive slope suggests that the value of Company 2’s stock is going up. Since the value of Company 1’s stock appears to be going down, and the value of Company
2’s stock appears to be going up, Della should buy
Company 2.

14. No. Past performance is no guarantee of the future performance of a stock.
Other factors that should be considered include the companies’ earnings data and how much debt they have. 15.

16. Sample answer using
(213, 26) and (298, 23): y ϭ Ϫ0.04x Ϫ 34.52

Precipitation (in.)

World Cities
40
35
30
25
20
15
10
5
0

200
400
600
Elevation (ft)

18. Sample answer: The predicted value differs from the actual value by more than 20%, possibly because no line fits the data very well.

40

Algebra 2

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19. Sample answer using (1975,
62.5) and (1995, 81.7): 96.1%

20. Sample answer: The predicted percent is almost certainly too high. Since the percent cannot exceed
100%, it cannot continue to increase indefinitely at a linear rate.

21. See students’ work.

22. Data can be used to write a linear equation that approximates the number of
Calories burned per hour in terms of the speed that a person runs. Answers should include the following.

Calories Burned
While Running

Calories

1000
800
600
400
200
0

5 6 7 8
Speed (mph)

9

• Sample answer using
(5, 508) and (8, 858): y ϭ 116.67x Ϫ 75.35
• about 975 calories;
Sample answer: The predicted value differs from the actual value by only about 2%.
23. D

24. A

25. 1988, 1993, 1998; 247,
360.5, 461

26. y ϭ 21.4x Ϫ 42,296.2

27. 354

28. about (1993, 356.17)

29. y ϭ 21.4x Ϫ 42,294.03

31. y ϭ 4x ϩ 6

32. y ϭ Ϫ x Ϫ

33. 3

34. 7

3
7

41

6
7

Algebra 2

Chapter 2

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29
3

36.

37. 5x 0 x Ͻ Ϫ7 or x Ͼ Ϫ16

38. 3

39. 11
41.

37
3

40. 0

2
3

42. 1.5

Lesson 2-6 Special Functions
Pages 92–95
3. Sample answer: f(x) ϭ 0 x Ϫ 10

2. Ϫ1

5. S

6. D ϭ all reals,
R ϭ all integers

7. D ϭ all reals,
R ϭ all integers

8. D ϭ all reals,
R ϭ all nonnegative reals

1. Sample answer: [[1.9]] ϭ 1

4. A

g (x )

h (x )

g (x ) ϭ ͠2x ͡
O

h (x ) ϭ |x Ϫ 4| x O

42

x

Algebra 2

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10. D ϭ all reals, R ϭ 5y 0 y Յ 26

9. D ϭ all reals,
R ϭ all nonnegative reals

g (x )

x

O

11. D ϭ all reals, R ϭ all reals

12. step function

13.

14. \$6

0

Time (hr)

15. C

16. A

17. S

18. S

19. A

20. P

21.

5
4
3
2
1
O

22.

y

60

x
180 300

1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0

1 2 3 4 5 6 7 8 9
Minutes

C

43

C

Algebra 2

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23. \$1.00

24. D ϭ all reals,
R ϭ all integers f (x )

x

O

f (x ) ϭ ͠x ϩ 3͡

25. D ϭ all reals,
R ϭ all integers

26. D ϭ all reals,
R ϭ all even integers

g (x )

f (x )

g (x ) ϭ ͠x Ϫ 2͡ x O

x

O

f (x ) ϭ 2͠x ͡

27. D ϭ all reals,
R ϭ {3a 0 a is an integer.}
12
9
6

O
Ϫ4 Ϫ3 Ϫ2 Ϫ1
Ϫ3
Ϫ6
Ϫ9
Ϫ12

28. D ϭ all reals,
R ϭ all integers

h (x )

g (x )

h (x ) ϭ Ϫ3͠x ͡ x 1 2 3 4

x

O

g (x ) ϭ ͠x ͡ ϩ 3

29. D ϭ all reals,
R ϭ all integers

30. D ϭ all reals,
R ϭ all nonnegative reals

f (x ) f (x ) ϭ ͠x ͡ Ϫ 1
O

x

44

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32. D ϭ all reals, R ϭ {y 0 y Ն 3}

31. D ϭ all reals,
R ϭ all nonnegative reals

g (x )

h (x )

h (x ) ϭ |Ϫx |

g (x ) ϭ |x | ϩ 3

x

O

x

O

33. D ϭ all reals, R ϭ {y 0 y Ն Ϫ4}

34. D ϭ all reals,
R ϭ all nonnegative reals

g (x )

h (x )

g (x ) ϭ |x | Ϫ 4 x O

O

h (x ) ϭ |x ϩ 3|

x

36. D ϭ all reals,
R ϭ all nonnegative reals

35. D ϭ all reals,
R ϭ all nonnegative reals

f (x )

f (x )

f (x ) ϭ |x ϩ 2| x O

|

f (x ) ϭ x Ϫ 1
4

|

O

x

38. D ϭ all reals, R ϭ {y 0 y Ն Ϫ3}

37. D ϭ all reals,
R ϭ all nonnegative reals

f (x )

f (x )

O

|

f (x ) ϭ x ϩ 1
2

|

O

x

x

45

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39. D ϭ {x 0 x Ͻ Ϫ2 or x Ͼ 2},
R ϭ {Ϫ1, 1}

40. D ϭ all reals,
R ϭ {y 0 y Յ 0 or y ϭ 2} f (x )

h (x )

O

x

x

O

41. D ϭ all reals, R ϭ {y 0 y Ͻ 2}

42. D ϭ all reals,
R ϭ all nonnegative whole numbers g (x )

f (x )
O

x
O

x

f (x ) ϭ ͠|x |͡

43. D ϭ all reals,
R ϭ all nonnegative whole numbers 44.

g (x )

O

2 if x Ͻ Ϫ1 f(x) ϭ • 2x if Ϫ1 Յ x Յ 1
Ϫx if x Ͼ 1

x

g (x ) ϭ |͠x ͡|

45. f (x) ϭ 0x Ϫ 2 0

46. {x 0 x Ն 0}

47.

48.

46

f (x) ϭ e

0 if 0 Յ x Յ 300
0.8 (x Ϫ 300) if x Ͼ 300

Algebra 2

Chapter 2

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50. A step function can be used to model the cost of a letter in terms of its weight.
Answers should include the following. • Since the cost of a letter must be one of the values
\$0.34, \$0.55, \$0.76, \$0.97, and so on, a step function is the best model for the cost of mailing a letter. The gas mileage of a car can be any real number in an interval of real numbers, so it cannot be modeled by a step function. In other words, gas mileage is a continuous function of time.

y
|x | ϩ |y | ϭ 3

x

Cost (\$)

O

2.10
1.80
1.50
1.20
0.90
0.60
0.30

0

1 2 3 4 5 6 7
Weight (oz)

51. B

52. D

53. Life Expectancy

54. Sample answer using
(10, 69.7) and (47, 76.5): y ϭ 0.18x ϩ 67.9

78
76
74
72
70

Years Since 1950

55. Sample answer: 78.7 yr

56. y ϭ 3x ϩ 10

47

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58. {x 0 x Ն 3}

57. y ϭ x Ϫ 2

Ϫ1 0 1 2 3 4 5 6

59. e y ` y Ͼ f
5
6

60. yes

Ϫ3 Ϫ2 Ϫ1 0 1 2 3

61. no

62. no

63. yes

64. no

65. yes

Chapter 2
Practice Quiz 2
Page 95
2
3

1. y ϭ Ϫ x ϩ

11
3

2. Houston Comets
250
200
150
100
50
0

65 70 75 80
Height (in.)

4. Sample answer: 168 Ib

3. Sample answer using
(66, 138) and (74, 178): y ϭ 5x Ϫ 192
5. D ϭ all reals,
R ϭ nonnegative reals f (x )

f (x ) ϭ |x Ϫ 1|
O

x

48

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Lesson 2-7

Graphing Inequalities
Pages 98–99
2. Substitute the coordinates of a point not on the boundary into the inequality. If the inequality is satisfied, shade the region containing the point. If the inequality is not satisfied, shade the region that does not contain the point. 1. y Յ Ϫ3x ϩ 4

3. Sample answer: y Ն 0 x 0

4.

y

y ϭ2 x O

6.

5.

y

O

x

x Ϫy ϭ0

7.

8.

y

y

y ϭ |2x |

x

O

O

x

x Ϫ 2y ϭ 5

49

Algebra 2

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10. 10c ϩ 13d Յ 40

y

x
O

11.

y ϭ 3|x | Ϫ 1

12. No; (3, 2) is not in the shaded region.

d
10c ϩ 13d ϭ 40

c
O

13.

14.

y

y
3 ϭ x Ϫ 3y

x
O

x

O

x ϩ y ϭ Ϫ5

15.

16.

y y ϭ 6x Ϫ 2 x O

17.

18.

y

y

y ϭ Ϫ4x ϩ 3

O

x

O

x

y Ϫ 2 ϭ 3x

50

Algebra 2

Chapter 2

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20.

y

y

y ϭ1 y ϩ1ϭ4 x O

21.

x

O

22.

y

y

4x Ϫ 5y Ϫ 10 ϭ 0

x

O

x

O

x Ϫ 6y ϩ 3 ϭ 0

23.

24.

y

y

y ϭ 1x ϩ 5
3

O

y ϭ 1x Ϫ 5

x

2

O

25.

x

26.

y

y

y ϭ |4x | y ϭ |x | O

x

O

51

x

Algebra 2

Chapter 2

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28.

y

y y ϭ |x Ϫ 1| Ϫ 2

x

O

O

x

y ϩ |x | ϭ 3

29.

30.

y

y y ϭ |x |

x ϩy ϭ1
O

x

x

O

x ϩ y ϭ Ϫ1 y ϭ Ϫ|x |

31. x Ͻ Ϫ2

32. y Ͻ 3x Ϫ 5 y y x ϭ Ϫ2

O
O

x

x y ϭ 3x Ϫ 5

33.

34. yes

y
350
250
0.4x ϩ 0.6y ϭ 90

150
50
O

50

150 250 350 x

52

Algebra 2

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36.

35. 4a ϩ 3s Ն 2000

800

s

600

4a ϩ 3s ϭ 2000

400
200

a
200 400 600 800

O

38. 1.2a ϩ1.8b Ն 9000

37. yes

b
6000

1.2a ϩ 1.8b ϭ 9000

4000
2000
O

a
2000 4000 6000 8000

40.

39. yes

y

|y | ϭ x
O

42. A

41. Linear inequalities can be used to track the performance of players in fantasy football leagues.

x

53

Algebra 2

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• Let x be the number of receiving yards and let y be the number of touchdowns. The number of points Dana gets from receiving yards is 5x and the number of points he gets from touchdowns is
100y. His total number of points is 5x ϩ 100y. He wants at least 1000 points, so the inequality 5x ϩ
100y Ն 1000 represents the situation.

y
12

5x ϩ 100y ϭ 1000

10
8
6
4
2
O
Ϫ50

100 200

300 x

• the first one
43. B

44.

[Ϫ10, 10] scl: 1 by [Ϫ10, 10] scl: 1

45.

46.

[Ϫ10, 10] scl: 1 by [Ϫ10, 10] scl: 1

[Ϫ10, 10] scl: 1 by [Ϫ10, 10] scl: 1

54

Algebra 2

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47.

48. D ϭ all reals,
R ϭ all integers f (x )

x

O
[Ϫ10, 10] scl: 1 by [Ϫ10, 10] scl: 1

f (x ) ϭ ͠x ͡ Ϫ 4

49. D ϭ all reals, R ϭ {y 0 y Ն Ϫ1}

50. D ϭ all reals,
R ϭ all nonnegative reals

g (x )

h (x )

g (x ) ϭ |x | Ϫ 1
O

x
O

51.

x

52. Sample answer using
(4, 6000) and (6, 8000): y ϭ 1000x ϩ 2000

Sales vs. Experience
10,000
Sales (\$)

h (x ) ϭ |x Ϫ 3|

8000
6000
4000
2000
0

1

2

3 4
Years

5

6

7

53. Sample answer: \$10,000

54. 8

55. 3

56.

55

1
2

Algebra 2

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Chapter 3 Systems of Equations and Inequalities
Lesson 3-1 Solving Systems of Equations by Graphing
Pages 112–115
1. Two lines cannot intersect in exactly two points.

2. Sample answer: x ϩ y ϭ 4, x Ϫ y ϭ 2

3. A graph is used to estimate the solution. To determine that the point lies on both lines, you must check that it satisfies both equations.

4.

y
(Ϫ2, 5)

y ϭ Ϫx ϩ 3

y ϭ 2x ϩ 9

x

O

5.

6.

y
3x ϩ 2y ϭ 10

4x Ϫ 2y ϭ 22
O

(2, 2)
2x ϩ 3y ϭ 10

y

x

x
(4, Ϫ3)

6x ϩ 9y ϭ Ϫ3

O

8. inconsistent

7. consistent and independent y y
2x ϩ 4y ϭ 8

yϭxϩ4 x yϭ6Ϫx
O

O

x ϩ 2y ϭ 2

x

56

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9. consistent and dependent

10. y ϭ 0.08x ϩ 3.2, y ϭ 0.10x ϩ 2.6

y

x

O

x Ϫ 2y ϭ 8
1
xϪyϭ4
2

11. The cost is \$5.60 for both stores to develop 30 prints.

12. You should use Specialty
Photos if you are developing less than 30 prints, and you should use The Photo Lab if you are developing more than
30 prints.

13.

14.

y

y x O

y ϭ Ϫ3x ϩ 1

y ϭ 2x Ϫ 4

15.

y ϭ 3x Ϫ 8

x

O
(1, Ϫ2)

(0, Ϫ8)

16.

y

y
2x ϩ 3y ϭ 12 (3, 2)

x ϩ 2y ϭ 6
(4, 1)

O

y

x

O

x

2x ϩ y ϭ 9

17.

yϭxϪ8

2x Ϫ y ϭ 4

18.

x ϩ 2y ϭ 11

y

(7, 6)

7x Ϫ 1 ϭ 8y

(5, 3)
3x Ϫ 7y ϭ Ϫ6

O

x

O

x

5x Ϫ 11 ϭ 4y

57

Algebra 2

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20.

y

x

(1.5, 5)

4x Ϫ 2y ϭ Ϫ4

(3.5, 0)

2x ϩ 3y ϭ 7

y

O

x

O

8x Ϫ 3y ϭ Ϫ3

2x Ϫ 3y ϭ 7

21.

22.

y

1 x ϩ 2y ϭ 5
4

y yϪ 1xϭ6
3

(4, 2)

x

O

(Ϫ9, 3)

2x Ϫ y ϭ 6

O x
2
x ϩ y ϭ Ϫ3
3

23.

24.

y
1
xϪyϭ0
2

y

4 xϩ 1yϭ3
3
5

x

O

x

O
(Ϫ4, Ϫ2)

25. inconsistent yϭxϩ4 (3, Ϫ5)

2 xϪ 3yϭ5
3
5

1 x ϩ 1 y ϭ Ϫ2
4
2

26. consistent and independent

y

y yϭxϩ3 x

O

x

O

yϭxϪ4

y ϭ 2x ϩ 6

58

Algebra 2

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27. consistent and independent

28. consistent and dependent

y

y
3x ϩ y ϭ 3

xϩyϭ4

x

O

x
O
Ϫ4x ϩ y ϭ 9

6x ϩ 2y ϭ 6

29. inconsistent

30. consistent and dependent y y yϪxϭ5 4x Ϫ 2y ϭ 6

x

O

x
O

6x Ϫ 3y ϭ 9

2y Ϫ 2x ϭ 8

31. consistent and independent

32. inconsistent

y

y

2y ϭ x
2y ϭ 5 Ϫ x

x

O

x

8y ϭ 2x ϩ 1

O
6y ϭ 7 Ϫ 3x

33. consistent and independent

34. consistent and dependent

y

y
1.6y ϭ 0.4x ϩ 1

0.8x Ϫ 1.5y ϭ Ϫ10

O

1.2x ϩ 2.5y ϭ 4
O

x

0.4y ϭ 0.1x ϩ 0.25

x

59

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35. inconsistent yϪ 1xϭ2
3

36. consistent and independent

y

y
4
x Ϫ y ϭ Ϫ2
3

O

x

x

O

3y Ϫ x ϭ Ϫ2
2y Ϫ 4x ϭ 3

37. (Ϫ3, 1)

38. (1, 3), (2, Ϫ1), (Ϫ2, Ϫ3) y y Ϫ 2x ϭ 1

4x ϩ y ϭ 7

x

O

2y Ϫ x ϭ Ϫ4

40. (120, 80)

39. y ϭ 52 ϩ 0.23x, y ϭ 80

120
Cost (\$)

y ϭ 80
80

y ϭ 52 ϩ 0.23x

40

0

40

80
120
Miles

160

41. Deluxe Plan

42. Supply, 200,000; demand,
300,000; prices will tend to rise. 43. Supply, 300,000; demand,
200,000; prices will tend to fall. 44. 250,000; \$10.00

60

Algebra 2

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46. 2015
Population (Thousands)

45. y ϭ 304x ϩ 15,982, y ϭ 98.6x ϩ18,976

24,000

y ϭ 98.6x ϩ 18,976

20,000
16,000

y ϭ 304x ϩ 15,982

12,000
0

a b a
48b.
b a 48c. b 4

d e d e d
,
e

c b 8
12
16
Years After 1999

20

f e 47. FL will probably be ranked third by 2020. The graphs intersect in the year 2015, so
NY will still have a higher population in 2010, but FL will have a higher population in 2020.

48a. ϭ ,

49. You can use a system of equations to track sales and make predictions about future growth based on past performance and trends in the graphs. Answers should include the following.
• The coordinates (6, 54) represent that 6 years after 1999 both the instore sales and online sales will be \$54,000.
• The in-store sales and the online sales will never be equal and in-store sales will continue to be higher than online sales.

50. A

51. C

52. (3.40, Ϫ2.58)

53. (Ϫ5.56, Ϫ12)

54. (4, 3.42)

55. no solution

56. (Ϫ9, 3.75)

61

ϭ

ϭ

c b f e Algebra 2

Chapter 3

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58.

57. (2.64, 42.43)

y y ϭ 5 ϩ 3x

O

59.

60.

y

x

y

2x ϩ y ϭ Ϫ4

x

O

O

x

2y Ϫ 1 ϭ x

61. A

62. C

63. S

64. {Ϫ13, 13}

65. {Ϫ15, 9}

66. л

67. {Ϫ2, 3}

68. eϪ5, f

69. {9}

70. 8 ϩ 2n

71. x 2 Ϫ 6

72. 41a ϩ 52

73.

z
3

7
2

74. x ϩ 2

ϩ1

75. 9y ϩ 1

76. Ϫ3x ϩ 6y

77. 12x ϩ 18y Ϫ 6

78. 15x ϩ 10y ϩ 10

79. x ϩ 4y

Lesson 3-2 Solving Systems of Equations Algebraically
Pages 119–122
1. See students’ work; one equation should have a variable with a coefficient of 1.

2. There are infinitely many solutions. 62

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3. Vincent; Juanita subtracted the two equations incorrectly;
Ϫy Ϫ y ϭϪ2y, not 0.

4. (4, 8)

5. (1, 3)

6. (4, Ϫ1)

7. (5, 2)

8. (9, 7)
10. no solution

9. (6, Ϫ20)
1
3

2
3

11. a3 , 2 b

12. C

13. (9, 5)

14. (2, 7)

15. (3, Ϫ2)

16. (Ϫ6, 8)

17. no solution

18. (1, 1)

19. (4, 3)

20. (Ϫ1, 8)

21. (2, 0)

22. (3, Ϫ1)

23. (10, Ϫ1)

24. (Ϫ7, 9)

25. (4, Ϫ3)

26. (6, 5)

27. (Ϫ8, Ϫ3)

28. (7, Ϫ1)

29. no solution

30. (Ϫ5, 8)

1
2

31. aϪ ,

3 b 2

1
3

32. a , 2b

33. (Ϫ6, 11)

34. infinitely many

35. (1.5, 0.5)

36. (2, 4)

37. 8, 6

38. 2, 12

39. x ϩ y ϭ 28,
16x ϩ 19y ϭ 478

40. 18 members rented skis and 10 members rented snowboards. 41. 4 2-bedroom, 2 3-bedroom

42. (Ϫ5, Ϫ2), (4, 4),
(Ϫ2, Ϫ8), (1, 10)

43. x ϩ y ϭ 30,
700x ϩ 200y ϭ 15,000

44. 18 printers, 12 monitors

45. 2x ϩ 4y ϭ 100, y ϭ 2x

46. 10 true/false, 20 multiplechoice

47. Yes; they should finish the test within 40 minutes.

48. a ϩ s ϭ 40, 11a ϩ 4s ϭ 335

63

Algebra 2

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49. 25 min of step aerobics,
15 min of stretching

50. (4, 6)

51. You can use a system of equations to find the monthly fee and rate per minute charged during the months of January and February.
Answers should include the following. • The coordinates of the point of intersection are
(0.08, 3.5).
• Currently, Yolanda is paying a monthly fee of
\$3.50 and an additional
8¢ per minute. If she graphs y ϭ 0.08x ϩ 3.5 (to represent what she is paying currently) and y ϭ 0.10x ϩ 3 (to represent the other long-distance plan) and finds the intersection, she can identify which plan would be better for a person with her level of usage.

52. C

53. A

54. inconsistent y yϭxϩ2 x O

yϭxϪ1

64

Algebra 2

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56. consistent and independent

55. consistent and dependent

y

y
4y Ϫ 2x ϭ 4
3x ϩ y ϭ 1
O

x

O

yϪ 1xϭ1
2

57.

x

y ϭ 2x Ϫ 4

58.

y

y

x

O

x

O

xϩyϭ3
5y Ϫ 4x ϭ Ϫ20

59.

60. 7x Ϫ y ϭ Ϫ4; 7, Ϫ1, Ϫ4

y

3x ϩ 9y ϭ Ϫ15
O

x

61. x Ϫ y ϭ 0; 1, Ϫ1, 0

62. 3x ϩ 5y ϭ 2; 3, 5, 2

63. 2x Ϫ y ϭ Ϫ3; 2, Ϫ1, Ϫ3

64. x Ϫ 2y ϭ 6; 1, Ϫ2, 6

65. 3x ϩ 2y ϭ 21; 3, 2, 21

66. 0.6 ampere

67. yes

68. no

69. no

70. yes

65

Algebra 2

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Chapter 3
Practice Quiz 1
Page 122
1.

2.

y

y

y ϭ 3x ϩ 10

(Ϫ1, 7)

(3, 2)
2x ϩ 3y ϭ 12

x

O

y ϭ Ϫx ϩ 6

2x Ϫ y ϭ 4

x

O

4. (4, Ϫ1)

3. (2, 7)
5. Hartsfield, 78 million; O’Hare,
72.5 million

Lesson 3-3

Solving Systems of Inequalities by Graphing
Pages 125–127

1. Sample answer: y Ͼ x ϩ 3, y Ͻ x Ϫ 2

2. true

3a. 4
3b. 2
3c. 1
3d. 3

4.

y

yϭ2 x O

xϭ4

5.

6.

y

y xϩyϭ2 y ϭ Ϫ2x ϩ 4
O

x ϭ Ϫ1 yϭxϪ2 ©Glencoe/McGraw-Hill

x

O

x

66

xϭ3

Algebra 2

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8. (Ϫ3, Ϫ3), (2, 2), (5, Ϫ3)

y y ϭ 2x ϩ 1 xϭ1 O

x

x ϩ 2y ϭ Ϫ3

9. (Ϫ4, 3), (1, Ϫ2), (2, 9), (7, 4)

10.

10 m

bϭ2

Muffins

8
6

2.5b ϩ 3.5m ϭ 28

4
2

mϭ3 b 0

11. Sample answer: 3 packages of bagels, 4 packages of muffins; 4 packages of bagels, 4 packages of muffins; 3 packages of bagels, 5 packages of muffins 12.

13.

2

14.

4

6
8
Bagels

12

y

yϭ3 x O

xϭ2

y

y

x

O

10

yϭ2Ϫx x x ϭ Ϫ1

O

y ϭ Ϫ4

yϭxϩ4

67

Algebra 2

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16.

y

y

yϭ2
3x ϩ 2y ϭ 6

y ϭ Ϫ2 x

O

x

O
4x Ϫ y ϭ 2

yϭxϪ3

17.

18.

y
4x Ϫ 3y ϭ 7

yϭ 1xϩ1
2

x

O

y

2y Ϫ x ϭ Ϫ6

x

O

y ϭ 2x Ϫ 3

19. no solution

20.

y yϭ1 xϭ3 x O

x ϭ Ϫ3

21.

y ϭ Ϫ1

22. no solution

y xϭ2 x ϩ 3y ϭ 6
O

x

x ϭ Ϫ4

23.

24. (0, 0), (0, 4), (8, 0)

y
2x Ϫ y ϭ 4
2x ϩ 4y ϭ Ϫ7
O

x x Ϫ 3y ϭ 2

68

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25. (Ϫ3, Ϫ4), (5, Ϫ4), (1, 4)

26. (0, 4), (3, 0), (3, 5)

27. (Ϫ6, Ϫ9), (2, 7), (10, Ϫ1)

28. (Ϫ11, Ϫ3), (Ϫ1, Ϫ3),
1
2

(6, 4), a6, 5 b
29. (Ϫ4, 3), (Ϫ2, 7),
1
3

30. 16 units2

1
3

(4, Ϫ1), a7 , 2 b
31. 64 units2

32.
Hours Raking Leaves

16
14
12

x ϩ y ϭ 15

10
8
6
4
2
0

10x ϩ 12y ϭ 120
2

x

4 6 8 10 12 14 16
Hours Cutting Grass

34. category 4; 13-18 ft

33. s Ն 111, s Յ 130, h Ն 9, h Յ 12
16

y

h

Storm Surge (ft)

14

h ϭ 12
12
10

s ϭ 130

hϭ9

8

s ϭ 111 s 0

80

100
120
140
Wind Speed (mph)

160

69

Algebra 2

Chapter 3

PQ245-6457F-P03[056-080].qxd

35.

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y
14

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36. Sample answer: 2 pumpkin,
8 soda; 4 pumpkin, 6 soda;
8 pumpkin, 4 soda

2x ϩ 1.5y ϭ 24

12
10

x ϩ 2.5y ϭ 26

8
6
4
2
0

2

4

6 8 10 12 14
Pumpkin

x

37. 6 pumpkin, 8 soda

38. 42 units2

39. The range for normal blood pressure satisfies four inequalities that can be graphed to find their intersection. Answers should include the following.
• Graph the blood pressure as an ordered pair; if the point lies in the shaded region, it is in the normal range. • High systolic pressure is represented by the region to the right of x ϭ 140 and high diastolic pressure is represented by the region above y ϭ 90.

40. B

41. Sample answer: y Յ 6, y Ն 2, x Յ 5, x Ն 1

42. (Ϫ3, 8)

43. (6, 5)

44. (8, Ϫ5)

45.

46. infinitely many

y

y
2x ϩ y ϭ Ϫ3

y ϭ 2x ϩ 1
(Ϫ2, Ϫ3)

O

x
O

yϭ Ϫ1xϪ4
2

x

6x ϩ 3y ϭ Ϫ9

70

Algebra 2

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1
2

48. y ϭ x ϩ 6

y
Ϫx ϩ 8y ϭ 12

(4, 2)

x

O

2x Ϫ y ϭ 6

49. Ϫ5

50. Ϫ12

51. 8

52. 27

53. 5

54. Ϫ8.25

Lesson 3-4 Linear Programming
Pages 132–135
1. sometimes

2. Sample answer: y Ն Ϫx, y Ն x Ϫ 5, y Յ 0

3.

4.

y

y

(1, 4)
(5, 2)

(Ϫ3, 1)

(1, 2)
O

O

x

x

vertices: (Ϫ3, 1), Q , 1R;

vertices: (1, 2), (1, 4), (5, 2); max: f (5, 2) ϭ 4, min: f (1, 4) ϭ Ϫ10

( 5 , 1)
3

5
3

min: f (Ϫ3, 1) ϭ Ϫ17; no maximum

71

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6.

y

y
(7, 8.5)

(1, 3)

(2, 6)

(6, 3)

(0, 1)
O

(10, 1)

x x O (2, 0)

vertices: (0, 1), (1, 3), (6, 3),
(10, 1); max: f (10, 1) ϭ 31, min: f (0, 1) ϭ 1

(7, Ϫ5)

vertices: (2, 0), (2, 6), (7, 8.5),
(7, Ϫ5); max: f(7, 8.5) ϭ 81.5, min: f(2, 0) ϭ 16
7.

8.

y

y

(Ϫ2, 4)

(Ϫ1, 2)

(2, 3)

(4, 1)

x

O

x

O

(Ϫ3, Ϫ1)

(Ϫ2, Ϫ3)

(3, Ϫ2)

(2, Ϫ3)

vertices: (Ϫ3, Ϫ1), (Ϫ1, 2),
(2, 3), (3,Ϫ2); max: f(3, Ϫ2) ϭ
5, min: f(Ϫ1, 2) ϭ Ϫ3

vertices: (Ϫ2, 4), (Ϫ2, Ϫ3),
(2, Ϫ3), (4, 1); max: f(2, Ϫ3) ϭ
5; min: f (Ϫ2, 4) ϭ Ϫ6
10.

28

Leather Tote Bags

9. c Ն 0, l Ն 0, c ϩ 3l Յ 56,
4c ϩ 2l Յ 104

24
20
16

(20, 12)

12
8
4
0

2
3

(0, 18 2 )
3

(0, 0)
4

(26, 0)
8

12 16 20 24
Canvas Tote Bags

11. (0, 0), (26, 0), (20, 12), a0, 18 b

12. f(c, l) ϭ 20c ϩ 35l

13. 20 canvas tote bags and
12 leather tote bags

28 c

14. \$820

72

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16.

y

y

(3, 5)

(6, 13)

x

O
(0, Ϫ4)
(3, Ϫ4)
(0, 1)

vertices: (0, Ϫ4), (3, 5),
(3, Ϫ4); max: f (3, Ϫ4) ϭ 7, min: f (3, 5) ϭ Ϫ2

(6, 1)

x

O

vertices: (0, 1), (6, 1), (6, 13); max: f (6, 13) ϭ 19; min: f (0, 1) ϭ 1
17.

18.

y

y

(5, 8)

(2, 3)
(2, 1)

(1, 4)

(4, 4)
(4, 1)

x

O
(1, 2)

(5, 2)

x

O

vertices: (2, 1), (2, 3), (4, 4),
(4, 1); max: f (4, 4) ϭ 16; min: f (2, 1) ϭ 5

vertices: (1, 4), (5, 8),
(5, 2), (1, 2); max: f (5, 2) ϭ
11, min: f (1, 4) ϭ Ϫ5
19.

y

20.

(3, 5)

y
12
8

(Ϫ3, Ϫ1)

O

4

x
Ϫ4

O
Ϫ4

vertices: (Ϫ3, Ϫ1), (3, 5); min: f (Ϫ3, Ϫ1) ϭ Ϫ9; no maximum

(6, 12)
(2, 8)
(2, 2)
4

8

x

(6, Ϫ6)

Ϫ8

vertices: (2, 2), (2, 8), (6, 12),
(6, Ϫ6); max: f(6, 12) ϭ 30, min: f(6, Ϫ6) ϭ Ϫ24
73

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22.

y

(0, 2)
O
(0, 0)

y
(0, 7)
(4, 3)

(2, 1)

x
(3, 0)
(0, 0)
O

vertices: (0, 0), (0, 2), (2, 1),
(3, 0); max: f (0, 2) ϭ 6; min: f (3, 0) ϭ Ϫ12
23.

(2, 0) x

vertices: (0, 0), (0, 7), (4, 3),
(2, 0); max: f (4, 3) ϭ 14; min: f (0, 7) ϭ Ϫ14
24.

y

y
(8, 6)
(0, 4)

(3, 0)

x

O

(4, 0)

(0, Ϫ3)

vertices: (0, 4), (4, 0), (8, 6); max: f (4, 0) ϭ 4; min: f (0, 4) ϭ Ϫ8

vertices: (3, 0), (0, Ϫ3); min: f (0, Ϫ3) ϭ Ϫ12; no maximum
25.

26.

y

(0, 2)

y

(0, 2)

(4, 3)

(4, 3)
(2, 0)

x

O

x

O

O

x

(7 ,Ϫ 1)
3
3 vertices: (0, 2), (4, 3), (2, 0); max: f (4, 3) ϭ 13, min: f (2, 0) ϭ 2

vertices: (0, 2), (4, 3),
7 1 a , Ϫ b; max: f (4, 3) ϭ 25,
3

3

min: f (0, 2) ϭ 6

74

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28.

y

y

(2, 5)
(2, 2)
(3, 0)

(4, 1)

(0, 1)

x

x

O

(0, 0) O

vertices: (0, 0), (0, 1), (2, 2),
(4, 1), (5, 0); max: f (5, 0) ϭ
15, min: f (0, 1) ϭ Ϫ5

vertices: (2, 5), (3, 0); min: f (3, 0) ϭ 3, no maximum
29.

y

30a. Sample answer: f (x, y) ϭ Ϫ2x Ϫy
30b. Sample answer: f (x, y) ϭ 3y Ϫ 2x
30c. Sample answer: f (x, y) ϭ x ϩ y
30d. Sample answer: f (x, y) ϭ Ϫx Ϫ 3y
30e. Sample answer: f (x, y) ϭ x ϩ 2y

(4, 4)
(2, 3)

(2, 1)
O

(5, 3)
(4, 1)

(5, 0)

x

vertices: (2, 1), (2, 3), (4, 1),
(4, 4), (5, 3); max: f(4, 1) ϭ 0, min: f (4, 4) ϭ Ϫ12
31. g Ն 0, c Ն 0, 1.5g ϩ 2c Յ
85, g ϩ 0.5c Յ 40

Graphing Calculators

32.

50

g

40
30
20
10
0

(0, 42.5)
(0, 0)

(30, 20)
(40, 0)

10 20 30 40
CAS Calculators

c
50

33. (0, 0), (0, 42.5), (30, 20),
(40, 0)

34. f (g, c) ϭ 50g ϩ 65c

35. 30 graphing calculators,
20 CAS calculators

36. \$2800

37. See students’ work.

38. c Ն 0, s Ն 0, c ϩ s Յ 4500,
4c ϩ 5s Յ 20,000

75

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39. (0, 0), (0, 4000), (2500, 2000),
(4000, 0)

40. 2500 acres of corn,
2000 acres soybeans;
\$125,000

S
4000
3000
2000

(0, 4000)

(2500, 2000)

1000
(0, 0)
0

(4500, 0)
2000

4000

c

41. 4000 acres corn, 0 acres soybeans; \$130,500

42. 3 chocolate chip, 9 peanut butter 43. There are many variables in scheduling tasks. Linear programming can help make sure that all the requirements are met. Answers should include the following.
• Let x ϭ the number of buoy replacements and let y ϭ the number of buoy repairs. Then, x Ն 0, y Ն 0, x Յ 8 and x ϩ 2.5y Յ 24.
The captain would want to maximize the number of buoys that a crew could repair and replace, so f (x, y) ϭ x ϩ y.
• Graph the inequalities and find the vertices of the intersection of the graphs.
The coordinate (0, 24) maximizes the function. So the crew can service the maximum number of buoys if they replace 0 and repair 24 buoys.

44. A

76

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46.

45. C

y

2y ϩ x ϭ 4

x

O

yϭxϪ4

48. (Ϫ5, 8)

47. л y 3x Ϫ 2y ϭ Ϫ6

x

O

y ϭ 3x Ϫ 1
2

49. (2, 3)
51. c ϭ average cost each year;
15c ϩ 3479 ϭ 7489
55. Multiplicative Inverse
57. 9
59. 16
61. 8

50. (5, 1)
52. about \$267 per year
54.
56.
58.
60.

Associative Property (ϫ)
Distributive Property
5
Ϫ3

62. Ϫ4

Chapter 3
Practice Quiz 2
Page 135
1.

y

2.

yϪxϭ0

y yϭxϩ3 x

O

y ϭ 3x Ϫ 4

yϩxϭ4

O

77

x

Algebra 2

Chapter 3

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4.

4x ϩ y ϭ 16

y
(1, 6)
(0, 4)

x ϩ 3y ϭ 15

(0, 0)

x

O

O

(3, 0)

x

vertices: (0, 0), (0, 4), (1, 6),
(3, 0); max: f (1, 6) ϭ 8, min: f (0, 0) ϭ 0
5.

y
(5, 6)

(Ϫ1, 3)

(5, 1)

x

O
(1, Ϫ3)

vertices: (1, Ϫ3), (Ϫ1, 3),
(5, 6), (5, 1); max: f (5, 1) ϭ
17, min: f (Ϫ1, 3) ϭ Ϫ13

Lesson 3-5
Solving Systems of Equations in Three Variables
Pages 142–144
1. You can use elimination or substitution to eliminate one of the variables. Then you can solve two equations in two variables.

2. No; the first two equations do represent the same plane, however they do not intersect the third plane, so there is no solution of this system.

3. Sample answer: x ϩ y ϩ z ϭ 4, 2x Ϫ y ϩ z ϭ Ϫ9, x ϩ 2y Ϫ z ϭ 5; Ϫ3 ϩ 5 ϩ
2 ϭ 4, 2(Ϫ3) Ϫ 5 ϩ 2 ϭ Ϫ9,
Ϫ3 ϩ 2(5) Ϫ 2 ϭ 5

4. (6, 3, Ϫ4)

78

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5. (Ϫ1, Ϫ3, 7)

6. infinitely many

7. (5, 2, Ϫ1)

8. no solution
10. 6c ϩ 3s ϩ r ϭ 42,

9. (4, 0, 8)

1
2

c ϩ s ϩ r ϭ 13 , r ϭ 2s
1
2

12. (3, 4, 7)

11. 4 lb chicken, 3 lb sausage,
6 lb rice
13. (Ϫ2, 1, 5)

14. (2, Ϫ3, 6)

15. (4, 0, Ϫ1)

16. no solution

17. (1, 5, 7)

18. (1, 2, Ϫ1)

19. infinitely many

20. a , , b

1
3

1 3 9
2 2 2

1 1
2 4

21. a , Ϫ , b

22. (8, 3, Ϫ6)

23. (Ϫ5, 9, 4)

24. 3, 12, 5

25. 8, 1, 3

26. 1-\$100, 3-\$50, and
6-\$20 checks

27. enchilada, \$2.50; taco, \$1.95; burrito, \$2.65

28. \$7.80

29. x ϩ y ϩ z ϭ 355, x ϩ 2y ϩ
3z ϭ 646, y ϭ z ϩ 27

30. 88 3-point goals, 115 2-point goals, 152 1-point free throws

4
3
4 2 x 3

1
3
1 x 3

32. You can write a system of three equations in three variables to find the number of each type of medal. Answers should include the following.
• You can substitute b ϩ 6 for g and b Ϫ 8 for s in the equation g ϩ s ϩ b ϭ 97.
This equation is now in terms of b. Once you find b, you can substitute again to find g and s. The U.S.
Olympians won 39 gold medals, 25 silver medals, and 33 bronze medals.

31. a ϭ , b ϭ , c ϭ 3; yϭ ©Glencoe/McGraw-Hill

ϩ

ϩ3

79

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• Another situation involving three variables is winning times of the first, second, and third place finishers of a race.
33. D

34. A

35. 120 units of notebook paper and 80 units of newsprint

36.

y yϭx ϩ2

x

O

y ϭ 7 Ϫ 2x

37.

38.

y

y

3x ϩ y ϭ 3

O

3x ϩ y ϭ 1

O

x

x

2y Ϫ x ϭ Ϫ4

4y Ϫ 2x ϭ 4

39. Sample answer using (7, 15) and (14, 22): y ϭ x ϩ 8

41. x ϩ 3y

42. Ϫ2z ϩ 8

43. 9s ϩ 4t

44. 18a ϩ 16b

80

Algebra 2

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Chapter 4 Matrices
Lesson 4-1 Introduction to Matrices
Pages 156–158 column matrix, B R, 2 ϫ 1;

2. Sample answers: row matrix, [1 2 3], 1 ϫ 3;

1. The matrices must have the same dimensions and each element of one matrix must be equal to the corresponding element of the other matrix.

square matrix, B zero matrix, B

3. Corresponding elements are elements in the same row and column positions.

4. 1 ϫ 5

5. 3 ϫ 4

6. (5, 6)

7. (3, 3)

8.
High
Low

B

1 2
R,
3 4
1
2

0 0
R,
0 0

2 ϫ 2;

2ϫ2

88 88 90 86 85
R
54 54 56 53 52
Fri

9. 2 ϫ 5

10. 2 ϫ 3

11. 3 ϫ 1

12. 4 ϫ 3

13. 3 ϫ 3

14. 2 ϫ 5

15. 3 ϫ 2

Sat Sun Mon Tue

16. (2.5, 1, 3)

17. a3, Ϫ b
1
3

18. (5, 3)

19. (3, Ϫ5, 6)

20. (2, Ϫ5)

21. (4, Ϫ3)

22. (1.5, 3)

23. (14, 15)

24. (Ϫ2, 7)

25. (5, 3, 2)

26.
Child
Senior

81

7.50
C 4.50
5.50

3.75
3.75 S
3.75

Evening Matinee Twilight

5.50
4.50
5.50

Algebra 2

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28.

27. 3 ϫ 3

Cost

Catalina Grill
Oyster Club
Casa di Pasta
Mason’s
Steakhouse

29. Sample answer: Mason’s
Steakhouse; it was given the highest rating possible for service and atmosphere, location was given one of the highest ratings, and it is moderately priced.

30.

31.

32.

**
≥ ***
****
**

Weekday

Double
Suite

60 70 75
B
R
79 89 95

Weekend

Atmosphere Location

*
**
***
****

*
*
***
****

60 79
C 70 89 S
75 95

*
** ¥
***
***

Weekday Weekend
Single

Single Double Suite

Service

1
2
4
G 7
11
16
22

3
5
8
12
17
23
30

6
9
13
18
24
31
39

10
14
19
25
32
40
49

15
20
26
33
41
50
60

21
27
34
42W
51
61
72

33. row 6, column 9

34. Matrices are used to organize information so it can be read and compared more easily. Answers should include the following.
• If you want the least expensive vehicle, the compact SUV has the best price; the large SUV has the most horsepower, towing capacity and cargo space, and the standard
SUV has the best fuel economy. • Sample answer: Matrices are used to report stock prices in the newspaper.

35. B

36. C

37. (7, 5, 4)

38. (7, 3, Ϫ9)

82

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39. aϪ3, 5, Ϫ11b
4 3

40.

y y ϭ Ϫ2x ϩ 15

yϭx ϩ2

yϭ3

x

O

vertices: (1, 3), (6, 3), a , max: f a ,

13 19 b 3 3

ϭ

83
,
3

13 19 b; 3 3

min: f (1, 3)ϭ11
42.

41.

3

ϩ 12

y

vertices: (3, 1), a ,
15
2

vertices: (2, 1), (6, 3); min: f(2, 1) ϭ 1, no maximum

44. step function

6
Cost (\$)

43.

3 17 a , b; 2 2
.
15 5 max: f a , b ϭ 35,
2 2
3 17 min: f a , b ϭ Ϫ1
2 2

5 b, 2

5
4
3
2
1
0

1

2 3 4
Hours

5

83

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45. \$4.50

46. 2

47. 2

48. 0

49. 20

50. 3

51. Ϫ10

52. 6.2

53. Ϫ18

54. 17

55. Ϫ3

56. 75

57.

3
2

Lesson 4-2

Operations with Matrices
Pages 163–166

1. They must have the same dimensions. 2. Sample answer: [Ϫ3 1],
[3 Ϫ1]

5. B

4. impossible

4 4
3. C 4 4 S
4 4

1 10
R
Ϫ7 5

7. B
9. B

Ϫ22 8
R
3 24

Ϫ21
29
R
12 Ϫ22

16,763
14,620
11. Males ϭ E14,486
9041
5234

16,439
14,545
Females ϭ E12,679
7931
5450

6. B

18 Ϫ3 15 6
R
21
9 Ϫ6 24

8. B

10 6
R
Ϫ1 7

10. B

Ϫ3 30
R
26 11

549,499
477,960
455,305U,
321,416
83,411

1,006,372
883,123
12. E 795,785U
579,002
216,646

456,873
405,163
340,480U
257,586
133,235
84

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10
14. C Ϫ4 S
5

13. No; many schools offer the same sport for males and females, so those schools would be counted twice.

16. B

15 0
4
R
0 13 Ϫ5

Ϫ4
8 Ϫ2
17. C 6 Ϫ10 Ϫ16 S
Ϫ14 Ϫ12
4
15. impossible

Ϫ13
19. C Ϫ3 S
23

1.5 3
R
21. B
4.5 9
23. C

1

Ϫ52

3

2
10
3

2
1
3

Ϫ2 Ϫ1
25. C 4 Ϫ1 S
Ϫ7 Ϫ4
38
4
27. C 32 Ϫ6 S
18 42
29. D 1
2

4

18. [15 Ϫ29 65 Ϫ2]
1.8 9.08
20. C 3.18 31.04 S
10.41 56.56
22. C

S

13 10
7S
24. C 4
7 Ϫ5

9
1
Ϫ2
2

3
2

Ϫ2

S

0 16
26. C Ϫ8 20 S
28 Ϫ4

Ϫ12 Ϫ13
28. C 3 Ϫ8 S
13
37

5T

120 97 64 75
30. Friday: C 80 59 36 60 S ,
72 84 29 48

2
3

112 87 56 74
Saturday: C 84 65 39 70 S
88 98 43 60

6 Ϫ1

Ϫ4 Ϫ15

85

Algebra 2

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232 184 120 149
31. C 164 124 75 130 S
160 182 72 108

Ϫ8 Ϫ10 Ϫ8 Ϫ1
32. C 4
6
3 10 S
16
14 14 12
15
41
34. E35U
27
51

245
228
33. E319U
227
117

36. Residents:
Before 6 3.00 4.50
B
R
After 6
2.00 3.50

35. 1996, floods; 1997, floods;
1998, floods; 1999, tornadoes; 2000, lightning

Nonresidents:
Before 6 4.50 6.75
B
R
After 6
3.00 5.25

37. B

1.50 2.25
R
1.00 1.75

Residents
3.00 4.50
B
R
Nonresidents 4.50 6.75

38. Before 6:00:

Residents
2.00 3.50
B
R
Nonresidents 3.00 5.25
After 6:00:

39. B

1.00 1.00
R
1.50 1.50

40.
0.5 0.75
3
1 1.5
6
2B
Rϭ B
R
1
4 0.1
2
8 0.2
42. D

41. You can use matrices to track dietary requirements and add them to find the total each day or each week. Answers should include the following.
566 18 7
• Breakfast ϭ C 482 12 17 S ,
530 10 11

86

Algebra 2

Chapter 4

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785 22 19
Lunch ϭ C 622 23 20 S ,
710 26 12

1257 40 26
Dinner ϭ C 987 32 45 S
1380 29 38
• Add the three matrices:
2608 80 52
£ 2091 67 82 § .
2620 65 61

43. A

44. 2 ϫ 2

45. 1 ϫ 4

46. 2 ϫ 4

47. 3 ϫ 3

48. 3 ϫ 2

49. 4 ϫ 3

50. (3, Ϫ4, 0)
52. a , 6, Ϫ b
1
4

51. (5, 3, 7)
53. (2, 5)

1
6

54. (Ϫ3, 1)

55. (6, Ϫ1)

56. 0.30p ϩ 0.15s Յ 6

57.

58. No, it would cost \$6.30.

59. Multiplicative Inverse

60. Associative Prop. (ϩ)

61. Distributive Property

62. Commutative Prop. (ϫ)

87

Algebra 2

Chapter 4

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Lesson 4-3 Multiplying Matrices
Pages 171–174

1 2
7 8
C3 4S ؒ B
R
9 10
5 6

2. Never; the inner dimensions will never be equal.

3. The Right Distributive
Property says that
1A ϩ B2C ϭ AC ϩ BC, but
AC ϩ BC CA ϩ CB since the Commutative Property does not hold for matrix multiplication in most cases.

4. 3 ϫ 2

5. undefined
15 Ϫ5 20
7. B
R
24 Ϫ8 32

6. [19 15]

9. B

24
R
41

8. not possible
2 Ϫ1
Ϫ4 1
3 2 ϭB R ؒ ¢B
RؒB
R≤
3
5
8 0
Ϫ1 2

10. yes
A(BC)
ϭB

ϭB

2 Ϫ1
Ϫ13 Ϫ6
RؒB
R
3
5
24 16

Ϫ50 Ϫ28
R
81
62
(AB)C

2 Ϫ1
Ϫ4 1
3 2 ϭ ¢B
Rؒ B
R≤ؒ B
R
3
5
8 0
Ϫ1 2

ϭB

ϭB

Ϫ16 2
3 2
RؒB
R
28 3
Ϫ1 2

Ϫ50 Ϫ28
R
81
62

88

Algebra 2

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11. [45

55

13. 4 ϫ 2
15. undefined

350 280
65], C 320 165 S
180 120

12. \$74,525
14. 2 ϫ 2
18. 3 ϫ 5
8 Ϫ11
20. B
R
22
12
16. 1 ϫ 5

17. undefined

22. B

Ϫ39
R
18

19. [6]
1 Ϫ25
2
R
23. B
29
1 Ϫ30
21. not possible

24
16
25. C Ϫ32 Ϫ5 S
Ϫ48 Ϫ11

0 64 Ϫ40
26. C 9 11 Ϫ11 S
Ϫ3 39 Ϫ23
24. not possible

ϭ 3¢B

27. yes
AC ϩ BC
1 Ϫ2
5
1 ϭB Rؒ B

4
3
2 Ϫ4
B

28. yes c (AB)

Ϫ5 2
5
1
Rؒ B
R
4 3
2 Ϫ4

ϭB

ϭB

1
9
Ϫ21 Ϫ13
RϩB
R
26 Ϫ8
26 Ϫ8

Ϫ20 Ϫ4
R
52 Ϫ16
(A ϩ B)C

ϭ ¢B

ϭB

ϭB

ϭ 3B

1 Ϫ2
Ϫ5
R ϩ B
4
3
4

ϭB

Ϫ13 Ϫ4
R
Ϫ8 17

Ϫ39 Ϫ12
R
Ϫ24
51
A(cB)

ϭB

1 Ϫ2
Ϫ5 2
R ؒ ¢3 B
R≤
4
3
4 3
1 Ϫ2 Ϫ15 6
R
RB
12 9
4
3

ϭB

2
5
1
R≤ ؒ B
R
3
2 Ϫ4

Ϫ4 0
5
1
RؒB
R
8 6
2 Ϫ4

Ϫ20
Ϫ4
R
52 Ϫ16

1 Ϫ2
Ϫ5 2
RؒB
R≤
4
3
4 3

ϭB

89

Ϫ39 Ϫ12
R
Ϫ24
51

Algebra 2

Chapter 4

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ϭB

1 Ϫ2
Ϫ5 2
5
1
RؒB
RؒB
R
4
3
4 3
2 Ϫ4

5 1
1 Ϫ2
Ϫ5 2 ϭB R ؒ ¢B
RϩB
R≤
2 Ϫ4
4 3
4 3

29. no
C (A ϩ B)

30. no
ABC
ϭB

5
1
Ϫ4 0 ϭB RؒB
R
2 Ϫ4
8 6

ϭB

Ϫ73
3
R
Ϫ6 Ϫ76
CBA

ϭB

Ϫ12
6
R
Ϫ40 Ϫ24
AC ϩ BC
1 Ϫ2
5
1 ϭB RؒB

4
3
2 Ϫ4

ϭB

5
1
Ϫ5 2
1 Ϫ2
RؒB
RؒB
R
2 Ϫ4
4 3
4
3

ϭB

Ϫ5 2
5
1
B
RؒB
R
4 3
2 Ϫ4

ϭB

1
9
Ϫ21 Ϫ13 ϭB RϩB
R
26 Ϫ8
26 Ϫ8

ϭB

Ϫ20 Ϫ4
R
52 Ϫ16

e f
R
g h

a b
R
c d

72
68
36. D
90
86

and

where bg ϭ cf, a ϭ d,

and e ϭ h
96.50
99.50
37. D
T
118
117
39. \$431

49
63
1.00
T, B
R
56
0.50
62

38. Juniors
40. \$24,900

41. \$26,360

31 81
R
Ϫ58 28

34. \$31,850

35. any two matrices B
B

Ϫ21 13
1 Ϫ2
RؒB
R
Ϫ26 Ϫ8
4
3

22
32. C 25 S
18

290 165 210
31. C 175 240 190 S
110 75
0
14,285
33. C 13,270 S
4295

Ϫ13 Ϫ4
5
1
RؒB
R
Ϫ8 17
2 Ϫ4

42. \$1460
90

Algebra 2

Chapter 4

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43. a ϭ 1, b ϭ 0, c ϭ 0, d ϭ 1; the original matrix

44. Sports statistics are often listed in columns and matrices. In this case, you can find the total number of points scored by multiplying the point matrix, which doesn’t change, by the record matrix, which changes for each season. Answers should include the following.
• P ؒ R ϭ [479]
• Basketball and wrestling use different point values in scoring.

45. B
12 Ϫ6
47. B
R
Ϫ3 21
Ϫ20 2
49. B
R
Ϫ28 12

46. A
48. impossible
50. (7, Ϫ4)

51. (5, Ϫ9)

52. (2, Ϫ5, Ϫ7)

53. \$2.50; \$1.50

54.

55. 8; Ϫ16

56. 2; Ϫ5

91

3
;
2

3

Algebra 2

Chapter 4

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57.

58.

59.

60.

Chapter 4
Practice Quiz 1
Page 174

(5, Ϫ1)
120 80 64 75
4. B
R,
65 105 77 53

1. (6, 3)

2.

3. (1, 3, 5)

5.
7.
9.

B

232 159 120 149
R
134 200 159 103

112 79 56 74
R
69 95 82 50

6. B

Ϫ3 5
R
3 13

B

4 3
R
1 3

8. B

Ϫ10
20 25
R
0 Ϫ20 35

10. B

15 Ϫ8 Ϫ10
R
Ϫ7 23
16

not possible

B

92

Algebra 2

Chapter 4

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Lesson 4-4

1.

Transformation reflection rotation translation p

Shape same same same p

Size same same same 3. Sample answer:
Ϫ4 Ϫ4 Ϫ4
B
R
1
1
1
5. A¿(4, 3), B¿(5, Ϫ6), dilation changes

same

Transformations with Matrices
Pages 178–181

p

2. B

Ϫ3 Ϫ3 Ϫ3
R
Ϫ2 Ϫ2 Ϫ2

Isometry yes 4. B

yes yes no

6.

3
3
3
R
Ϫ1 Ϫ1 Ϫ1

C¿(Ϫ3, Ϫ7)

7. B

0 5 5 0
R
4 4 0 0

8. A¿(0, 12), B¿(15, 12),
C¿(15, 0), D¿(0, 0)

12. B

11. B

14.

13. D¿(Ϫ3, 6), E¿(Ϫ2, Ϫ3),
F¿(Ϫ10, Ϫ4)

15. B

0
1.5 Ϫ2.5
R
2 Ϫ1.5
0

Ϫ4 Ϫ4 Ϫ4
R
2
2
2

10. A¿(0, Ϫ4), B¿(Ϫ5, Ϫ4),
C¿(Ϫ5, 0), D¿(0, 0)

9. A¿(0, Ϫ4), B¿(5, Ϫ4),
C¿(5, 0), D¿(0, 0)

16. A¿(0, 6), B¿(4.5, Ϫ4.5),
C¿(Ϫ7.5, 0)

93

Algebra 2

Chapter 4

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18. B

17.

19. X¿(Ϫ1, 1), Y¿(Ϫ4, 2),
Z¿(Ϫ1, 7)

21. B
23.

2 5 4 1
R
4 4 1 1

F'
25.

20.

22. D¿(4, Ϫ2), E¿(4, Ϫ5),
F¿(1, Ϫ4), G¿(1, Ϫ1)

y D

G
O G'

24. E¿(6, Ϫ2), F¿(8, Ϫ9)

E

F

x

D'
26. B

2
4
2 Ϫ3
R ؒ (Ϫ1) ϭ
3 Ϫ3 Ϫ5 Ϫ2

E'

J(Ϫ5, 3), K(7, 2), L(4, Ϫ1)

1
2
7
R
Ϫ1 Ϫ4 Ϫ1

94

B

Ϫ2 Ϫ4 Ϫ2 3
R
Ϫ3
3
5 2

Algebra 2

Chapter 4

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27.

28. 180Њ rotation

30. B

4 Ϫ4 Ϫ4 4
R
Ϫ4 Ϫ4
4 4

29. B
31. B

4 Ϫ4 Ϫ4 4
R
Ϫ4 Ϫ4
4 4

4 4 Ϫ4 Ϫ4
R
Ϫ4 4
4 Ϫ4

3
35. B R
4

32. The figures in Exercise 29 and Exercise 30 have the same coordinates, but the figure in Exercise 31 has different coordinates.
34. (Ϫ3.75, Ϫ2.625)

33. (Ϫ1.5, Ϫ1.5), (Ϫ4.5, Ϫ1.5),
(Ϫ6, Ϫ3.75), (Ϫ3, Ϫ3.75)

36. (6.5, 6.25)

37. (Ϫ8, 7), (Ϫ7, Ϫ8), and
(8, Ϫ7) by B

1
0
R,
0 Ϫ1

38. The object is reflected over the x-axis, then translated
6 units to the right.

39. Multiply the coordinates

40. No; since the translation does not change the y-coordinate, it does not matter whether you do the translation or the reflection over the x-axis first. However, if the translation did change the y-coordinate, then order would be important.

result to B R.

6
0

41. (17, Ϫ2), (23, 2)

42. There is no single matrix to achieve this. However, you could reflect the object over the y-axis and then translate it 2(3) or 6 units to the right.

95

Algebra 2

Chapter 4

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43. Transformations are used in computer graphics to create special effects. You can simulate the movement of an object, like in space, which you wouldn’t be able to recreate otherwise. Answers should include the following.
• A figure with points (a, b),
(c, d), (e, f ), (g, h), and (i, j) could be written in a 2 ϫ 5 matrix B

a c e g i
R
b d f h j

44. B

and

multiplied on the left by the
2 ϫ 2 rotation matrix.
• The object would get smaller and appear to be moving away from you.
45. A

46. 2 ϫ 2

47. undefined

48. 2 ϫ 5

20
10 Ϫ24
50. C 31 Ϫ46 Ϫ9 S
Ϫ10
3
7
52.

11
24 Ϫ7
8S
49. C 18 Ϫ13
33 Ϫ8 21
51.

D ϭ 53, 4, 56, R ϭ 5Ϫ4, 5, 66; yes ©Glencoe/McGraw-Hill

D ϭ {all real numbers},
R ϭ {all real numbers}; yes

96

Algebra 2

Chapter 4

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54. 0 x 0 Ն 4

53.

D ϭ 5x 0 x Ն 06,
R ϭ 5all real numbers6; no

55. 0 x 0 Ͻ 2.8

56. 0 x ϩ 1 0 Ͼ 2

59. 6

60. 5

61. 28

62.

10
3

64.

5
3

57. 0 x Ϫ 1 0 Ͻ 1

63.

2
3

58. 513 mi

9
4

1. Sample answer: B

2 1
R
8 4

Lesson 4-5 Determinants
Pages 185–188
2. Khalid; the value of the determinant is the difference of the products of the diagonals.
3 1
4 3
B
R, B
R
6 5
1 3

3. It is not a square matrix.

Ϫ2
3 5
6. † 0 Ϫ1 4 † ϭ
9
7 2

5. Cross out the column and row that contains 6. The minor is the remaining 2 ϫ 2 matrix. Ϫ2 `

Ϫ1 4
0 4
0 Ϫ1
` Ϫ 3`
` ϩ 5`
`
7 2
9 2
9
7

ϭ Ϫ2(Ϫ2 Ϫ 28) Ϫ
3(0 Ϫ 36) ϩ 5(0 Ϫ (Ϫ9)) ϭ Ϫ2(Ϫ30) Ϫ 3(Ϫ36) ϩ 5(9) ϭ 60 ϩ 108 ϩ 45 ϭ 213

97

Algebra 2

Chapter 4

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Ϫ2
3 5 Ϫ2
3
† 0 Ϫ1 4 † 0 Ϫ1
9
7 2 9
7
4 108

0

Ϫ45 Ϫ56
0
Ϫ2
3 5 Ϫ2
3
† 0 Ϫ1 4 † 0 Ϫ1
9
7 2 9
7
4 ϩ 108 ϩ 0 Ϫ (Ϫ45) Ϫ
(Ϫ56) Ϫ 0 ϭ 213
8. 0

7. Ϫ38

10. Ϫ28

9. Ϫ40
11. Ϫ43

12. 0

13. 45

14. 26 units2

15. 20

16. Ϫ22

17. Ϫ22

18. 0

19. Ϫ29

20. Ϫ14

21. 63

22. Ϫ6

23. 32

24. Ϫ37

25. 32

26. 11.3

27. Ϫ58

28. 0

29. 62

30. 60

31. 172

32. Ϫ265

33. Ϫ22

34. 21

35. Ϫ5

36. 49

37. Ϫ141

38. Ϫ123

39. Ϫ6

40.

41. 14.5 units2

42. 12

43. about 26 ft2

44. 2875 mi2

98

5
,
3

Ϫ1

Algebra 2

Chapter 4

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1 1 1
45. Sample answer: † 1 1 1 †
1 1 1

46. Multiply each member in the top row by its minor and position sign. In this case the minor is a 3 ϫ 3 matrix.
Evaluate the 3 ϫ 3 matrix using expansion by minors again. 47. If you know the coordinates of the vertices of a triangle, you can use a determinant to find the area. This is convenient since you don’t need to know any additional information such as the measure of the angles.
Answers should include the following. • You could place a coordinate grid over a map of the Bermuda Triangle with one vertex at the origin. By using the scale of the map, you could determine coordinates to represent the other two vertices and use a determinant to estimate the area.
• The determinant method is advantageous since you don’t need to physically measure the lengths of each side or the measure of the angles between the vertices. 48. C

49. C

50. 63.25

51. Ϫ36.9

52. Ϫ25.21

53. Ϫ493
55. Ϫ3252

99

54. 0
Ϫ2 1
2
56. B
R
1 2 Ϫ3

Algebra 2

Chapter 4

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57. A¿(Ϫ5, 2.5), B¿(2.5, 5),
C¿(5, Ϫ7.5)

58.

60. B

2
26
R
Ϫ9 Ϫ12
62. undefined
7 69
64. B
R
Ϫ5 16
66. y ϭ x Ϫ 2

59. [Ϫ4]
61. undefined
63. [14

Ϫ8]

65. 138,435 ft
4
3

67. y ϭ Ϫ x

68. y ϭ 2x ϩ 1

1
2

69. y ϭ x ϩ 5

70. (0, Ϫ3)

71. (1, 9)

72. (2, 1)

73. (Ϫ1, 1)

74. (2, 5)

75. (4, 7)

Lesson 4-6 Cramer’s Rule
Pages 192–194
1. The determinant of the coefficient matrix cannot be zero. 2. Sample answer: 2x ϩ y ϭ 5 and 6x ϩ 3y ϭ 8

3. 3x ϩ 5y ϭ Ϫ6, 4x Ϫ 2y ϭ 30

4. (5, 1)

5. (0.75, 0.5)

6. (Ϫ6, Ϫ8)

8. aϪ5, , Ϫ b
2
3

7. no solution
9. a6, Ϫ , 2b
1
2

1
2

10. s ϩ d ϭ 4000,
0.065s ϩ 0.08d ϭ 297.50

100

Algebra 2

Chapter 4

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11. savings account, \$1500; certificate of deposit, \$2500

12. (2, Ϫ1)

13. (Ϫ12, 4)

14. (3, 5)

15. (6, 3)

16. (2.3, 1.4)

17. (Ϫ0.75, 3)

18. (Ϫ0.75, 0.625)
20. a , Ϫ1b
2
3

19. (Ϫ8.5625, Ϫ19.0625)
21. (4, Ϫ8)

22. (3, 10)

23. a , b
2 5
3 6

24. (Ϫ1.5, 2)

25. (3, Ϫ4)

26. (Ϫ1, 3, 4)
28. aϪ ,

11 39
,
19 19

27. (2, Ϫ1, 3)
29. a

141
,
29

31. aϪ

102 244 b ,
29 29

Ϫ b
14
19

30. (11, Ϫ17, 14)

Ϫ

155 143 673 b ,
,
28 70 140

32. r ϩ s ϭ 8, 7r ϩ 5s ϭ 50

33. race car, 5 plays; snowboard,
3 plays

34. 8s ϩ 13c ϭ 604.79,

35. silk, \$34.99; cotton, \$24.99

36. p ϩ r ϩ c ϭ 5, 2r Ϫ p ϭ 0,
3.2p ϩ 2.4r ϩ 4c ϭ 16.8

37. peanuts, 2 lb; raisins, 1 lb; pretzels, 2 lb

38. If the determinant is zero, there is no unique solution to the system. There is either no solution or there are infinitely many solutions.
Sample answer: 2x ϩ y ϭ 4 and 4x ϩ 2y ϭ 8 has a det ϭ
0; there are infinitely many solutions of this system.
2x ϩ y ϭ 4 and
4x ϩ 2y ϭ 10 has a det ϭ 0; there are no solutions of this system. ©Glencoe/McGraw-Hill

1
2

5 s ϩ 14c ϭ 542.30

101

Algebra 2

Chapter 4

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39. Cramer’s Rule is a formula for the variables x and y where (x, y) is a solution for a system of equations.
Answers should include the following. • Cramer’s Rule uses determinants composed of the coefficients and constants in a system of linear equations to solve the system.
• Cramer’s Rule is convenient when coefficients are large or involve fractions or decimals. Finding the value of the determinant is sometimes easier than trying to find a greatest common factor if you are solving by using elimination or substituting complicated numbers.

40. B

41. 111Њ, 69Њ

45. B

42. 16

43. 40

44. Ϫ53

47.

1 1 1
R
3 3 3

46. A¿(1, 5), B¿(Ϫ2, 2),
C¿(Ϫ1, Ϫ1)
48.

y y ϭ 3x ϩ 5
(Ϫ2, Ϫ1)

O

x

y ϭ Ϫ2x Ϫ 5

(Ϫ2, Ϫ1)

102

Algebra 2

Chapter 4

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49.

50.

y xϩy ϭ7

y
2x Ϫ 4y ϭ 12

x

O

(4, 3)

O

1 x Ϫ y ϭ Ϫ1
2

x Ϫ 2y ϭ 10

x

51. c ϭ 10h ϩ 35
72
9
53. B
R
66 Ϫ23

52. [Ϫ4 32]
21
54. B R
43

(4, 3)

1. B
3.

1
4
1 Ϫ2
R
2 Ϫ1 Ϫ4 Ϫ1

no solution

Chapter 4
Practice Quiz 2
Page 194

2. A¿(Ϫ1, 2), B¿(Ϫ4, Ϫ1),
C¿(Ϫ1, Ϫ4), D¿(2, Ϫ1)
4. 22

5. Ϫ58

6. Ϫ105

7. 26

8. (1, Ϫ2)
10. (1, 2, 1)

9. (4, Ϫ5)

103

Algebra 2

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Lesson 4-7
1
0
1. D
0
0

0
1
0
0

0
0
1
0

0
0
T
0
1

Identity and Inverse Matrices
Pages 198–201

2. Exchange the values for a and d in the first diagonal in the matrix. Multiply the values for b and c by Ϫ1 in the second diagonal in the matrix. Find the determinant of the original matrix. Multiply the negative reciprocal of the determinant by the matrix with the above mentioned changes. 3. Sample answer: B

3 3
R
3 3

5. yes

6. B

2 5
R
3 8

4. no

4
1
B
27 Ϫ7

8. Ϫ

7. no inverse exists

10. yes

9. See students’ work.
11. yes

12. no

13. no

14. yes

15. yes

Ϫ1
R
Ϫ5

16. true

18. true
1 1 0
R
20. B
5 0 5

17. true

22. Ϫ B

19. false
1 1
B
7 4

Ϫ1
R
3

1
3

21. no inverse exists
23.
25.

1 Ϫ6
B
4 Ϫ2

Ϫ7
R
Ϫ3

1 Ϫ2
R
Ϫ2
1

7
1
B
34 Ϫ2

3
R
4

24. no inverse exists
26.

104

Algebra 2

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6
1
B
12 Ϫ5

1
1
B
32 Ϫ6

27. Ϫ

29.

31. 10 C

3
4

1
Ϫ
5

0
R
Ϫ2

5
R
2

5
8

Ϫ

3
10

30. 4 C

S

28. no inverse exists

S

1
4

3
4

1
Ϫ
6

1
2

32a. no
32b. Sample answer: y C
A

B

x

O A'

A''

B''

34. B

33a. yes

35. B

0 Ϫ4 4 8
R
0
4 12 8

37. dilation by a scale factor of

0 Ϫ2 2 4
R
0
2 6 4
C''

B'

C'

36. dilation by a scale factor of 2
38. B D

1
2

Ϫ1

1
2

0

0

1
2

T;

the graph of the

inverse transformation is the original figure.

105

Algebra 2

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39. MEET_IN_THE_LIBRARY

40. AT_SIX_THIRTY

41. BRING_YOUR_BOOK

42. See students’ work.

43. a ϭ Ϯ1, d ϭ Ϯ1, b ϭ c ϭ 0

44. A matrix can be used to code a message. The key to the message is the inverse of the matrix. Answers should include the following.
• The inverse matrix undoes the work of the matrix. So if you multiply a numeric message by a matrix it changes the message.
When you multiply the changed message by the inverse matrix, the result is the original numeric message. • You must consider the dimensions of the coding matrix so that you can write the numeric message in a matrix with dimensions that can be multiplied by the coding matrix. 46. D

45. A
Ϫ5 Ϫ9
R
47. B
Ϫ6 Ϫ11
49. C

3
5

Ϫ

1
5

1
5

2
Ϫ
5

51. F Ϫ

1

Ϫ1
1
3
7
3

S

2
3
8
3

Ϫ

53. (2, Ϫ4)

50. C

S

48. no inverse exists
2
5

1
3
2
5
16

1
2
2
1
Ϫ
8

1
4

0

5
32

0V

3
5

3
16

52. FϪ

1
3

1
3

Ϫ

54. (0, 7)

106

1
V
4

1
16

Ϫ

1
32

Ϫ

Algebra 2

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55. (Ϫ5, 4, 1)

56. 52

57. Ϫ14

58. 0

59. 1

60. Ϫ3

61. Ϫ5

62.

63.

1
3
3
8

5
2

64. Ϫ

66. 27

65. 7.82 tons/in2

1
2

1
2

67. 5

68. Ϫ

69. 3

70. 296

71. 300

72. Ϫ1

73. Ϫ2

74. 6

75. 4

76. Ϫ27

77. Ϫ34

Lesson 4-8

Using Matrices to Solve Systems of Equations
Pages 205–207
4. B

1 Ϫ1 x Ϫ3
RؒB RϭB R
1
3 y 5

2. Sample answer: x ϩ 3y ϭ 8 and 2x ϩ 6y ϭ 16

1. 2r Ϫ 3s ϭ 4, r ϩ 4s ϭ Ϫ2
3. Tommy; a 2 ϫ 1 matrix cannot be multiplied by a
2 ϫ 2 matrix.

3 Ϫ5
2
a
6. C 4
7
1S ؒ CbS
2
0 Ϫ1 c 2
3
g
8
5. B
RؒB RϭB R
Ϫ4 Ϫ7 h Ϫ5

9 ϭ C 3S
12
8. (1.5, Ϫ4)

7. (5, Ϫ2)

10. (1, 1.75)

9. (Ϫ3, 5)

107

Algebra 2

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12. B
14. B

4 Ϫ7 x 2
RؒB RϭB R
3
5 y 9

3 Ϫ7 m Ϫ43
RؒB RϭB
R
15. B
6
5 n Ϫ10

2
3 Ϫ5 a 16. C 7
0
3S ؒ CbS
3 Ϫ6
1
c
1
ϭ C 7S
Ϫ5

3 Ϫ5 2 x 17. C 1 Ϫ7 3 S ؒ C y S
4 0 Ϫ3 z 1 Ϫ1
0
x
18. C Ϫ2 Ϫ5 Ϫ6 S ؒ C y S
9 10 Ϫ1 z 9 ϭ C 11 S
Ϫ1

8 ϭ C Ϫ27 S
54

3
Ϫ5
6 r 19. C 11 Ϫ12 16 S ؒ C s S
Ϫ5
8
Ϫ3
t

20. (5, Ϫ2)

21 ϭ C 15 S
Ϫ7
21. (3, 4)

22. (Ϫ2, 3)
24. a , Ϫ3b
1
2

23. (6, 1)

25. aϪ , 4b
1
3

26. (2, Ϫ3)

27. (Ϫ2, Ϫ2)

28. (7, 3)

30. aϪ1, b
9
2

29. (0, 9)
31. a , b
3 1
2 3

32. 27 h of flight instruction and
23 h in the simulator
34. 80 mL of the 60% solution, and
120 mL of the 40% solution

33. 2010

3 Ϫ1 x 0
RؒB RϭB
R
1
2
y
Ϫ21

5 Ϫ6 a Ϫ47
RؒB RϭB
R
3
2
b
Ϫ17

13. B

11. h ϭ 1, c ϭ 12

108

Algebra 2

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35. The solution set is the empty set or infinite solutions.

36. The food and territory that two species of birds require form a system of equations. Any independent system of equations can be solved using a matrix equation. Answers should include the following.
• Let a represent the number of nesting pairs of
Species A and let b represent the number of nesting pairs of Species B.
Then, 140a ϩ 120b ϭ
20,000 and 500a ϩ 400b ϭ
69,000.
a
• B Rϭ b 400
1
B
4000 Ϫ500

Ϫ

Ϫ120
20,000
RؒB
R;
140
69,000

a ϭ 70 and b ϭ 85, so the area can support 70 pairs of Species A and 85 pairs of Species B.

37. D

38. 17 small, 24 medium, 11 large

39. (Ϫ6, 2, 5)

40. (1, Ϫ3, 2)

42. C

S

43. B

4 Ϫ5
R
Ϫ7
9
45. (4, Ϫ2)

Ϫ1

1
Ϫ
2

41. (0, Ϫ1, 3)

3
4

1

44. no inverse exists
46. (4.27, Ϫ5.11)

47. (Ϫ6, Ϫ8)

48. about 114.3 ft

49. {Ϫ4, 10}

50. {Ϫ5, 1}

51. {2, 7}

109

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Chapter 5 Polynomials
Lesson 5-1 Monomials
Pages 226–228
(2x 2)3 ϭ 8x 6 since
(2x 2)3 ϭ 2x 2 ؒ 2x 2 ؒ 2x 2 ϭ
2x ؒ x ؒ 2x ؒ x ؒ 2x ؒ x ϭ 8x 6

2. Sometimes; in general, x y ؒ x z ϭ x yϩz, so x y ؒ x z ϭ x yz when y ϩ z ϭ yz, such as when y ϭ 2 and z ϭ 2.

3. Alejandra; when Kyle used the
Power of a Product Property in his first step, he forgot to put an exponent of Ϫ2 on a.
Also, in his second step,

4. x 10

1
4

(Ϫ2)Ϫ2 should be , not 4.
5. 16b 4

6. 1

7. Ϫ6y 2

8. Ϫ

ab 4
9

9. 9p 2q 3

10.

1 w z

9 c d

12.

1
4x 6

11.

2 2

12 6

13. 4.21 ϫ 105

14. 8.62 ϫ 10Ϫ4

15. 3.762 ϫ 103

16. 5 ϫ 100

17. about 1.28 s

18. a 8

19. b 4

20. n16

21. z10

22. 16x 4

23. Ϫ8c 3

24. an

25. Ϫy 3z 2

26.

27. Ϫ21b5c 3

28. ab

29. Ϫ24r 7s 5

30. 24x 4y 4

31. 90a4b4

32. Ϫ

28x 4 y2 1
4y 4

110

Algebra 2

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34.

m 4n 9
3

35. Ϫ

cd 4
5

36.

a 2c 2
3b 4

a4
16b 4

37.

8y 3 x6 38.

1 x y

39.

1
3 6 v w

40.

a 4b 2
2

41.

2x 3y 2
5z 7

42. 6

2 2

43. 7

44. 4.623 ϫ 102

45. 4.32 ϫ 104

46. 1.843 ϫ 10Ϫ4

47. 6.81 ϫ 10Ϫ3

48. 5.0202 ϫ 108

49. 6.754 ϫ 108

50. 1.245 ϫ 1010

51. 6.02 ϫ 10Ϫ5

52. 4.5 ϫ 102

53. 6.2 ϫ 1010

54. 4.225 ϫ 109

55. 1.681 ϫ 10Ϫ7

56. 6.08 ϫ 109

57. 2 ϫ 10Ϫ7 m

58. 1.67 ϫ 1025

59. about 330,000 times

60. 10010 ϭ (102)10 or 1020, and
10100 Ͼ 1020, so 10100 Ͼ 10010.

61. Definition of an exponent

62. (ab)m m factors
6447448
ϭ ab ؒ ab ؒ p ؒ ab m factors m factors
64748 64748 ϭaؒaؒpؒaؒbؒbؒpؒb ϭ a mb m

111

Algebra 2

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63. Economics often involves large amounts of money.
Answers should include the following. • The national debt in 2000 was five trillion, six hundred seventy-four billion, two hundred million or 5.6742 ϫ
1012 dollars. The population was two hundred eighty-one million or 2.81 ϫ 108.
• Divide the national debt by the population.
5.6742 ϫ 1012
2.81 ϫ 108

64. D

Ϸ

\$2.0193 ϫ 104 or about \$20,193 per person.
66. (1, 2)
Ϫ2 Ϫ5
68. c d 1
2

65. B
67. (Ϫ3, 3)
69. C

1
2

Ϫ

3
2 S

70. Ϫ6

1 Ϫ2

71. 7

72. (2, 3, Ϫ1)

73. (2, 0, 4)

74.

Median Age (yr)

Median Age of Vehicles y 8
7
6
5
4
0

0

10
20
30
Years Since 1970

75. Sample answer using (0, 4.9) and (28, 8.3): y ϭ 0.12x ϩ 4.9

76. Sample answer: 9.7 yr

77. 7

78. Ϫ3

79. 2x ϩ 2y

x

80. 3x Ϫ 3z

112

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81. 4x ϩ 8

82. Ϫ6x ϩ 10

83. Ϫ5x ϩ 10y

84. 3y Ϫ 15

Lesson 5-2 Polynomials
Page 231–232
1. Sample answer: x 5 ϩ x 4 ϩ x 3

2. 4

3.

4. yes, 1

x x x

2

x
2

x

x
2

x

2

x

x

x

x

x

x

5. yes, 3

6. no

7. 10a Ϫ 2b

8. Ϫ3x 2 Ϫ 7x ϩ 8
10. 10p 3q 2 Ϫ 6p 5q 3 ϩ 8p3q 5

9. 6xy ϩ 18x
11. y 2 Ϫ 3y Ϫ 70

12. x 2 ϩ 9x ϩ 18

13. 4z 2 Ϫ 1

14. 4m 2 Ϫ 12mn ϩ 9n 2

15. 7.5x 2 ϩ 12.5x ft 2

16. yes, 2

17. yes, 3

18. no

19. no

20. yes, 6

21. yes, 7

22. 4x 2 ϩ 3x Ϫ 7

23. Ϫ3y Ϫ 3y 2

24. r 2 Ϫ r ϩ 6

25. 10m 2 ϩ 5m Ϫ 15

26. 4x 2 Ϫ 3xy Ϫ 6y 2

27. 7x 2 Ϫ 8xy ϩ 4y 2

28. 4b 2c Ϫ 4bdz

29. 12a 3 ϩ 4ab

30. 15a3b3 Ϫ 30a4b3 ϩ 15a5b6

31. 6x 2y 4 Ϫ 8x 2y 2 ϩ 4xy 5

32. 6x 3 ϩ 9x 2y Ϫ 12x 3y 2

33. 2a4 Ϫ 3a3b ϩ 4a4b4

34. 46.75 Ϫ 0.018x

35. Ϫ0.001x 2 ϩ 5x Ϫ 500

36. \$5327.50

37. p 2 ϩ 2p Ϫ 24

38. a 2 ϩ 9a ϩ 18

39. b 2 Ϫ 25

40. 36 Ϫ z 2

41. 6x 2 ϩ 34x ϩ 48

42. 8y 2 ϩ 16y Ϫ 42

113

Algebra 2

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43. a 6 Ϫ b 2

44. 2m 4 Ϫ 7m 2 Ϫ 15

45. x 2 Ϫ 6xy ϩ 9y 2

46. 1 ϩ 8c ϩ 16c 2

47. d 2 Ϫ 2 ϩ

1 d4 48. xy 3 ϩ y ϩ

1 x 49. 27b 3 Ϫ 27b 2c ϩ 9bc 2 Ϫ c 3

50. x 3 Ϫ y 3

51. 9c 2 Ϫ 12cd ϩ 7d 2

52. Ϫ18x 2 ϩ 27x Ϫ 10

53. R 2 ϩ 2RW ϩ W 2

54. 14; Sample answer:
(x 8 ϩ 1)(x 6 ϩ 1) ϭ x 14 ϩ x 8 ϩ x 6 ϩ 1

55. The expression for how much an amount of money will grow to is a polynomial in terms of the interest rate. Answers should include the following.
• If an amount A grows by r percent for n years, the amount will be A(1 ϩ r )n after n years. When this expression is expanded, a polynomial results.
• 13,872(1 ϩ r )3, 13,872r 3 ϩ
41,616r 2 ϩ 41,616r ϩ
13,872
• Evaluate one of the expressions when r ϭ 0.04.
For example,
13,872(1 ϩ r)3 ϭ
13,872(1.04)3 or \$15,604.11 to the nearest cent. The value given in the table is
\$15,604 rounded to the nearest dollar.

56. D

57. B

58. Ϫ64d 6

59. 20r 3t 4

60.

61.

b2
4a 2

xz 2 y2 62. (1, 4)

114

Algebra 2

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63.

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64.

y

y

x ϩ y Ͼ Ϫ2 x O

x

O

y ϭ Ϫ1x ϩ 2
3

65.

66. x 2

y

2x ϩ y ϭ 1

x

O

68. xy 2

67. 2y 3
69. 3a 2

Lesson 5-3 Dividing Polynomials
Pages 236–238
(x 2 ϩ x ϩ 5) Ϭ (x ϩ 1)

2. The divisor contains an x 2 term. 3. Jorge; Shelly is subtracting in the columns instead of adding.

4. 6y Ϫ 3 ϩ 2x

5. 5b Ϫ 4 ϩ 7a

6. x Ϫ 12

7. 3a3 Ϫ 9a2 ϩ 7a Ϫ 6

8. z 4 ϩ 2z 3 ϩ 4z 2 ϩ 5z ϩ 10

9. x 2 Ϫ xy ϩ y 2

10. x 2 ϩ 11x Ϫ 34 ϩ

11. b3 ϩ b Ϫ 1

12. 2y ϩ 5

13. 3b ϩ 5

14. B

15. 3ab Ϫ 6b 2

16. 5y Ϫ

17. 2c 2 Ϫ 3d ϩ 4d 2

60 xϩ2 18. 4n 2 ϩ 3mn Ϫ 5m

115

6y 2 x ϩ 3xy 2

Algebra 2

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2 b 19. 2y 2 ϩ 4yz Ϫ 8y 3z 4

20. Ϫa2b ϩ a Ϫ

21. b 2 ϩ 10b

22. x Ϫ 15

23. n 2 Ϫ 2n ϩ 3

24. 2c 2 ϩ c ϩ 5 ϩ

25. x 3 Ϫ 5x 2 ϩ 11x Ϫ 22 ϩ

39 xϩ2 6 cϪ2 26. 6w 4 ϩ 12w 3 ϩ 24w 2 ϩ
30w ϩ 60

27. x 2

28. x 2 ϩ 3x ϩ 9

29. y 2 Ϫ y Ϫ 1

30. m 2 Ϫ 7

31. a3 Ϫ 6a 2 Ϫ 7a ϩ 7 ϩ

3 aϩ1 32. 2m 3 ϩ m 2 ϩ 3m Ϫ

34. 3c 4 Ϫ c 3 ϩ 2c 2 Ϫ 4c ϩ

33. x 4 Ϫ 3x 3 ϩ 2x 2 Ϫ 6x ϩ
19 Ϫ

5 mϪ3 56 xϩ3 9Ϫ

13 cϩ2 4 bϩ1 35. g ϩ 5

36. 2b 2 Ϫ b Ϫ 1 ϩ

37. t 4 ϩ 2t 3 ϩ 4t 2 ϩ 5t ϩ 10

38. y 4 Ϫ 2y 3 ϩ 4y 2 Ϫ 8y ϩ 16

39. 3t 2 Ϫ 2t ϩ 3

40. h 2 Ϫ 4h ϩ 17 Ϫ

41. 3d 2 ϩ 2d ϩ 3 Ϫ
43. x 3 Ϫ x Ϫ

2
3d Ϫ 2

51
2h ϩ 3

42. x 2 ϩ x Ϫ 1

6
2x ϩ 3

44. 2x 3 ϩ x 2 Ϫ 1 ϩ

2
3x ϩ 1

Ϫ3x ϩ 7 x2 ϩ 2

45. x Ϫ 3

46. x 2 Ϫ 1 ϩ

47. x ϩ 2

48. x Ϫ 3

49. x 2 Ϫ x ϩ 3

50. 2y 2 Ϫ 3y ϩ 1

116

Algebra 2

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51. \$0.03x ϩ 4 ϩ

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1000 x 52. Let x be the number.
Multiplying by 3 results in 3x.
The sum of the number, 8, and the result of the multiplication is x ϩ 8 ϩ 3x or 4x ϩ 8. Dividing by the sum of the number and
2 gives

4x ϩ 8 xϩ2 or 4. The end

result is always 4.
53. 170 Ϫ

170 t ϩ1

54. 85 people

2

55. x 3 ϩ x 2 ϩ 6x Ϫ 24 ft

56. x Ϫ 2 s

57. x 2 ϩ 3x ϩ 12 ft /s

58. Sample answer: r 3 Ϫ 9r 2 ϩ
27r Ϫ 28 and r Ϫ 3

59. Division of polynomials can be used to solve for unknown quantities in geometric formulas that apply to manufacturing situations.
Answers should include the following. • 8x in. by 4x ϩ s in.
• The area of a rectangle is equal to the length times the width. That is, A ϭ /w .
• Substitute 32x 2 ϩ x for A,
8x for /, and 4x ϩ s for w.
Solving for s involves dividing 32x 2 ϩ x by 8x.
A ϭ /w
2
32x ϩ x ϭ 8x (4x ϩ s)

60. A

32x 2 ϩ x
8x
1
4x ϩ
8
1
8

ϭ 4x ϩ s ϭ 4x ϩ s ϭs The seam is
61. D

1
8

inch.
62. Ϫx 2 Ϫ 4x ϩ 14
117

Algebra 2

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63. y 4z 4 Ϫ y 3z 3 ϩ 3y 2z

64. y 2 ϩ 2y Ϫ 15

65. a 2 Ϫ 2ab ϩ b 2

66. 5 ϫ 102 s or 8 min 20 s

67. y ϭ Ϫx ϩ 2

68. y ϭ x Ϫ

69. 9

70. 12

71. 4

72. 3

73. 6

74. 5

2
3

4
3

Chapter 5
Practice Quiz 1
Page 238
1. 6.53 ϫ 108

2. 7.2 ϫ 10Ϫ3

3. Ϫ108x 8y 3

4.

5.

x2 z6 a3 b4c 3

6. 2x ϩ 5y

7. 3t 2 ϩ 2t Ϫ 8
9. m 2 Ϫ 3 Ϫ

8. n3 Ϫ n 2 Ϫ 5n ϩ 2

19 mϪ4 10. d 2 ϩ d Ϫ 3

Lesson 5-4 Factoring Polynomials
Pages 242–244
1. Sample answer: x 2 ϩ 2x ϩ 1

3. sometimes

2. Sample answer: If a ϭ 1 and b ϭ 1, then a 2 ϩ b 2 ϭ 2 but
1a ϩ b2 2 ϭ 4.

5. a(a ϩ 5 ϩ b)

6. (x ϩ 7)(3 Ϫ y)

7. (y ϩ 2)(y Ϫ 4)

8. (z Ϫ 6)(z ϩ 2)

4. Ϫ6x(2x ϩ 1)

9. 3(b Ϫ 4)(b ϩ 4)

10. (4w ϩ 13)(4w Ϫ 13)

11. (h ϩ 20)(h 2 Ϫ 20h ϩ 400)

12.

118

xϪ4 xϪ7 Algebra 2

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2y yϪ4 14. x ϩ y cm

15. 2x(y 3 Ϫ 5)

16. 6ab 2(a ϩ 3b)

17. 2cd 2(6d Ϫ 4c ϩ 5c 4d )

18. prime

19. (2z Ϫ 3)(4y Ϫ 3)

20. (3a ϩ 1)(x Ϫ 5)

21. (x ϩ 1)(x ϩ 6)

22. (y Ϫ 1)(y Ϫ 4)

23. (2a ϩ 1)(a ϩ 1)

24. (2b Ϫ 1)(b ϩ 7)

25. (2c ϩ 3)(3c ϩ 2)

26. (3m ϩ 2)(4m Ϫ 3)

27. 3(n ϩ 8)(n Ϫ 1)

28. 3(z ϩ 3)(z ϩ 5)

29. (x ϩ 6)2

30. (x Ϫ 3)2

31. prime

32. 3(m ϩ n)(m Ϫ n)

33. (y 2 ϩ z)(y 2 Ϫ z)

34. 3(x ϩ 3y)(x Ϫ 3y)

35. (z ϩ 5)(z 2 Ϫ 5z ϩ 25)

36. (t Ϫ 2)(t 2 ϩ 2t ϩ 4)

37. (p 2 ϩ 1)(p ϩ 1)(p Ϫ 1)

38. (x 2 ϩ 9)(x ϩ 3)(x Ϫ 3)

39. (7a ϩ 2b)(c ϩ d )(c Ϫ d )

40. (8x ϩ 3)(x ϩ y ϩ z)

41. (a Ϫ b)(5ax ϩ 4by ϩ 3cz)

42. (a ϩ 3b)(3a ϩ 5)(a Ϫ 1)

43. (3x Ϫ 2)(x ϩ 1)

44. (2y ϩ 1)(y ϩ 4)

45. 30 ft by 40 ft

46.

xϩ1 xϪ4 xϪ5 xϪ2 47.

xϩ5 xϪ6 48.

49.

xϪ4 x ϩ 2x ϩ 4

50. x

2

51. x ϩ 2

52. x Ϫ 1 s

53. 16x ϩ 16 ft /s

54. x Ϫ 8 cm

119

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55. (8pn ϩ 1)2

56. Factoring can be used to find possible dimensions of a geometric figure, given the area. Answers should include the following.
• Since the area of the rectangle is the product of its length and its width, the length and width are factors of the area. One set of possible dimensions is
4x Ϫ 2 by x ϩ 3.
• The complete factorization of the area is 2(2x Ϫ 1)
(x ϩ 3), so the factor of
2 could be placed with either 2x Ϫ 1 or x ϩ 3 when assigning the dimensions. 57. B

58. C

59. yes

60. no; (x ϩ 2)(x 2 Ϫ 2x ϩ 4)

61. no; (2x ϩ 1)(x Ϫ 3)

62. yes

63. t 2 Ϫ 2t ϩ 1

64. y ϩ 3

65. x 2 ϩ 2

66. x 3 ϩ x 2 Ϫ 2x ϩ 2 ϩ

67. 4x 2 ϩ 3xy Ϫ 3y 2

68. 14x 2 ϩ 26x Ϫ 4
70. c

69. [Ϫ2]

1
3x Ϫ 2

Ϫ36 7 d 18 4

72. yes

71. 15 in. by 28 in.
73. no

74. Distributive Property

75. Associative Property (ϩ)

76. rational

77. irrational

78. rational

79. rational

80. irrational

81. rational

120

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Lesson 5-5 Roots of Real Numbers
Pages 247–249
1. Sample answer: 64

2. If all of the powers in the result of an even root have even exponents, the result is nonnegative without taking absolute value.

3. Sometimes; it is true when x Ͼ 0.

4. 8.775

5. Ϫ2.668

6. 2.632

7. 4

8. 2

9. Ϫ3

10. not a real number

11. x

12. 0 y 0

15. about 3.01 mi

14. 0 4x ϩ 3y 0
16. 11.358

17. Ϫ12.124

18. 0.933

19. 2.066

20. 3.893

21. Ϫ7.830

22. 4.953

23. 3.890

24. 4.004

25. 4.647

26. 26.889

27. 59.161

28. 15

29. Ϯ13

30. not a real number

31. 18

32. Ϫ3

33. Ϫ2

34.

13. 6 0 a 0 b 2

1
4

1
5

36. 0.5

37. Ϫ0.4

38. z 2

35.

39. Ϫ 0 x 0
41. 8a 4

40. 7 0 m3 0

43. Ϫc 2

44. 25g 2

42. 3r

46. 5x 2 0 y 3 0

45. 4z 2

121

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49. 3p 6 0 q 3 0

48. 13x 4y 2

53. p ϩ q

52. 0 4x Ϫ y 0

57. not a real number

56. 0 2a ϩ 1 0
58. 2

59. Ϫ5

60. about 127.28 ft

61. about 1.35 m

62. about 11,200 mրs

63. x ϭ 0 and y Ն 0, or y ϭ 0 and x Ն 0

64. The speed and length of a wave are related by an expression containing a square root. Answers should include the following.
3.00 knots, and 4.24 knots
• As the value of / increases, the value of s increases.

65. B

66. D

67. 7xy 2(y Ϫ 2xy 3 ϩ 4x 2)

68. (a ϩ 3)(b Ϫ 5)

69. (2x ϩ 5)(x ϩ 5)

70. (c Ϫ 6)(c 2 ϩ 6c ϩ 36)

47. 6x 2z 2

50. 2ab

51. Ϫ3c 3d 4

54. Ϫ 0 x ϩ 2 0

55. 0 z ϩ 4 0

71. 4x 2 ϩ x ϩ 5 ϩ
73. c

810 2320 d 1418 2504

8 xϪ2 72. x 3 Ϫ x 2 ϩ x
74. (Ϫ2, 2)

75. (1, Ϫ3)

76. (9, 4)

77. x 2 ϩ 11x ϩ 24

78. y 2 ϩ 3y Ϫ 10

79. a 2 Ϫ 7a Ϫ 18

80. a 2 ϩ 3ab ϩ 2b 2

81. x 2 Ϫ 9y 2

82. 6w 2 Ϫ 7wz Ϫ 5z 2

122

Algebra 2

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1a

Page 123 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:

ϭ 1a only

22 ϩ 23 ϩ 22

Lesson 5-6 Radical Expressions
Pages 254–256
1. Sometimes;

1

n

n

when a ϭ 1.
3. The product of two conjugates yields a difference of two squares.
Each square produces a rational number and the difference of two rational numbers is a rational number. 4. 1527

4
5. 2x 0 y 0 2x

214y
4y

8. 225

9. 2a 2b 2 23

12. 3 ϩ 323 Ϫ 25 Ϫ 215

3
11. 22 22

6.

13. 2 ϩ 25

3

7. Ϫ24235

4
10. 523 ϩ 323
3
20. 2y 22

19. 5x 2 22

14. about 49 mph
22. 2ab 2 210a

21. 3 0 x 0 y 22y
15. 923

16. 622

3
17. 322

4
18. 226

23. 6y 2z 27
25.

26
2

1 c 0d
3

a 2 2b b2 4
0 2c

26.

31. 36 27
3

27.
29.

33.

1 wz 2

2r 4 2t t5 4

28.

26
2

30.

210
5

32. Ϫ60230

38. 425 ϩ 23 26
34.

36. 522

35. 323
37. 723 Ϫ 222

2 54
3

5
2wz 2

3
24. 4mn23mn 2

3

123

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40. 6 ϩ 326 ϩ 227 ϩ 242

28 ϩ 7 23
13

39. 25 Ϫ 522 ϩ 526 Ϫ 223
Ϫ1 Ϫ 23
2

5 26 Ϫ 3 22
22

12 ϩ 7 22
23

41. 13 Ϫ 2222

42. 8 Ϫ 2215

43.

44.

45.
47.

48. 2x ϩ 1

2x 2 Ϫ 1 xϪ1 46.

49. 6 ϩ 1622 yd, 24 ϩ 622 yd2

50. The square root of a difference is not the difference of the square roots. 51. 0 ft /s

52. d ϭ v

53. about 18.18 m

54. 80 ft /s or about 55 mph

55. x and y are nonnegative.

56. The formula for the time it takes an object to fall a certain distance can be written in various forms involving radicals. Answers should include the following.
• By the Quotient Property

24.9h
4.9

2g

2g

of Radicals, t ϭ
Multiply by

22d
2g

.

22dg
.
g

to

rationalize the denominator.
The result is h ϭ
• about 1.12 s
57. B

58. D

59. 12z 4

60. 6ab 3

61. 0 y ϩ 2 0
63.

62.

Ϫ2 Ϫ4
64. £ 9 15 §
3 Ϫ5

xϩ7 xϪ4 124

Algebra 2

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1
4
d
Ϫ5 Ϫ4

66. 16, Ϫ15

67. consistent and independent

68. \$4.20

69. Ϫ5

70. 2

71. Ϫ2, 4

72. Ϫ , 1

7
3

73. 5x 0 x Ͼ 66
75.

1
4

77.

74. 5x 0 x Ն Ϫ76
76.

1
2

5
6

78.

13
12

79.

13
24

80.

19
30

81.

3
8

82. Ϫ

5
12

Chapter 5
Practice Quiz 2
Page 256
1. x 2y (3x ϩ y ϩ 1)

2. prime

3. a(x ϩ 3)2

4. 8(r Ϫ 2s 2)(r 2 ϩ 2rs 2 ϩ 4s 4)
6. Ϫ4a 2b 3

5. 6 0 x 0 0 y 3 0

7. 0 2n ϩ 3 0

8.
10.

9. Ϫ1 Ϫ17

125

x 2 2y y2 8 Ϫ 3 22
2

Algebra 2

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2. In radical form, the expression would be 2Ϫ16, which is not a real number because the index is even and the radicand is negative.

Lesson 5-7 Rational Exponents
Pages 260–262
1. Sample answer: 64

3. In exponential form 2bm is
1
equal to (b m)n . By the Power of a 1Power Property, m m
(b m )n ϭ b n . But, b n is also
1
equal to (b n )mby the Power of a Power Property. This last n expression is equal to ( 2b)m . n n
Thus, 2bm ϭ (2b)m.

3
4. 27

n

3
3
5. 2x 2 or (2x)2
1

5

1

6. 264

7

8. 5

7. 63x 3y 3
9.

1
3

10. 9
11

12. a12

11. 2

2

13. x

14.

19. 23
3
2

15. a b

5
21. 26

17.

z3
2z

16.

2
3

m 3 n3 mn 18. 23x
1

2

2
3

3
22. 24

1

z (x Ϫ 2y)2 x Ϫ 2y

3
24. x 2 2x 2

5
5
23. 2c 2 or (2c)2

20. \$5.11

1

1

26. 623

25. 232

1

1

2

1

27. 2 z 2

28. 53 x 3 y 3

29. 2

30. 6

31.

1
5

32.

126

1
27

Algebra 2

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1
8

1
9

34. Ϫ

35. 81

36. 4096

37.

2
3

38. 27

39.

4
3

40.

42. x 3

41. y 4
1

1

43. b 5

44. a9
5
x6
46.
x

1

45.

w5 w 1

1

48. r 2

47. t 4

54. 23
15

53. 25
5

49.
51.

50.

y 2 Ϫ 2y 2 yϪ4 6
55. 17 217
3

57. 25x 2y 2 xy 1z z x ϩ 3x 2 ϩ 2 xϪ1 1

ab 2 c 2 c 6
56. 5255

62. 26

61. 12

3

4

58. b29a 2b
3

60.

3

1

2

64. x Ϫ x 3z 3

63. 216 Ϫ 5
3

2c 16 c 52.

a12
6a

59.

1
2

1

1

65. 22 ϩ 32

66. 2 ؒ 3 3

67. 880 vibrations per second

68. about 262 vibrations per second ©Glencoe/McGraw-Hill

127

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70. Rewrite the equation so that the bases are the same on each side.
1
9x ϭ 3xϩ 2
1
(32 )x ϭ 3xϩ 2
1
32x ϭ 3xϩ 2
Since the bases are the same and this is an equation, the exponents must be equal.
1
2

Solve 2x ϭ x ϩ . The result
1
2

is x ϭ .
71. The equation that determines the size of the region around a planet where the planet’s gravity is stronger than the Sun’s can be written in terms of a fractional exponent. Answers should include the following.
• The radical form of the
2
Mp
.
2
B Ms

B Ms

equation is r ϭ D 5 a r ϭ D5

Mp

b

2

72. C

or

Multiply the

2
3
Mp Ms
ؒ 3
2
B Ms Ms

fraction under the radical by
3
2
Mp Ms
5
B Ms

3
Ms
3
Ms

.

r ϭ D5

5
2 3
2Mp Ms

ϭ D5 ϭD ϭ

5
5
2Ms

5
2 3
D2MpMs

Ms

The simplified radical form is rϭ 5
2 3
D2Mp Ms

Ms

.

• If Mp and Ms are constant, then r increases as D increases because r is a linear function of D with positive slope.

128

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74. 2x 0 y 0 2x

75. 36 22
73. C

76. 222

78. 4 0 x Ϫ 5 0

77. 8
79.

1 2 x 2

80. 1440

81. x Ϫ 2

82. 2x Ϫ 3

83. x ϩ 22x ϩ 1

84. 4x Ϫ 122x ϩ 9

Lesson 5-8

2. The trinomial is a perfect square in terms of 1x . x Ϫ 61x ϩ 9 ϭ (1x Ϫ 3)2, so the equation can be written as (1x Ϫ 3)2 ϭ 0.
Take the square root of each side to get 1x Ϫ 3 ϭ 0. Use the Addition Property of
Equality to add 3 to each side, then square each side to get x ϭ 9.

Radical Equations and Inequalities
Pages 265–267

1. Since x is not under the radical, the equation is a linear equation, not a radical equation. The solution is xϭ 23 Ϫ 1
.
2

2x ϩ 2x ϩ 3 ϭ 3

4. 2

5. Ϫ9

6. no solution

7. 15

8. 18
3
2

9. 31

10. Ϫ Յ x Յ 39

11. 0 Յ b Ͻ 4

12. about 13.42 cm

13. 16

14. 49

15. no solution

16. no solution

17. 9

18. 5

19. Ϫ1

20.

21. Ϫ20

22. 5

129

27
2

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23. no solution

24. 9

25. x Ͼ 1

26. Ϫ2 Յ x Յ 1

27. x Յ Ϫ11

28. y Ͼ 4

29. no solution

30. 4

31. 3

32. no solution

33. 0 Յ x Յ 2

34. 0 Յ a Ͻ 3

35. b Ն 5

36. c Ͼ Ϫ

37. 3

38. 16

39. 1152 lb

40. t ϭ

41. 34 ft

42. 21.125 kg

4␲ 2r 3
B GM
79
16

43. Since 1x ϩ 2 Ն 0 and
12x Ϫ 3 Ն 0, the left side of the equation is nonnegative.
Therefore, the left side of the equation cannot equal Ϫ1.
Thus, the equation has no solution. 44. If a company’s cost and number of units manufactured are related by an equation involving radicals or rational exponents, then the production level associated with a given cost can be found by solving a radical equation. Answers should include the following.
3
• C ϭ 102n 2 ϩ 1500
2

• 10,000 ϭ 10n 3 ϩ 1500

C ϭ 10,000

8500 ϭ 10n
2

850 ϭ n 3
3

8502 ϭ n

2
3

Subtract 1500 from each side.
Divide each side by 10.
Raise each side to the
3
power.
2

24,781.55 Ϸ n

Use a calculator. Round down so that the cost does not exceed \$10,000.
The company can make at most 24,781 chips.

130

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50. 6 0 x 3 0 y 22y

45. D

46. C

2 100
10
3
7

1

47. 5

48. (x ϩ 7)2
2

49. (x 2 ϩ 1)3
3

51.

52. 28 Ϫ 1023
54. 4 ϩ x

53. x ϩ y ϭ 7, 30x ϩ 20y ϭ 160;
(2, 5) y 30x ϩ 20y ϭ 160
(2, 5)

x ϩy ϭ7

x

O

55. 1 Ϫ y

56. 2 ϩ 4x

57. Ϫ11

58. 4 ϩ 6z ϩ 2z 2

59. Ϫ3 Ϫ 10x Ϫ 8x 2

Lesson 5-9 Complex Numbers
Pages 273–275
5. 5i 0 xy 0 22

1a. true
1b. true

2. all of them

3. Sample answer: 1 ϩ 3i and
1 Ϫ 3i

4. 6i
6. 12

7. Ϫ18023

8. i
10. 42 Ϫ 2i

9. 6 ϩ 3i
11.

7
17

Ϫ

11 i 17

12. Ϯ3i

13. Ϯ2i22

14. Ϯi25

15. 3, Ϫ3

16. 5, 4

17. 10 ϩ 3j amps

18. 12i

131

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21. 10a 2 0 b 0 i

20. 8x 2i

23. Ϫ12

24. Ϫ48i

25. Ϫ75i

26. i

27. 1

28. Ϫ1

29. Ϫi

30. 9 ϩ 2i

31. 6

32. 2

33. 4 Ϫ 5i

34. 25

35. 6 Ϫ 7i

36. 8 ϩ 4i

37. Ϫ8 ϩ 4i

38.

2
5

40.

39
17

19. 9i

39.

10
17

41.

2
5

Ϫ

22. Ϫ1322

6 i 17

1
5

ϩ i

6
5

ϩ i ϩ 14 i 17

5 23 i 14

42. Ϫ163 Ϫ 16i

2 22 i 3

43. 20 ϩ 15i

44.

11
14

Ϫ

51. Ϯ2i 23

46. (j ϩ 4)x 2 ϩ (3 Ϫ i )x ϩ 2 Ϫ 4i

49. Ϯ4i

50. Ϯi 26

55. Ϯ

56. 4, 5

1
3

45. Ϫ Ϫ

52. Ϯi 23

53. Ϯ2i 210

54. Ϯ3i 25

47. (5 Ϫ 2i )x 2 ϩ (Ϫ1 ϩ i )x ϩ 7 ϩ i

48. Ϯi

25 i 2

7
2

58. Ϫ , Ϫ3

57. 4, Ϫ3
59.

5
,
3

61.

67 19
,
11 11

60. 3, 1

4

62. 5 Ϫ 2j ohms
64. 4 ϩ 2j amps

63. 13 ϩ 18j volts

132

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65. Case 1: i Ͼ 0
Multiply each side by i to get i 2 Ͼ 0 ؒ i or Ϫ1 Ͼ 0. This is a contradiction. Case 2: i Ͻ 0
Since you are assuming i is negative in this case, you must change the inequality symbol when you multiply each side by i. The result is again i 2 Ͼ 0 ؒ i or Ϫ1 Ͼ 0, a contradiction. Since both possible cases result in contradictions, the order relation “Ͻ” cannot be applied to the complex numbers. 66. Some polynomial equations have complex solutions.
Answers should include the following. • a and c must have the same sign.
• Ϯi

67. C

68. C

69. Ϫ1, Ϫi, 1, i, Ϫ1, Ϫi, 1, i, Ϫ1

70. Examine the remainder when the exponent is divided by 4.
If the remainder is 0, the result is 1. If the remainder is
1, the result is i. If the remainder is 2, the result is
Ϫ1. And if the remainder is
3, the result is Ϫi.

71. 12

72. 11

73. 4

74. x 15

7

1

75. y
77. c

1
3

76.

2
1 Ϫ2 d 3 Ϫ2
1

a4 a 78. c

133

1
0
d
0 Ϫ1

Algebra 2

Chapter 5

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2 1 Ϫ2 d Ϫ3 2 Ϫ1

80.

y

B'

C'

x

O

A'

81. sofa: \$1200, love seat: \$600, coffee table: \$250

82.

y

y ϭ Ϫ2x Ϫ 2
O

x

yϭxϩ1

83.

84.

y

1
10

xϩyϭ1 x Ϫ 2y ϭ 4

O

x

85. 0

134

Algebra 2

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Chapter 6 Quadratic Functions and Inequalities
Lesson 6-1 Graphing Quadratic Functions
Pages 290–293
2a. (2, 1); x ϭ 2
2b. (Ϫ3, Ϫ2); x ϭ Ϫ3

1. Sample answer: f (x ) ϭ
3x 2 ϩ 5x Ϫ 6; 3x 2, 5x, Ϫ6
3a. up; min.
3b. down; max.
3c. down; max.
3d. up; min.

4a. 0; x ϭ 0; 0
4b.

x f(x)
Ϫ1 Ϫ4
0
0
1 Ϫ4

p

4c.

f (x)
O (0, 0)

x

f (x) ϭ Ϫ4x 2

5a. 0; x ϭ Ϫ1; Ϫ1

6a. Ϫ1; x ϭ 2; 2

5b.

6b.

x f(x)
Ϫ3
3
Ϫ2
0
Ϫ1 Ϫ1
0
0
1 3

p

5c.

6c.

f(x)

x
0
1
2
3
4

p

f(x)
Ϫ1
2
3
2
Ϫ1

f(x)
(2, 3)

f (x) ϭ x 2 ϩ 2x

O

O

x

f (x) ϭ Ϫx 2 ϩ 4x Ϫ 1

x

(Ϫ1, Ϫ1)

135

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7a. 3; x ϭ Ϫ4; Ϫ4

8a. 1; x ϭ 1; 1

7b.

8b.

x f(x)
Ϫ1
7
1
0
1 Ϫ1
2
1
3
7

8c.

f(x)

x f(x) Ϫ6
Ϫ9
Ϫ5 Ϫ12
Ϫ4 Ϫ13
Ϫ3 Ϫ12
Ϫ2
Ϫ9

p

7c.

f(x)
Ϫ10

Ϫ8

x

O

Ϫ4

p

Ϫ4 f (x) ϭ 2x 2 Ϫ 4x ϩ 1

Ϫ8
2

O

f (x) ϭ x ϩ 8x ϩ 3

Ϫ12
(Ϫ4, Ϫ13)

5
3

5
3

9a. 0; x ϭ Ϫ ; Ϫ
9b.

x
Ϫ3
Ϫ2

(1, Ϫ1)

x

10. max.; 7

f(x)
Ϫ3
Ϫ8

5
3

25
3

Ϫ

Ϫ

Ϫ1
0

Ϫ7
0

9c.

f(x)
4
Ϫ2

Ϫ4

O

2

x

Ϫ4

(

)

Ϫ 5 , Ϫ 25
3

3

Ϫ8 f (x) ϭ 3x 2 ϩ 10x
Ϫ12

25
4

12. min.; 0

11. min.; Ϫ

136

Algebra 2

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13. \$8.75

14a. 0; x ϭ 0; 0
14b.

x f(x)
Ϫ2 8
Ϫ1 2
0 0
1 2
2 8

p

14c.

f (x)

f (x ) ϭ 2x 2

(0, 0)

x

O

15a. 0; x ϭ 0; 0

16a. 4; x ϭ 0; 0

15b.

16b.

x f(x)
Ϫ2 Ϫ20
Ϫ1 Ϫ5
0
0
1 Ϫ5
2 Ϫ20

p

15c.

O f (x ) ϭ Ϫ5x

(0, 0)

p

16c.

f(x)
2

x f(x)
Ϫ2 8
Ϫ1 5
0 4
1 5
2 8 f(x) 12

x

8
(0, 4) f (x ) ϭ x 2 ϩ 4
Ϫ4

Ϫ2

O

17a. Ϫ9; x ϭ 0; 0

18b.

4x

18a. Ϫ4; x ϭ 0; 0

17b.

2

x
Ϫ2
Ϫ1
0
1
2

p

f(x)
Ϫ5
Ϫ8
Ϫ9
Ϫ8
Ϫ5

137

x
Ϫ2
Ϫ1
0
1
2

p

f(x)
4
Ϫ2
Ϫ4
Ϫ2
4
Algebra 2

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17c.

18c.

f(x)

f(x)

4
Ϫ4

O

Ϫ2

4x

2

x

O

Ϫ4
(0, Ϫ9)

f (x ) ϭ 2x 2 Ϫ 4

f (x ) ϭ x 2 Ϫ 9

19a. 191; x ϭ 0; 0

20a. 4; x ϭ 2; 2

19b.

20b.

(0, Ϫ4)

x f(x)
Ϫ2 13
4
Ϫ1
0
1
1
4
2 13

p

19c.

20c.

f (x)

f (x ) ϭ 3x 2 ϩ 1

x
0
1
2
3
4

p

f(x)
4
1
0
1
4

f(x)

(0, 1)

f (x ) ϭ x 2 Ϫ 4x ϩ 4 x O

O

x

(2, 0)

21a. 9; x ϭ 4.5; 4.5

22a. Ϫ5; x ϭ 2; 2

21b.

22b.

21c.

x
3
4
4.5
5
6

p

f(x)
Ϫ9
Ϫ11
Ϫ11.25
Ϫ11
Ϫ9

p

f(x)
Ϫ5
Ϫ8
Ϫ9
Ϫ8
Ϫ5

22c.

f(x)

f(x) x O

2
O

x
0
1
2
3
4

4

8

12

x

Ϫ4
Ϫ8
Ϫ12

f (x ) ϭ x 2 Ϫ 9x ϩ 9 f (x ) ϭ x 2 Ϫ 4x Ϫ 5

(4 1 , Ϫ111 )
2
4

(2, Ϫ9)

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23a. 36; x ϭ Ϫ6; Ϫ6

24a. Ϫ1; x ϭ Ϫ1; Ϫ1

23b.

24b.

x f(x)
Ϫ8 4
Ϫ7 1
Ϫ6 0
Ϫ5 1
Ϫ4 4

p

23c.

x
Ϫ3
Ϫ2
Ϫ1
0
1

p

f(x)
8
Ϫ1
Ϫ4
Ϫ1
8

24c.

f(x)
6

f(x) f (x ) ϭ 3x 2 ϩ 6x Ϫ 1

4
2

f (x ) ϭ x ϩ 12x ϩ 36

Ϫ16 Ϫ12

Ϫ8 Ϫ4
(Ϫ6, 0)

x

O

2
O x

(Ϫ1, Ϫ4)

2
3

2
3

25a. Ϫ3; x ϭ 2, 2

26a. 0; x ϭ Ϫ , Ϫ

25b.

26b.

x
0
1
2
3
4

25c.

p

f(x)
Ϫ3
3
5
3
Ϫ3

f(x)

x
Ϫ2
Ϫ1

f(x)
Ϫ4
1

2
3

4
3

0
1

0
Ϫ7

Ϫ

26c.

(2, 5) f (x ) ϭ Ϫ2x 2 ϩ 8x Ϫ 3

O

5
4

5
4

28a. Ϫ1; x ϭ 0; 0

27a. 0; x ϭ Ϫ ; Ϫ

x

f (x ) ϭ Ϫ3x 2 Ϫ 4x

x

O

f(x)

(Ϫ 2 , 4 )
3 3

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27b.

x
Ϫ3
Ϫ2

28b.

f(x)
3
Ϫ2

5
4

x
Ϫ2

25
8

Ϫ

Ϫ

Ϫ1
0

0

Ϫ3
0

27c.

p

Ϫ1

p

1
2

f(x)
1
1
2

Ϫ

Ϫ1
1
2

Ϫ

1

28c.

f(x)

f(x) f (x ) ϭ 0.5x 2 Ϫ 1

f (x ) ϭ 2x 2 ϩ 5x
O

x
O

(Ϫ 5 , Ϫ 25)
4
8
29a. 0; x ϭ Ϫ6; Ϫ6

30a.

29b.

30b.

29c.

x
Ϫ8
Ϫ7
Ϫ6
Ϫ5
Ϫ4

f(x)
8
8.75
9
8.75
8

p

x
(0, Ϫ1)

(Ϫ6, 9)

8

9
;
2

x ϭ Ϫ3, Ϫ3

x
Ϫ5
Ϫ4
Ϫ3
Ϫ2
Ϫ1

p

f(x)
2
0.5
0
0.5
2

30c.

f(x)

f(x)

4
Ϫ8

Ϫ4

O

x

(Ϫ3, 0)

Ϫ4 f (x ) ϭ 1 x 2 ϩ 3x ϩ 9

f (x ) ϭ Ϫ0.25x 2 Ϫ 3x

2

140

O

x

2

Algebra 2

Chapter 6

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8
9

1 1
3 3

31a. Ϫ ; x ϭ ;
31b.

x

32. min.; 0

f(x)
7
9
8
Ϫ
9

Ϫ1
0
1
3

Ϫ1

1

Ϫ

5
9
7
1
9

2
31c.

f(x)

2

2

8

f (x ) ϭ x Ϫ 3 x Ϫ 9

O

(

x

)

1
, Ϫ1
3

33. max.; Ϫ9

34. min.; Ϫ14

35. min.; Ϫ11

36. max.; 5

37. max.; 12

38. min.;

7
8

9
2

39. max.; Ϫ

40. max.; 5

41. min.; Ϫ11

42. max.; 5

43. min.; Ϫ10

1
3

44. x ϭ 40; (40, 40)

45. 40 m

46. 300 ft, 2.5 s

47. The y-intercept is the initial height of the object.

48. 120 Ϫ 2x

49. 60 ft by 30 ft

50. 1800 ft2

51. \$11.50

52. \$2645

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53. 5 in. by 4 in.

54. c; The x-coordinate of the
0
vertex of y ϭ ax 2 ϩ c is Ϫ
2a
or 0, so the y-coordinate of the vertex, the minimum of the function, is a(0)2 ϩ c or c ; Ϫ12.5

55. If a quadratic function can be used to model ticket price versus profit, then by finding the x-coordinate of the vertex of the parabola you can determine the price per ticket that should be charged to achieve maximum profit.
Answers should include the following. • If the price of a ticket is too low, then you won’t make enough money to cover your costs, but if the ticket price is too high fewer people will buy them.
• You can locate the vertex of the parabola on the graph of the function. It occurs when x ϭ 40.
Algebraically, this is found

56. C

by calculating x ϭ Ϫ

b
2a

which, for this case, is
Ϫ4000
xϭ or 40. Thus the
2(Ϫ50)

ticket price should be set at \$40 each to achieve maximum profit.
57. C

58. Ϫ2.08

59. 3.20

60. 0.88

61. 3.38

62. 0.43

63. 1.56

64. Ϫ1

65. Ϫ1 ϩ 3i

66. 9 Ϫ 5i

142

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67. 23

68. Ϫ13

69. 4

70. [10 Ϫ4 5]
Ϫ28
20 Ϫ44 d 72. c
8 Ϫ16
36

71. [5 Ϫ13 8]
6

73. C 14

0

Ϫ24

2
3

Ϫ8

Ϫ

S

74.

y y ϭ Ϫ3x yϪxϭ4 (Ϫ1, 3)

O

x

(Ϫ1, 3); consistent and independent 75. 5

76. 8

77. Ϫ2

78. Ϫ1

Lesson 6-2

Solving Quadratic Equations by Graphing
Pages 297–299
2. Sample answer: f (x) ϭ 3x 2 ϩ
2x Ϫ 1; 3x 2 ϩ 2x Ϫ 1 ϭ 0

1a. The solution is the value that satisfies an equation.
1b. A root is a solution of an equation. 1c. A zero is the x value of a function that makes the function equal to 0.
1d. An x-intercept is the point at which a graph crosses the x-axis. The solutions, or roots, of a quadratic equation are the zeros of the related quadratic function. You can find the zeros of a quadratic function by finding the x-intercepts of its graph.

143

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3. The x-intercepts of the related function are the solutions to the equation. You can estimate the solutions by stating the consecutive integers between which the x-intercepts are located.

4. Ϫ4, 1

5. Ϫ2, 1

6. Ϫ4

7. Ϫ7, 0

8. Ϫ4, 6

9. Ϫ7, 4

10. Ϫ5

11. between Ϫ2 and Ϫ1, 3

12. between Ϫ1 and 0; between
1 and 2

13. Ϫ2, 7

14. 0, 6

15. 3

16. Ϫ2, 1

17. 0

18. Ϫ , 3

19. no real solutions

20. 0, 3

21. 0, 4

22. between Ϫ5 and Ϫ4; between 0 and 1

23. between Ϫ1 and 0; between
2 and 3

24. Ϫ4, 5

25. 3, 6

26. Ϫ7

27. 6

28. Ϫ1 , 3

1
2

1
2

1
2

1
2

29. Ϫ , 2

1
2

1
2

30. Ϫ4, 1

˛

1
2

31. Ϫ2 , 3

32. between Ϫ4 and Ϫ3; between 0 and 1

33. between 0 and 1; between 3 and 4

34. between Ϫ1 and 0, between
2 and 3

35. between Ϫ3 and Ϫ2; between 2 and 3

36. no real solutions

37. no real solutions

38. Ϫ8, Ϫ9

˛

144

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40. Let x be the first number.
Then, Ϫ9 Ϫ x is the other number. x(Ϫ9 Ϫ x) ϭ 24
2
Ϫx Ϫ 9x Ϫ 24 ϭ 0

39. Let x be the first number.
Then, 7 Ϫ x is the other number. x(7 Ϫ x) ϭ 14
2
Ϫx ϩ 7x Ϫ 14 ϭ 0

y

y
2

y ϭ Ϫx ϩ 7x Ϫ 14
O

O

x

x

2

y ϭ Ϫx Ϫ 9x Ϫ 24

Since the graph of the related function does not intersect the x-axis, this equation has no real solutions. Therefore, no such numbers exist.

Since the graph of the related function does not intersect the x-axis, this equation has no real solutions. Therefore, no such numbers exist.
41. Ϫ2, 14

42. 4 s

43. 3 s

44. about 12 s

45. about 35 mph

46. about 8 s

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48. Answers should include the following. • h (t )
2

47. Ϫ4 and Ϫ2; The value of the function changes from negative to positive, therefore the value of the function is zero between these two numbers. h (t ) ϭ Ϫ16t ϩ 185

180

160
140
120
100
80
60
40
20
0

1

2 3 4 5 t

• Locate the positive x-intercept at about 3.4.
This represents the time when the height of the ride is 0. Thus, if the ride were allowed to fall to the ground, it would take about
3.4 seconds.
49. A

50. B

51. Ϫ1

52. Ϯ3

53. 3, 5

54. Ϫ9, 1

55. Ϯ1.33

56. no real solutions

57. 4, x ϭ 3; 3

58. Ϫ1; x ϭ 1; 1 f (x)

f(x)

(1, 3)

f (x) ϭ Ϫ4x 2 ϩ 8x Ϫ 1

x

O

O

x

f (x) ϭ x 2 Ϫ 6x ϩ 4
(3, Ϫ5)

146

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60.

1
5

62.

59. 4; x ϭ Ϫ6; Ϫ6

1
13

f(x)

3
5

ϩ i

8 f (x) ϭ 1 x 2 ϩ 3x ϩ 4
4

4
O
Ϫ12 Ϫ8

x

Ϫ4
Ϫ4

(Ϫ6, Ϫ5)

61.

10
13

ϩ

2 i 13

ϩ

5 i 13

63. 24

64. Ϫ8

65. Ϫ60

66. \$500

67. x(x ϩ 5)

68. (x Ϫ 10)(x ϩ 10)

69. (x Ϫ 7)(x Ϫ 4)

70. (x Ϫ 9)2

71. (3x ϩ 2)(x ϩ 2)

72. 2(3x ϩ 2)(x Ϫ 3)

Lesson 6-3

Solving Quadratic Equations by Factoring
Pages 303–305

1. Sample answer: If the product of two factors is zero, then at least one of the factors must be zero.

2. Sample answer: roots
6 and Ϫ5; x 2 Ϫ x Ϫ 30 ϭ 0

3. Kristin; the Zero Product
Property applies only when one side of the equation is 0.

4. {0, 11}

5. {Ϫ8, 2}

6. {Ϫ7, 7}

7. {3}

8. eϪ , 4 f
3
4

10. x 2 ϩ 3x Ϫ 28 ϭ 0

9. {Ϫ3, 4}
11. 6x 2 Ϫ 11x ϩ 4 ϭ 0

12. 15x 2 ϩ 14x ϩ 3 ϭ 0

13. D

14. {Ϫ8, 3}

15. {Ϫ4, 7}

16. {Ϫ5, 5}

17. {Ϫ9, 9}

18. {Ϫ6, 3}

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20. e 0, f
5
3

19. {Ϫ3, 7}
21. e 0, Ϫ f
3
4

22. {6}
24.

23. {8}
25.

e

1
,
4

eϪ2,

1 f 4

26. eϪ , Ϫ f

4f

1
2

3
2

27. eϪ , Ϫ f

28. eϪ , Ϫ f

29. e ,

30. {2, 4}

31. {Ϫ3, 1}

32. 0, Ϫ6, 5

33. 0, Ϫ3, 3

34. x 2 Ϫ 9x ϩ 20 ϭ 0

35. x 2 Ϫ 5x Ϫ 14 ϭ 0

36. x 2 ϩ x Ϫ 20 ϭ 0

37. x 2 ϩ 14x ϩ 48 ϭ 0

38. 2x 2 Ϫ 7x ϩ 3 ϭ 0

39. 3x 2 Ϫ 16x ϩ 5 ϭ 0

40. 12x 2 Ϫ x Ϫ 6 ϭ 0

41. 10x 2 ϩ 23x ϩ 12 ϭ 0

42.

43. Ϫ14, Ϫ16

44. 12 cm by 16 cm

45. B ϭ D 2 Ϫ 8D ϩ 16

46. 4; The logs must have a diameter greater than 4 in. for the rule to produce positive board feet values.

47. y ϭ (x Ϫ p)(x Ϫ q) y ϭ x 2 Ϫ px Ϫ qx Ϫ pq y ϭ x 2 Ϫ (p ϩ q)x Ϫ pq a ϭ 1, b ϭ Ϫ(p ϩ q), c ϭ Ϫpq axis of symmetry:

48. Ϫ1

2
3

8
3

3
2

3 9 f 4 4

1
4

2
3

s

b
2a
Ϫ(p ϩ q)
Ϫ
2(1) pϩq 2

xϭϪ xϭ xϭ

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The axis of symmetry is the average of the x-intercepts.
Therefore the axis of symmetry is located halfway between the x-intercepts.
49. Ϫ6

50. Answers should include the following. • Subtract 24 from each side of x 2 ϩ 5x ϭ 24 so that the equation becomes x 2 ϩ 5x Ϫ 24 ϭ 0. Factor the left side as (x Ϫ 3)
(x ϩ 8). Set each factor equal to zero. Solving each equation for x. The solutions to the equation are 3 and Ϫ8. Since length cannot be negative, the width of the rectangle is
3 inches, and the length is
3 ϩ 5 or 8 inches.
• To use the Zero Product
Property, one side of the equation must equal zero.

51. D

52. B

53. Ϫ5, 1

54. Ϫ

55. between Ϫ1 and 0; between
3 and 4

56. min.; Ϫ19

57. 322 Ϫ 223

58. 523

1
2

59. 33 ϩ 20 22

60. (Ϫ4, Ϫ4)
66. 5i 22

61. (3, Ϫ5)

62. a , 2b

63. 222

64. 225

1
3

67. 2i 23
65. 323

˛

68. 4i23

˛

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Chapter 6

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Chapter 6
Practice Quiz 1
Page 305
2. max.;

1. 4; x ϭ 2; 2 f(x) 4
O

37
4

or 9

1
4

f (x) ϭ 3x 2 Ϫ 12x ϩ 4
4

8

12

x

Ϫ4
Ϫ8

(2, Ϫ8)

4. e Ϫ5, f
1
2

1
2

3. 1 , 4
5. 3x 2 ϩ 11x Ϫ 4 ϭ 0

Lesson 6-4 Completing the Square
Pages 310–312
1. Completing the square allows you to rewrite one side of a quadratic equation in the form of a perfect square. Once in this form, the equation is solved by using the Square Root
Property.

2. Never; the value of c that makes ax 2 ϩ bx ϩ c a perfect square trinomial is

3. Tia; before completing the square, you must first check to see that the coefficient of the quadratic term is 1. If it is not, you must first divide the equation by that coefficient.

4. {Ϫ10, Ϫ4}

5. e

b

the square of and the
2
square of a number can never be negative.

4 Ϯ 22 f 3

9. 54 Ϯ 256
7.

9
, ax
4

10. 5Ϫ1 Ϯ i 256
6. 36; (x Ϫ 6)2

Ϫ b

3 2
2

8. {Ϫ6, 3}

150

Algebra 2

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11. e

3 Ϯ 233 f 4

16. 5Ϫ4 Ϯ 27, Ϫ4 Ϫ 276

12. Jupiter

13. Earth: 4.5 s, Jupiter: 2.9 s
15. {Ϫ2, 12}

Ϫ5 Ϯ 211 f 3

17. 53 Ϯ 2226
19. e

7 Ϯ 25 f 2

14. {3, Ϫ7}
18. e

20. e Ϫ ,
5
4

1 f 4

21. {Ϫ1.6, 0.2}

22. 25 ft

23. about 8.56 s

24. 64; (x ϩ 8)2

25. 81; (x Ϫ 9)2

26.

27.

49
; ax
4

Ϫ b

7 2
2

15 2 b 2

34. 5Ϫ1 Ϯ 276
30.

35. 52 Ϯ 236
25
, ax
16

Ϫ

28. 0.09; (x ϩ 0.3)2

29. 1.44; (x Ϫ 1.2)2
31.

225
; ax
4

ϩ b

5 2
4

16
; ax
9

Ϫ b

4 2
3

32. {3, 5}

33. {Ϫ12, 10}

36. 52 Ϯ i 6

5 Ϯ 213 f 6

37. {Ϫ3 Ϯ 2i}

38. e Ϫ , 1 f

39. e , 1 f

40. e

41. e

42. e

5
2

2 Ϯ 210 f 3

1
2

43. e

Ϫ5 Ϯ i 123 f 6

46. e Ϯ 23 f

47. e Ϯ 22 f

44. {Ϫ2, 0.6}
1
3

45. {0.7, 4}
3
4

49.

1 ϩ 25
2

48. 5

x
1
,
1 xϪ1

7 Ϯ i 247 f 4

50.

151

1
2

in. by 5

1
2

in.

Algebra 2

Chapter 6

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51. Sample answers: The golden rectangle is found in much of ancient Greek architecture, such as the Parthenon, as well as in modern architecture, such as in the windows of the United
Nations building. Many songs have their climax at a point occurring 61.8% of the way through the piece, with 0.618 being about the reciprocal of the golden ratio. The reciprocal of the golden ratio is also used in the design of some violins.

52a. n ϭ 0
52b. n Ͼ 0
52c. n Ͻ 0

53. 18 ft by 32 ft or 64 ft by 9 ft

54. To find the distance traveled by the accelerating race car in the given situation, you must solve the equation t 2 ϩ 22t ϩ 121 ϭ 246 or t 2 ϩ 22t Ϫ 125 ϭ 0.
Answers should include the following. • Since the expression t 2 ϩ 22t Ϫ 125 is prime, the solutions of t 2 ϩ 22t ϩ
121 ϭ 246 cannot be obtained by factoring.
• Rewrite t 2 ϩ 22t ϩ 121 as
(t ϩ 11)2. Solve (t ϩ 11)2 ϭ
246 by applying the
Square Root Property.
Then, subtract 11 from each side. Using a calculator, the two solutions are about 4.7 and
Ϫ26.7. Since time cannot be negative, the driver takes about 4.7 seconds to reach the finish line.

55. D

56. D

152

Algebra 2

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57. x 2 Ϫ 3x ϩ 2 ϭ 0

58. x 2 Ϫ 6x Ϫ 27 ϭ 0

59. 3x 2 Ϫ 19x ϩ 6 ϭ 0

60. 12x 2 ϩ 13x ϩ 3 ϭ 0

61. between Ϫ4 and Ϫ3; between 0 and 1

62. 6, 8

1
2

3

63. Ϫ4, Ϫ1

64. 57

65. (2, Ϫ5)

66. a , b
43 6
21 7

67. 0 x Ϫ (Ϫ257)0 ϭ 2

68. greatest: Ϫ255ЊC; least:
Ϫ259ЊC

69. 37

70. Ϫ16

71. 121

72. 0

Lesson 6-5

The Quadratic Formula and the Discriminant
Pages 317–319

2. The square root of a negative number is a complex number.

y

x

O

1b. Sample answer: y O

x

153

Algebra 2

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1c. Sample answer: y O

x

3. b 2 Ϫ 4ac must equal 0.

4a. 484
4b. 2 rational
4c.

2 Ϯ 22
2

1
,
4

5
2

Ϫ

5a. 8
5b. 2 irrational

6a. 0
6b. one rational

5c.

6c. Ϫ

1
2

Ϫ3 Ϯ i 23
2

10. 1 Ϯ 23
8. 0, Ϫ8

7a. Ϫ3
7b. two complex
7c.

Ϫ5 Ϯ i 22
2

9. Ϫ3, Ϫ2
11.

12. at about 0.7 s and again at about 4.6 s
Ϫ3 Ϯ 221
2

13. No; the discriminant of
Ϫ16t 2 ϩ 85t ϭ 120 is Ϫ455, indicating that the equation has no real solutions.

14a. 21
14b. 2 irrational

15a. 240
15b. 2 irrational
15c. 8 Ϯ 2215

16a. Ϫ16
16b. 2 complex
16c. 1 Ϯ 2i

17a. Ϫ23
17b. 2 complex

18a. 121
18b. 2 rational

17c.

18c. Ϫ ,

14c.

1 Ϯ i 223
2

1 2
4 3

19a. 49
19b. 2 rational

20a. 20
20b. 2 irrational
154

Algebra 2

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20c. Ϫ2 Ϯ 25

21c. Ϫ1 Ϯ 26
19c. Ϫ2,

1
3

22a. 0
22b. one rational

21a. 24
21b. 2 irrational

22c.

23a. 0
23b. one rational

1
3

9 Ϯ i 231
8
28
9

24a. Ϫ31
24b. 2 complex

5
2

24c.

23c. Ϫ

Ϫ1 Ϯ i 215
4

2 Ϯ 4 27
9

25a. Ϫ135

26a.

25b. 2 complex

26b. 2 irrational

25c.

26c.

Ϫ1 Ϯ 2 20.37
0.8

27a. 1.48
27b. 2 irrational
27c.

33.
35.

30. 2 Ϯ i 23

221
7

Ϫ3 Ϯ 215
2

29. Ϯi
31.

28. Ϫ2, 32

34. Ϫ3 Ϯ i 27
36. 4 Ϯ 27
32. Ϯ22

5 Ϯ 246
3

9
2

3
10

37. 0, Ϫ

38. 3 Ϯ 222

39. Ϫ2, 6

40.

41. This means that the cables do not touch the floor of the bridge, since the graph does not intersect the x-axis and the roots are imaginary.

42. domain: 0 Յ t Յ 25, range:
73.7 Յ A(t ) Յ 1201.2

43. 1998

44. about 40.2 mph

155

Ϸ Ϫ0.00288

Algebra 2

Chapter 6

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46. The person’s age can be substituted for A in the appropriate formula, depending upon their gender, and their average blood pressure calculated. See student’s work.
• If a woman’s blood pressure is given to be 118, then solve the equation 118 ϭ
0.01A2 ϩ 0.05A ϩ 107 to find the value of A. Use the
Quadratic Formula, substituting 0.01 for a, 0.05 for b, and Ϫ11 for c. This gives solutions of about
Ϫ35.8 or 30.8. Since age cannot be negative, the only valid solution for A is 30.8.

45a. k ϭ Ϯ6
45b. k Ͻ Ϫ6 or k Ͼ 6
45c. Ϫ6 Ͻ k Ͻ 6

49. Ϫ14, Ϫ4

50. 4 Ϯ 27

51.

52. Ϫ2, 0

1 Ϯ 2 22
2

47. D

48. C

2
,
3

53. Ϫ2, 7

54.

55. a 4b10

56. 10p6 0 q 0

57. 4b 2c 2

5

58. 7.98 ϫ 106

59.

60.

y

y xϭ1 xϩyϭ9 8 yϪxϭ4 6
4
2
Ϫ6 Ϫ4

O
Ϫ4
Ϫ6

yϭx

2 4 6 8 xϪy ϭ 3

x

x

O

y ϭ Ϫ1

61. no

62. yes; (x Ϫ 7)2

63. yes; (2x ϩ 3)2

64. yes; (5x ϩ 2)2

65. no

66. yes; (6x Ϫ 5)2

156

Algebra 2

Chapter 6

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Lesson 6-6

Analyzing Graphs of Quadratic Functions
Pages 325–328

1a.
1b.
1c.
1d.
1e.

y ϭ 2(x ϩ 1)2 ϩ 5 y ϭ 2(x ϩ 1)2 y ϭ 2(x ϩ 3)2 ϩ 3 y ϭ 2(x Ϫ 2)2 ϩ 3
Sample answer: y ϭ 4(x ϩ 1)2 ϩ 3
1f. Sample answer: y ϭ (x ϩ 1)2 ϩ 3
1g. y ϭ Ϫ2(x ϩ 1)2 ϩ 3

2. Substitute the x-coordinate of the vertex for h and the y -coordinate of the vertex for k in the equation y ϭ a(x Ϫ h)2 ϩ
k. Then substitute the x-coordinate of the other point for x and the y-coordinate for y into this equation and solve for a. Replace a with this value in the equation you wrote with h and k.

3. Sample answer: y ϭ 2(x Ϫ 2)2 Ϫ1

4. Jenny; when completing the square is used to write a quadratic function in vertex form, the quantity added is then subtracted from the same side of the equation to maintain equality.

5. (Ϫ3, Ϫ1); x ϭ Ϫ3; up

6. y ϭ (x ϩ 4)2 Ϫ 19,
(Ϫ4, Ϫ19); x ϭ Ϫ4; up

7. y ϭ Ϫ3(x Ϫ 3)2 ϩ 38;
(Ϫ3, 38); x ϭ Ϫ3; down

8.

y

y ϭ 3(x ϩ 3)2
O x

9.

10.

y

y

O

y ϭ Ϫ 2x 2 ϩ 16x Ϫ 31 x 1 y ϭ 3 (x Ϫ 1)2 ϩ 3

O

x

12. y ϭ Ϫ(x ϩ 3)2 ϩ 6

11. y ϭ 4(x Ϫ 2)2

157

Algebra 2

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1
2

13. y ϭ Ϫ (x ϩ 2)2 Ϫ 3

14. h(d ) ϭ Ϫ2d 2 ϩ 4d ϩ 6

15. (Ϫ3, 0); x ϭ Ϫ3 down

16. (1, 2); x ϭ 1; up

17. (0,Ϫ6); x ϭ 0 up

18. (0, 3); x ϭ 0; down

19. y ϭ Ϫ(x ϩ 2)2 ϩ 12;
(Ϫ2, 12); x ϭ Ϫ2; down

20. y ϭ (x Ϫ 3)2 Ϫ 8;
(3, Ϫ8); x ϭ 3; up

21. y ϭ Ϫ3(x Ϫ 2)2 ϩ 12;
(2, 12); x ϭ 2; down

22. y ϭ 4(x ϩ 3)2 Ϫ 36;
(Ϫ3, Ϫ36); x ϭ Ϫ3; up

23. y ϭ 4(x ϩ 1)2 Ϫ 7; (Ϫ1, Ϫ7); x ϭ Ϫ1; up

24. y ϭ Ϫ2(x Ϫ 5)2 ϩ 15;
(5, 15); x ϭ 5; down

25. y ϭ 3 ax ϩ b Ϫ ;

26. y ϭ 4 ax Ϫ b Ϫ 20;

1 2
2

1 aϪ ,
2

3 2
2

7
4

Ϫ b; x ϭ Ϫ ; up
7
4

3 a ,
2

1
2

27.

Ϫ20b; x ϭ , up
3
2

28.

y

y
O

y ϭ 4(x ϩ 3)2 ϩ 1

29.

y ϭ Ϫ(x Ϫ 5)2 Ϫ 3

x

x

O

30.

y

y

O

x

1 y ϭ 4 (x Ϫ 2)2 ϩ 4

31.

1 y ϭ 2 (x Ϫ 3)2 Ϫ 5

x

O

32.

y

y

O x

y ϭ x 2 Ϫ 8x ϩ 18
O

x

y ϭ x 2 ϩ 6x ϩ 2

158

Algebra 2

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33.

y

34.

y ϭ Ϫ4x 2 ϩ 16x Ϫ 11

y ϭ Ϫ5x 2 Ϫ 40x Ϫ 80

y

O

O

35.

y

x

x

36.

y ϭ Ϫ 1 x 2 ϩ 5x Ϫ 27
2
2

y

x

O

O

x

y ϭ 1 x 2 Ϫ 4x ϩ 15
3

37. Sample answer: the graph of y ϭ 0.4(x ϩ 3)2 ϩ 1 is narrower than the graph of y ϭ 0.2(x ϩ 3)2 ϩ 1.

38. Sample answer: the graphs have the same shape, but the graph of y ϭ 2(x Ϫ 4)2 ϩ 1 is
1 unit to the left and 5 units below the graph of y ϭ
2(x Ϫ 5)2 Ϫ 4

39. y ϭ 9(x Ϫ 6)2 ϩ 1

40. y ϭ 3(x ϩ 4)2 ϩ 3

2
3

41. y ϭ Ϫ (x Ϫ 3)2

42. y ϭ Ϫ3(x Ϫ 5)2 ϩ 4
5
2

1
3

43. y ϭ x 2 ϩ 5

44. y ϭ (x ϩ 3)2 Ϫ 2

45. y ϭ Ϫ2x 2

46. y ϭ (x ϩ 3)2 Ϫ 4

47. 34,000 feet; 32.5 s after the aircraft begins its parabolic flight 48. about 1.6 s

49. d (t ) ϭ Ϫ16t 2 ϩ 8t ϩ 50

50. about 2.0 s

51. Angle A; the graph of the equation for angle A is higher than the other two since 3.27 is greater than
2.39 or 1.53.

52. Angle B; the vertex of the equation for angle B is farther to the right than the other two since 3.57 is greater than 3.09 or 3.22.

4
3

159

Algebra 2

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53.

54. All quadratic equations are a transformation of the parent graph y ϭ x 2. By identifying these transformations when a quadratic function is written in vertex form, you can redraw the graph of y ϭ x 2.
Answers should include the following. • In the equation y ϭ a(x Ϫ h)2 ϩ k, h translated the graph of y ϭ x 2 h units to the right when h is positive and h units to the left when h is negative. The graph of y ϭ x 2 is translated k units up when k is positive and k units down when k is negative. When a is positive, the graph opens upward and when a is negative, the graph opens downward. If the absolute value of a is less than 1, the graph will be narrower than the graph of y ϭ x 2, and if the absolute value of a is greater than 1, the graph will be wider than the graph of y ϭ x 2.
• Sample answer: y ϭ 2(x ϩ 2)2 Ϫ 3 is the graph of y ϭ x 2 translated
2 units left and 3 units down. The graph opens upward, but is narrower than the graph of y ϭ x 2.

y ϭ ax 2 ϩ bx ϩ c y ϭ a ax 2 ϩ xb ϩ c b a

y ϭ a cx 2 ϩ x ϩ a b d ϩ b a

b 2
2a

c Ϫ aa b

b 2
2a

y ϭ a ax ϩ

b 2 b 2a

ϩcϪ

b2
4a

The axis of symmetry is x ϭ h or Ϫ

b
.
2a

55. D

56. B

160

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60. 5Ϫ5 Ϯ 2226

59. Ϫ23; 2 complex
61. 53 Ϯ 3i 6
63. 2t 2 ϩ 2t Ϫ

Ϫ2 Ϯ 213 f 2

58. 225; 2 rational

57. 12; 2 irrational

62. e
3
tϪ1

64. t 2 Ϫ 2t ϩ 1

65. n 3 Ϫ 3n 2 Ϫ 15n Ϫ 21

66. y 3 ϩ 1 Ϫ

67a. Sample answer using (1994,
76,302) and (1997, 99,448): y ϭ 7715x Ϫ 15,307,408
67b. 161,167

68. yes

69. no

4 yϩ3 70. yes

71. no

Chapter 6
Practice Quiz 2
Page 328
1. 5Ϫ7 Ϯ 2236

2. e

3. Ϫ11; 2 complex

4. 100; 2 rational

5. e

6. e

Ϫ9 Ϯ 5 25 f 2

7. y ϭ 1x Ϫ 22 2 Ϫ 5
2
3

8. y ϭ (x ϩ 4)2 ϩ 2; (Ϫ4, 2), x ϭ Ϫ4; up

9. y ϭ Ϫ1x Ϫ 62 2; 16, 02, x ϭ 6; down ©Glencoe/McGraw-Hill

2 Ϯ 2i 22 f 3
1 Ϯ 3i f 2

10. y ϭ 2(x ϩ 3)2 Ϫ 5; (Ϫ3, Ϫ5), x ϭ Ϫ3; up

161

Algebra 2

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Lesson 6-7

Graphing and Solving Quadratic Inequalities
Pages 332–335

1. y Ն 1x Ϫ 32 2 Ϫ 1

2. Sample answer: one number less than Ϫ3, one number between Ϫ3 and 5, and one number greater than 5
4.

3a. x ϭ Ϫ1, 5
3b. x Յ Ϫ1 or x Ն 5
3c. Ϫ1 Յ x Յ 5

y

O

5.

12
8
4
Ϫ4

Ϫ2

Ϫ4
Ϫ8
Ϫ12

y
12

y ϭ Ϫ2x 2 Ϫ 4x ϩ 3

x

y

O

Ϫ20

7.

6.

y

y ϭ x 2 Ϫ 10x ϩ 25

2

4x
O

x

y ϭ x 2 Ϫ 16

8. x Ͻ 1 or x Ͼ 5

y ϭ Ϫx 2 ϩ 5x ϩ 6

8
4
Ϫ2

O

2

4

12. 5x 0 Ϫ 23 Յ x Յ 236

6x

9. 5x 0 Ϫ1 Ͻ x Ͻ 76

10. 5x 0 x Ͻ Ϫ3 or x Ͼ 46

11. л

162

Algebra 2

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13. about 6.1 s

14.

15

y

5
Ϫ8

Ϫ4

Ϫ5

O

8x

4

Ϫ15

y ϭ x 2 ϩ 3x Ϫ 18

Ϫ25

15.

y

16.

y ϭ Ϫx 2 ϩ 7x ϩ 8

y

12
8
4
Ϫ4

O

17.

4

x

8

y ϭ x 2 ϩ 4x ϩ 4

18.

y

5
Ϫ8

O

x

O

y
8x

4

Ϫ4 O
Ϫ10

x

Ϫ20
Ϫ30

y ϭ x 2 ϩ 4x

y ϭ x 2 Ϫ 36

19.

20.

y

y ϭ Ϫx 2 Ϫ 3x ϩ 10

14

y

10
O

x

6
2
Ϫ6

y ϭ x 2 ϩ 6x ϩ 5

163

Ϫ4

Ϫ2 O

2x

Algebra 2

Chapter 6

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21.

y ϭ Ϫx 2 Ϫ 7x ϩ 10

22.

y

y y ϭ Ϫx 2 ϩ 10x Ϫ 23

20
12

x

O

4
Ϫ12 Ϫ8

23.

y

O
Ϫ4

Ϫ4

4x

24.

y ϭ Ϫx 2 ϩ 13x Ϫ 36

y
4

6
2
O

Ϫ2
2

6

10

x

2

O

4

x

Ϫ4

Ϫ4
Ϫ8

y ϭ 2x 2 ϩ 3x Ϫ 5

Ϫ8

25.

26. 5

y

x

O

y ϭ 2x 2 ϩ x Ϫ 3

27. Ϫ2 Յ x Յ 6

28. x Ͻ Ϫ3 or x Ͼ 3

29. x Ͻ Ϫ7 or x Ͼ Ϫ3

30. 5x 0 x Ͻ Ϫ3 or x Ͼ 66

33. 5x 0 x Յ Ϫ6 or x Ն 46

34. 5x 0 Ϫ4 Յ x Յ 36

31. 5x 0 Ϫ7 Ͻ x Ͻ 46

32. 5x 0 Ϫ1 Յ x Յ 56

35. 5x 0 x Յ Ϫ7 or x Ն 16

36. e x ` x ϭ

37. all reals

38. л

39. 5x 0 x ϭ 76

40. all reals

42. 5x 0 Ϫ4 Ͻ x Ͻ 1 or x Ͼ 36

41. л

43. 0 to 10 ft or 24 to 34 ft

1 f 3

44a. 0.98, 4.81; The owner will break even if he charges
\$0.98 or \$4.81 per square foot. 164

Algebra 2

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44b. 0.98 Ͻ r Ͻ 4.81; The owner will make a profit if the rent is between \$0.98 and \$4.81.
44c. 1.34 Ͻ r Ͻ 4.45; If rent is set between \$1.34 and \$4.45 per sq ft, the profit will be greater than \$10,000
44d. r Ͻ 1.34 or r Ͼ 4.45; If rent is set between \$0 and \$1.34 or above \$4.45 per sq ft, the profit will be less than \$10,000.
45. The width should be greater than 12 cm and the length should be greater than 18 cm

46. P(n) ϭ n[15 ϩ 1.5(60 Ϫ n)] Ϫ
525 or Ϫ1.5n 2 ϩ 105n Ϫ 525

47. 6

48. \$1312.50; 35 passengers

49.

50. Answers should include the following. • Ϫ16t 2 ϩ 42t ϩ 3.75 Ͼ 10
• One method of solving this inequality is to graph the related quadratic function h(t ) ϭ Ϫ16t 2 ϩ 42t ϩ
3.75 Ϫ 10. The interval(s) at which the graph is above the x-axis represents the times when the trampolinist is above 10 feet. A second method of solving this inequality would be to find the roots of the related quadratic equation Ϫ16t 2 ϩ
42t ϩ 3.75 Ϫ 10 ϭ 0 and then test points in the three intervals determined by these roots to see if they satisfy the inequality. The interval(s) at which the inequality is satisfied represent the times when the trampolinist is above
10 feet.

y y ϭ Ϫx 2 ϩ 4

O

x y ϭ x2 Ϫ 4

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52. A

51. C

53. 5x 0 all reals, x

54. 5x 0 Ϫ7 Ͻ x Ͻ 76

26

55. 5x 0 x Ͻ Ϫ9 or x Ͼ 36

56. 5x 0 x Յ Ϫ3.5 or x Ն Ϫ2.56

57. 5x 0 Ϫ1.2 Յ x Յ Ϫ0.46

58. no real solutions

60. y ϭ Ϫ21x Ϫ 42 2; 14, 02, x ϭ 4; down 59. y ϭ (x Ϫ 1)2 ϩ 8; (1, 8), x ϭ 1; up
Ϫ5 Ϯ i 23
2
1
2

61. y ϭ (x ϩ 6)2; (Ϫ6, 0), x ϭ Ϫ6; up
63.

64.

66. Ϫ6x 3 Ϫ 4x 2y ϩ 13xy 2

65. 4a 2b 2 ϩ 2a 2b ϩ 4ab 2 ϩ
12a Ϫ7b
67. xy 3 ϩ y ϩ
69. B

1 x 68. Ϫ15a 2 ϩ 14a Ϫ 3

Ϫ21 48
R
Ϫ13 22

70. 3Ϫ54 64

71. 0x Ϫ 0.008 0 Յ 0.002;
0.078 Յ x Յ 0.082

Ϫ3 Ϯ 2 26
3

62. Ϫ4, Ϫ8

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Chapter 7 Polynomial Functions
Lesson 7-1 Polynomial Functions
Pages 350–352
1. 4 ϭ 4x 0; x ϭ x 1

2. Sample answer: Evendegree polynomial functions with positive leading coefficients have graphs in which f (x ) S ϩϱ as x S ϩϱ and as x SϪϱ. Odd-degree polynomial functions with positive leading coefficients have graphs in which f (x ) S ϩϱ as x S ϩϱ and f (x ) S Ϫϱ as x S Ϫϱ.

3. Sample answer given.

4. Sometimes; a polynomial function with 4 real roots may be a sixth-degree polynomial function with
2 imaginary roots. A polynomial function that has
4 real roots is at least a fourth-degree polynomial.

f (x)

O

x

5. 6; 5

6. 5; Ϫ3

7. Ϫ21; 3

8. 4; 12

9. 2a 9 ϩ 6a 3Ϫ12

10. 100a 2 ϩ 20

11. 6a 3 Ϫ 5a 2 ϩ 8a Ϫ 45

12a. f (x ) S Ϫϱ as x S ϩϱ, f (x ) S ϩϱ as x S Ϫϱ
12b. odd
12c. 3

13a. f(x ) S ϩϱ as x S ϩϱ, f(x ) S ϩϱ as x S Ϫϱ
13b. even
13c. 0

14a. f (x ) S ϩϱ as x S ϩϱ, f (x ) S Ϫϱ as x S Ϫϱ
14b. odd
14c. 1

15. 109 lumens

16. 1; Ϫ1

17. 3; 1

18. No, the polynomial contains two variables, a and b.

19. 4; 6

20. 3; Ϫ5

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21. No, this is not a polynomial

22. Ϫ2; 4

1

because the term cannot c be written in the form x n, where n is a nonnegative integer. 23. 12; 18

24. 125; Ϫ37

25. 1008; Ϫ36

26. Ϫ166; 50

27. 86; 56

28. 100; 4

29. 7; 4

30. 27a3 ϩ 3a ϩ 1

31. 12a 2 Ϫ 8a ϩ 20

32. 3a4 Ϫ 2a 2 ϩ 5

33. 12a6 Ϫ 4a3 ϩ 5

34. x 3 ϩ 3x 2 ϩ 4x ϩ 3

35. 3x 4 ϩ 16x 2 ϩ 26

36. 6x 2 ϩ 44x ϩ 90

37. Ϫx 6 ϩ x 3 ϩ 2x 2 ϩ 4x ϩ 2

38. 9x 4 Ϫ 12x 2 Ϫ 8x ϩ 50

39a. f (x ) S ϩϱ as x S ϩϱ, f (x ) S Ϫϱ as x S Ϫϱ
39b. odd
39c. 3

40a. f (x) S ϩϱ as x S ϩϱ, f (x) S ϩϱ as x S Ϫϱ
40b. even
40c. 4

41a. f (x) S ϩϱ as x S ϩϱ, f (x) S Ϫϱ as x S Ϫϱ
41b. even
41c. 0

42a. f (x ) S ϩϱ as x S ϩϱ, f (x ) S ϩϱ as x S Ϫϱ
42b. odd
42c. 5

43a. f (x) S ϩϱ as x S ϩϱ, f (x) S Ϫϱ as x S Ϫϱ
43b. odd
43c. 1

44a. f (x ) S Ϫϱ as x S ϩϱ, f (x ) S Ϫϱ as x S Ϫϱ
44b. even
44c. 2

45. 5.832 units

46. even

47. f (x ) S Ϫϱ as x S ϩϱ; f(x ) S Ϫϱ as x S Ϫϱ

48. Sample answer: Decrease; the graph appears to be turning at x ϭ 30 indicating a relative maximum at that point. So attendance will decrease after 2000.

49.

1
2

50. Ϫ1, 0, 4

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3
2

51. f(x) ϭ x 3 Ϫ x 2 Ϫ 2x

52.

8

f (x )

f (x) ϭ 1 x 3 Ϫ 3 x 2Ϫ 2x
4 2
2
Ϫ4 Ϫ2

O

2

x

Ϫ4
Ϫ8

53. 4

54. 16 regions

55. 8 points

56. Many relationships in nature can be modeled by polynomial functions; for example, the pattern in a honeycomb or the rings in a tree trunk. Answers should include the following.
• You can use the equation to find the number of hexagons in a honeycomb with 10 rings and the number of hexagons in a honeycomb with 9 rings.
The difference is the number of hexagons in the tenth ring.
• Other examples of patterns found in nature include pinecones, pineapples, and flower petals.

57. C

58. C

60. 5x 0 x Յ Ϫ9 or x Ն 76

59. 5x 0 2 Ͻ x Ͻ 66

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61. e x ` Ϫ1 Յ x Յ f
4
5

62.

y y ϭ Ϫ2(x Ϫ 2)2 ϩ 3

x

O

63.

64.

y

y

2
O

Ϫ12 Ϫ8

x

y ϭ 1 x2 ϩ x ϩ 3

Ϫ2

y ϭ 1 (x ϩ 5)2 Ϫ 1
3

x

O
2

2

Ϫ4

65. 54 Ϯ 3226

66. e Ϫ ,
7
6

67. 23,450(1 ϩ p);
23,450(1 ϩ p)3

68.

5 f 6 y y ϭ x2 ϩ 4

x

O

69.

70.

y

8

y

4
O

x

Ϫ8

Ϫ4 O

4

8x

Ϫ4

1

y ϭ 2 x 2 ϩ 2x Ϫ 6

y ϭ Ϫx 2 ϩ 6x Ϫ 5

170

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Graphing Polynomial Functions
Pages 356–358

1. There must be at least one real zero between two points on a graph when one of the points lies below the x-axis and the other point lies above the x-axis.

2. 4

3.

4.

f (x)

x

p

f(x )
8

Ϫ3 Ϫ20
O

Ϫ2

x

Ϫ1
0

f (x ) ϭ x 3 Ϫ x 2 Ϫ 4x ϩ 4

1
2
3

5. x p

0

f (x )

4

6
4

Ϫ4

Ϫ2 O

0

2

4x

Ϫ4

0

f(x) ϭ x 3 Ϫ x 2 Ϫ 4x ϩ 4

10

6. between Ϫ1 and 0 f(x )

f (x)

8

Ϫ3 20
Ϫ2 Ϫ9
Ϫ1 Ϫ2
0
5
1
0
2 Ϫ5
3 26

f (x)

4
Ϫ4

Ϫ2 O

2

4x

x

O

Ϫ4

f (x ) ϭ x 3 Ϫ x 2 ϩ 1

Ϫ8

f (x ) ϭ x 4 Ϫ 7x 2 ϩ x ϩ 5

8.

7. between Ϫ2 and Ϫ1, between
Ϫ1 and 0, between 0 and 1, and between 1 and 2

8

f (x)

4

f (x)
Ϫ4

Ϫ2

O

2

4x

Ϫ4
O

Ϫ8

x

f (x ) ϭ x 3 ϩ 2x 2 Ϫ 3x Ϫ 5

Sample answer: rel. max. at x ϭ Ϫ2, rel. min.: at x ϭ 0.5

f (x ) ϭ x 4 Ϫ 4x 2 ϩ 2

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10.

f (x)

C (t )
12000

8
4
O

Ϫ2

2

Cable TV Systems

Ϫ4

10000

4x

Ϫ4

f (x ) ϭ x 4 Ϫ 8x 2 ϩ 10

Sample answer: rel. max. at x ϭ 0, rel. min. at x ϭ Ϫ2 and at x ϭ 2

8000
6000

C(t) ϭ Ϫ43.2t 2 ϩ 1343t ϩ 790

4000
2000
O

x

p

14a. f (x )

Ϫ5 25
Ϫ4
0
Ϫ3 Ϫ9
Ϫ2 Ϫ8
Ϫ1 Ϫ3
0
0
1 Ϫ5
2 Ϫ24

x f (x )

2

Ϫ2

4x

Ϫ4
Ϫ8

f (x) ϭ Ϫx 3 Ϫ 4x 2

8
4
Ϫ4

Ϫ2 O

2

4x

Ϫ4

f (x) ϭ x 3 Ϫ 2x 2ϩ 6

14b. between Ϫ2 and Ϫ1
14c. Sample answer: rel. max. at x ϭ 0, rel. max. at x ϭ 1

13b. at x ϭ Ϫ4 and x ϭ 0
13c. Sample answer: rel. max. at x ϭ 0, rel. min. at x ϭ Ϫ3

p

f (x )

f (x )

Ϫ2 Ϫ10
Ϫ1
3
0
6
1
5
2
6
3 15
4 38

4
O

t

8
12
16
Years Since 1985

12. The number of cable TV systems rose steadily from
1985 to 2000. Then the number began to decline.

11. rel. max. between x ϭ 15 and x ϭ 16, and no rel. min.; f(x ) S Ϫϱ as x S Ϫϱ, f(x ) S Ϫϱ as x S ϩϱ.
13a.

4

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f (x )

f (x )

Ϫ2 Ϫ18
Ϫ1 Ϫ2
0
2
1
0
2 Ϫ2
3
2
4 18

x

x

O

Ϫ2
3
Ϫ1 Ϫ5
0 Ϫ9
1 Ϫ3
2 19

f (x) ϭ x 3 Ϫ 3x 2ϩ 2

x

p

Ϫ1 75
0 16
1 Ϫ3
2
0
3
7
4
0
5 Ϫ39

x

f (x )

Ϫ2
Ϫ1
0
1
2
3
4
5

4
Ϫ4

Ϫ2

O

2

4x

Ϫ4
Ϫ8

f (x) ϭ Ϫ3x 3 ϩ 20x 2 Ϫ 36x ϩ 16

17b. between 0 and 1, at x ϭ 2, and at x ϭ 4
17c. Sample answer: rel. max. at x ϭ 3, rel. min. at x ϭ 1

4
Ϫ8

Ϫ4

O

4

8x

Ϫ4
Ϫ8

f (x ) ϭ x 3 ϩ 5x 2 Ϫ 9

16b. between Ϫ5 and Ϫ4, between Ϫ2 and Ϫ1, and between 1 and 2
16c. Sample answer: rel. max. at x ϭ Ϫ3, rel. min. at x ϭ 0
18a.

f (x )

f (x )

f (x )

Ϫ5 Ϫ9
Ϫ4
7
Ϫ3
9

15b. at x ϭ 1, between Ϫ1 and 0, and between 2 and 3
15c. Sample answer: rel. max. at x ϭ 0, rel. min. at x ϭ 2

17a.

p

16a.

p

f (x )
Ϫ29
Ϫ8
Ϫ1
Ϫ2
Ϫ5
Ϫ4
7
34

f (x ) x O

f (x ) ϭ x 3 Ϫ 4x 2ϩ 2x Ϫ 1

18b. between 3 and 4
18c. Sample answer: rel. max. at x ϭ 0.5, rel. min. at x ϭ 2.5

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19a.

p

x

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f (x )

Ϫ3 73
Ϫ2
8
Ϫ1 Ϫ7
0 Ϫ8
1 Ϫ7
2
8
3 73

x

f (x )

Ϫ4

O

Ϫ2

2

x

x

Ϫ4

f (x) ϭ x 4 Ϫ 8 Ϫ8

p

Ϫ4 Ϫ169
Ϫ3 Ϫ31
Ϫ2
7
Ϫ1
5
0 Ϫ1
1
1
2 Ϫ1
3 Ϫ43

8

x

f (x )

Ϫ2

O

2

4x

Ϫ4
Ϫ8

f (x ) ϭ Ϫx 4 ϩ 5x 2Ϫ 2x Ϫ 1

21b. between Ϫ3 and Ϫ2, between
Ϫ1 and 0, between 0 and 1, and between 1 and 2
21c. Sample answer: rel. max. at x ϭ Ϫ2 and at x ϭ 1.5, rel. min. at x ϭ 0

f (x )

8
Ϫ4

Ϫ2

O

x

2

Ϫ8
Ϫ16

f (x ) ϭ x 4 Ϫ 10x 2ϩ 9

p

f (x )

Ϫ3 Ϫ39
Ϫ2
5
Ϫ1
3
0 Ϫ3
1
5
2 21
3 15
4 Ϫ67

4
Ϫ4

16

20b. at x ϭ Ϫ3, x ϭ Ϫ1, x ϭ 1, and x ϭ 3
20c. Sample answer: rel. max. at x ϭ 0, rel. min. at x ϭ Ϫ2 and x ϭ 2
22a.

f (x )

f (x )

Ϫ3
0
Ϫ2 Ϫ15
Ϫ1
0
0
9
1
0
2 Ϫ15
3
0
4 105

4

19b. between Ϫ2 and Ϫ1 and between 1 and 2
19c. Sample answer: no rel. max., rel. min. at x ϭ 0

21a.

p

20a.

24

f (x )

16
8
Ϫ4

Ϫ2

O

2

4x

Ϫ8

f (x) ϭ Ϫx 4 ϩ x 3ϩ 8x 2Ϫ 3

22b. between Ϫ3 and Ϫ2, between
Ϫ1 and 0, between 0 and 1, and between 3 and 4
22c. Sample answer: rel. max. at x ϭ Ϫ1.5 and at x ϭ 2.5, rel. min. at x ϭ 0.

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24a.

p

x f (x )
Ϫ1 65
0
6
1 Ϫ1
2
2
3 Ϫ3
4 Ϫ10
5 11

x

f (x )
4
Ϫ2

O

2

4

Ϫ2
Ϫ1
0
1
2
3

x

Ϫ4
Ϫ8

f (x) ϭ x 4 Ϫ 9x 3 ϩ 25x 2 Ϫ 24x ϩ 6

p

f (x )

2

O

4

6x

Ϫ4
Ϫ8

f (x) ϭ 2x 4 Ϫ 4x 3 Ϫ 2x 2 ϩ 3x Ϫ 5

p

x f (x )

24

Ϫ2
Ϫ1

16

0

8
Ϫ4

Ϫ2

O

1

2

2

4x

3

f (x) ϭ x 5 ϩ 4x 4 Ϫ x 3 Ϫ 9x 2 ϩ 3

4

25b. between Ϫ4 and Ϫ3, between
Ϫ2 and Ϫ1, between Ϫ1 and
0, between 0 and 1, and between 1 and 2
25c. Sample answer: rel. max. at x ϭ Ϫ3 and at x ϭ 0, rel. min. at x ϭ Ϫ1 and at x ϭ 1

Ϫ2

f (x )

26a.

x f (x )
Ϫ4 Ϫ77
Ϫ3 30
Ϫ2
7
Ϫ1 Ϫ2
0
3
1 Ϫ2
2 55

45
Ϫ4
Ϫ5
Ϫ6
Ϫ7
40

4

24b. between Ϫ2 and Ϫ1, and between 2 and 3
24c. Sample answer: rel. max. at x ϭ 0.5; rel. min. at x ϭ
Ϫ0.5 and at x ϭ 1.5

23b. between 0 and 1, between 1 and 2, between 2 and 3, and between 4 and 5
23c. Sample answer: rel. max. at x ϭ 2, rel. min. at x ϭ 0.5 and at x ϭ 4
25a.

p

f (x )

5

40

Ϫ88
5

f (x )

20

Ϫ6
5
20

Ϫ4

Ϫ2

O

2

4x

Ϫ20

Ϫ3
Ϫ10
269

Ϫ40

f (x) ϭ x 5 Ϫ 6x 4 ϩ 4x 3 ϩ 17x 2 Ϫ 5x Ϫ 6

26b. between Ϫ2 and Ϫ1, between
Ϫ1 and 0, between 0 and 1, between 2 and 3, and between 4 and 5
26c. Sample answer: rel. max. at x ϭ Ϫ1 and at x ϭ 2, rel. min. at x ϭ 0 and at x ϭ 3.5

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27. highest: 1982; lowest: 2000

28. Rel. max. between 1980 and
1985 and between 1990 and
1995, rel. min. between 1975 and 1980 and between 1985 and 1990; as the number of years increases, the percent of the labor force that is unemployed decreases.

29. 5

30. Sample answer: increase, based on the past fluctuations of the graph

31.

x
0 2 4 6 8 10
B(x) 25 34 40 45 50 54
G(x) 26 33 39 44 49 53

32. The growth rate for both boys and girls increases steadily until age 18 and then begins to level off, with boys averaging a height of
71 in. and girls a height of
60 in.

x 12 14 16 18 20
B(x) 59 64 68 71 71
G(x) 56 59 61 61 60

Average Height (in.)

y

B (x )

70
65
60
55

G (x )

50
45
40
35
30
25
20
0

2

4

6

8 10 12 14 16 18 x
Age (yrs)

33. 0 and between 5 and 6

34. 5 s

35. 3 s

36.

y

O

176

x

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y

37.

38.

x

O

O

x

O

y

39.

y

40. The turning points of a polynomial function that models a set of data can indicate fluctuations that may repeat. Answers should include the following.
• To determine when the percentage of foreign-born citizens was at its highest, look for the rel. max. of the graph, which is at t ϭ 5.
The lowest percentage is found at t ϭ 75, the rel. min. of the graph.
• Polynomial equations best model data that contain turning points, rather than a constant increase or decrease like linear equations. x

41. D

42. B

43. Ϫ1.90; 1.23

44. 3.41; 0.59

45. 0; Ϫ1.22, 1.22

46. 0.52; Ϫ0.39, 1.62

47. 24a3 Ϫ 4a 2 Ϫ 2

48. 10c 2 Ϫ 25c ϩ 20

49. 8a4 Ϫ 10a 2 ϩ 4

50. 3x 3 Ϫ 10x 2 ϩ 11x Ϫ 6

51. 2x 4 ϩ 11x 2 ϩ 16

52. 4x 4 Ϫ 9x 3 ϩ 28x 2 Ϫ 33x ϩ 20

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54.

y

y ϭ x 2 Ϫ 4x ϩ 6

x

O

x

O

y

y ϭ Ϫx 2 ϩ 6x Ϫ 3

55.

56. (7, Ϫ4)

y

x

O
2

y ϭ x Ϫ 2x

57. (Ϫ3, Ϫ2)

58. (4, Ϫ2)

59. (1, 3)

60. 1 ft

61. (x ϩ 5)(x Ϫ 6)

62. (2b Ϫ 1)(b Ϫ 4)

63. (3a ϩ 1)(2a ϩ 5)

64. (2m ϩ 3)(2m Ϫ 3)

65. (t Ϫ 3)(t 2 ϩ 3t ϩ 9)

66. (r 2 ϩ 1)(r ϩ 1)(r Ϫ 1)

Lesson 7-3 Solving Equations Using
Pages 362–364
16x 4 Ϫ 12x 2 ϭ 0;
4[4(x 2)2 Ϫ 3x 2] ϭ 0

2. The solutions of a polynomial equation are the points at which the graph intersects the x-axis.

3. Factor out an x and write the equation in quadratic form so you have x[(x 2)2 Ϫ 2(x 2) ϩ
1] ϭ 0. Factor the trinomial and solve for x using the
Zero Product Property. The solutions are Ϫ1, 0, and 1.

4. not possible

178

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5. 84(n 2 )2 Ϫ 62(n 2)

6. 0, Ϫ5, Ϫ4

7. Ϫ4, Ϫ1, 4, 1

8. 6, Ϫ3 ϩ 3i13, Ϫ3 Ϫ 3i13

9. 64

10. 8 feet

11. 2(x 2)2 ϩ 6(x 2) Ϫ 10

12. not possible

13. 11(n 3 )2 ϩ 44(n 3)

14. b[7(b 2)2 Ϫ 4(b 2 ) ϩ 2)]

15. not possible

16. 6 (x 5 )2 Ϫ 4 (x 5 ) Ϫ 16 ϭ 0

17. 0, Ϫ4, Ϫ3

18. 0, Ϫ1, Ϫ5

19. Ϫ13, 13, Ϫi 13, i 13
21. 2, Ϫ2, 212, Ϫ212

22. 12, Ϫ12, 3, Ϫ3

23. Ϫ9,

24. 8, Ϫ4 ϩ 4i13, Ϫ4 Ϫ 4i13

1

9 ϩ 9i 13 9 Ϫ 9i 13
,
2
2

1

20. 0, Ϫ4, 4, Ϫ4i, 4i
30. 8, i 23, Ϫi 23

25. 81, 625

26. Ϫ343, Ϫ64

27. 225, 16

28. 400

29. 1, Ϫ1, 4
31. w ϭ 4 cm, / ϭ 8 cm, h ϭ 2 cm

32. x 4 Ϫ 7x 2 ϩ 9 ϭ 27

33. 3 ϫ 3 in.

34. 6 ϫ 6 in.

35. h 2 ϩ 4, 3h ϩ 2, h ϩ 3

36. The height increased by 3, the width increased by 2, and the length increased by 4.

37. Write the equation in quadratic form, u 2 Ϫ 9x ϩ
8 ϭ 0, where u ϭ |a Ϫ 3|.
Then factor and use the Zero
Product Property to solve for a; 11, 4, 2, and Ϫ5.

38. Answers should include the following. • Solve the cubic equation
4x 3 ϩ (Ϫ164x 2) ϩ
1600x ϭ 3600 in order to determine the dimensions of the cut square if the desired volume is 3600 in3.
Solutions are 10 in. and
31Ϫ 2601
2

in.
• There can be more than one square cut to produce the same volume because the height of the box is not specified and 3600 has a variety of different factors.

179

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39. D

40.

p

1
18

42.

41. x f (x )

f (x )

Ϫ2 Ϫ21
Ϫ1 Ϫ1
0
5
1
3
2 Ϫ1
3 Ϫ1
4
9
5 35

O

x

Ϫ1 15
0 Ϫ3
1
1
2
3
3
3
4 25

x

f (x) ϭ x 3 Ϫ 4x 2 ϩ x ϩ 5

1715
;
3

x

O

f (x) ϭ x 4 Ϫ 6x 3 ϩ 10x 2 Ϫ x Ϫ 3

44. 262; 2
Ϫ2 Ϫ3
3
46. B
R
1 Ϫ3 Ϫ1 y 48.

43. 17; 27
45.

p

f (x )

f (x )

135

47. A¿(Ϫ1, Ϫ2), B¿(3, Ϫ3),
C¿(1, 3)

C'

A
O

A'
B

C x
B'
64 xϩ2 49. x 2 ϩ 5x Ϫ 4

50. 4x 2 Ϫ 16x ϩ 27 Ϫ

51. x 3 ϩ 3x 2 Ϫ 2

52. x 3 ϩ 2x 2 Ϫ 10x ϩ15 Ϫ

180

Algebra 2

21 xϩ1 Chapter 7

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Chapter 7
Practice Quiz 1
Page 364
1. 2a3 Ϫ 6a 2 ϩ 5a Ϫ 1

2. f(x ) S Ϫϱ as x S ϩϱ, f(x ) S ϩϱ as x S Ϫϱ; odd; 3

3. Sample answer: maximum at x ϭ Ϫ2, minimum at x ϭ 0.5

4. (6x 3)2 ϩ 3(6x 3) ϩ 5 or
3
3
36(2x)2 ϩ 18(2x) ϩ 5

8

1

1

f (x )

4
Ϫ4

O

Ϫ2

4x

2

Ϫ4
Ϫ8

f (x) ϭ x 3 ϩ 2x 2 Ϫ 4x Ϫ 6

5. Ϫ3, 3, Ϫi13, i13

Lesson 7-4

The Remainder and Factor Theorems
Pages 368–370

1. Sample answer: f(x ) ϭ x 2 Ϫ 2x Ϫ 3

2. 4

3. dividend: x 3 ϩ 6x ϩ 32; divisor: x ϩ 2; quotient: x 2 Ϫ 2x ϩ 10; remainder: 12

4. 7, Ϫ91

5. 353, 1186

6. x ϩ 1, x Ϫ 3

7. x Ϫ 1, x ϩ 2

8. 2x ϩ 1, x Ϫ 4

9. x Ϫ 2, x 2 ϩ 2x ϩ 4

10. \$2.894 billion

11. \$2.894 billion

12. Sample answer: Direct substitution, because it can be done quickly with a calculator. 13. Ϫ9, 54

14. 37, Ϫ19

15. 14, Ϫ42

16. 55, 272

181

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17. Ϫ19, Ϫ243

18. 267, 680

19. 450, Ϫ1559

20. 422, 3110

21. x ϩ 1, x ϩ 2

22. x Ϫ 4, x ϩ 2

23. x Ϫ 4, x ϩ 1

24. x Ϫ 3, x Ϫ 1

25. x ϩ 3, x Ϫ

1
2

26. x Ϫ 1, x ϩ

or 2x Ϫ 1

4
3

or 3x ϩ 4

27. x ϩ 7, x Ϫ 4

28. x Ϫ 1, x ϩ 6

29. x Ϫ 1, x 2 ϩ 2x ϩ 3

30. 2x Ϫ 3, 2x ϩ 3, 4x 2 ϩ 9

31. x Ϫ 2, x ϩ 2, x 2 ϩ 1

32.

33. 3

34. 8

35. 1, 4

36. Ϫ3

37.

38. Yes; 2 ft lengths. The binomial x Ϫ 2 is a factor of the polynomial since f (2) ϭ 0.

5 1 Ϫ14 69 Ϫ140 100
5 Ϫ45 120 Ϫ100
1 Ϫ9 24 Ϫ20
0

8

1 Ϫ4 Ϫ29 Ϫ24
8
32
24;
1 4
3
0
(x ϩ 3)(x ϩ 1)

39. 7.5 ft/s, 8 ft/s, 7.5 ft/s

40. 0; The elevator is stopped.

41. By the Remainder Theorem, the remainder when f(x ) is divided by x Ϫ 1 is equivalent to f(1), or a ϩ b ϩ c ϩ d ϩ e. Since a ϩ b ϩ c ϩ d ϩ e ϭ 0, the remainder when f (x ) is divided by x Ϫ 1 is 0. Therefore, x Ϫ 1 is a factor of f(x ).

42. \$31.36

43. \$16.70

44. B(x) ϭ 2000x 5 Ϫ 340(x 4 ϩ x 3 ϩ x 2 ϩ x ϩ 1)

182

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45. No, he will still owe \$4.40.

46. Using the Remainder
Theorem you can evaluate a polynomial for a value a by dividing the polynomial by x Ϫ a using synthetic division. Answers should include the following.
• It is easier to use the
Remainder Theorem when you have polynomials of degree 2 and lower or when you have access to a calculator.
• The estimated number of international travelers to the
U.S. in 2006 is 65.9 million.

47. D

48. x Ϫ 2, x ϩ 2, x ϩ 1, x 2 ϩ 1

49. (x 2)2 Ϫ 8(x 2) ϩ 4

50. 9(d 3)2 ϩ 5(d 3) Ϫ 2

51. not possible

52. Sample answer: rel. max, at x ϭ 0.5, rel. min. at x ϭ 3.5 f (x )
16
f (x) ϭ x 3 Ϫ 6x 2 ϩ 4x ϩ 3
8
Ϫ2

O

2

2␲ 2mrFc
Fc

4

x

Ϫ8

Ϫ16

54. T ϭ

53. Sample answer: maximum at x ϭ Ϫ1, rel. max. at x ϭ 1.5, rel. min. at x ϭ 1 f (x )
8
4
Ϫ4

Ϫ2

O

2

4x

Ϫ4

f (x) ϭ Ϫx 4 ϩ 2x 3 ϩ 3x 2 Ϫ 7x ϩ 4

183

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55. (4, Ϫ2)

Ϫ7 Ϯ 117
2

56. (Ϫ3, Ϫ1)

9 Ϯ 157
6

Ϫ3 Ϯ i 17
4

57. A

58. C

59. S

60.

61.

62.

Lesson 7-5 Roots and Zeros
Pages 375–377
1. Sample answer: p(x ) ϭ x 3 Ϫ
6x 2 ϩ x ϩ 1; p(x ) has either
2 or 0 positive real zeros, 1 negative real zero, and 2 or
0 imaginary zeros.

2. An odd-degree function approaches positive infinity in one direction and negative infinity in the other direction, so the graph must cross the x-axis at least once, giving it at least one real root.

3. 6

4. 2i, Ϫ2i; 2 imaginary

5. Ϫ7, 0, and 3; 3 real

6. 2 or 0; 1; 2 or 0

7. 2 or 0; 1; 2 or 4

8. Ϫ4, 1 ϩ 2i, 1 Ϫ 2i
5 Ϯ i 271
;
4

10. 2i, Ϫ2i, 3

9. 2, 1 ϩ i, 1 Ϫ i

12. f (x ) ϭ x 3 Ϫ 2x 2 ϩ 16x Ϫ 32

11. 2 ϩ 3i, 2 Ϫ 3i, Ϫ1
8
3

13. Ϫ ; 1 real

14.

15. 0, 3i, Ϫ3i; 1 real, 2 imaginary

16. 3i, 3i, Ϫ3i, and Ϫ3i; 4 imaginary 17. 2, Ϫ2, 2i, and Ϫ2i; 2 real,
2 imaginary

18. Ϫ2, Ϫ2, 0, 2, and 2, 5 real

19. 2 or 0; 1; 2 or 0

20. 2 or 0; 1; 2 or 0

21. 3 or 1; 0; 2 or 0

22. 1; 3 or 1; 2 or 0

23. 4, 2, or 0; 1; 4, 2, or 0

24. 5, 3, or 1; 5, 3, or 1; 0, 2, 4,
6, or 8

25. Ϫ2, Ϫ2 ϩ 3i, Ϫ2 Ϫ 3i

26. 4, 1 ϩ i, 1 Ϫ i

184

2 imaginary

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i
,
2

27. 2i, Ϫ2i,

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i
2

28. 5i, Ϫ5i, 7

Ϫ

3
2

1
,
2

29. Ϫ , 1 ϩ 4i, 1 Ϫ 4i

30.

31. 4 Ϫ i, 4 ϩ i, Ϫ3

32. 3 Ϫ i, 3 ϩ i, 4, Ϫ1

33. 3 Ϫ 2i, 3 ϩ 2i, Ϫ1, 1

34. 5 Ϫ i, 5 ϩ i, Ϫ1, 6

35. f(x ) ϭ x 3 Ϫ 2x 2 Ϫ 19x ϩ 20

36. f(x ) ϭ x 4 Ϫ 10x 3 ϩ 20x 2 ϩ
40x Ϫ 96

37. f(x ) ϭ x 4 ϩ 7x 2 Ϫ 144

38. f(x ) ϭ x 5 Ϫ x 4 ϩ 13x 3 Ϫ
13x 2 ϩ 36x Ϫ 36

39. f(x ) ϭ x 3 Ϫ 11x 2 ϩ 23x Ϫ 45

40. f(x) ϭ x 3 Ϫ 10x 2 ϩ 32x Ϫ 48

41a.

42. (3 Ϫ x)(4 Ϫ x)(5 Ϫ x) ϭ 24

f (x )

x

O

41b.

4 ϩ 5i, 4 Ϫ 5i

f (x )

x

O

41c.

f (x )

O

x

185

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43. 1 ft

44. V(r ) ϭ ␲r 3 ϩ 17␲r 2

45. radius ϭ 4 m, height ϭ 21 m

46. 1; 2 or 0; 2 or 0

47. Ϫ24.1, Ϫ4.0, 0, and 3.1

48. Nonnegative roots represent times when there is no concentration of dye registering on the monitor.

[Ϫ30, 10] scl: 5 by [Ϫ20, 20] scl: 5

49. Sample answer: f(x) ϭ x 3 Ϫ
6x 2 ϩ 5x ϩ 12 and g(x) ϭ
2x 3 Ϫ 12x 2 ϩ 10x ϩ 24 each have zeros at x ϭ 4, x ϭ Ϫ2, and x ϭ 3.

50. One root is a double root.
Sample graph: f (x )

O

186

x

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51. If the equation models the level of a medication in a patient’s bloodstream, a doctor can use the roots of the equation to determine how often the patient should take the medication to maintain the necessary concentration in the body.
Answers should include the following. • A graph of this equation reveals that only the first positive real root of the equation, 5, has meaning for this situation, since the next positive real root occurs after the medication level in the bloodstream has dropped below 0 mg. Thus according to this model, after 5 hours there is no significant amount of medicine left in the bloodstream. • The patient should not go more than 5 hours before taking their next dose of medication. 52. A

53. C

54. Ϫ127, 41

55. Ϫ254, 915

56. 36 in.

57. min.; Ϫ13

58. max.; 32

59. min.; Ϫ7

60. 5ab 2(3a Ϫ c 2)

61. (6p Ϫ 5)(2p Ϫ 9)
Ϫ3
2
63. C 3 Ϫ4 S
Ϫ2
9

62. 4y (y ϩ 3)2
11
5
64. C 7
0S
4 Ϫ5

187

Algebra 2

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29 Ϫ8
65. £ 8

16 Ϫ16
1
2

2
3

66. y Ն Ϫ x Ϫ 1

5
2

68. Ϯ

67. Ϯ , Ϯ1, Ϯ , Ϯ5
1
9

1
,
14

1
16

1
3

1
7

2
7

1
2

1
8

1
4

1
2

Ϯ , Ϯ , Ϯ , Ϯ1, Ϯ2

70. Ϯ , Ϯ , Ϯ , Ϯ , Ϯ1, Ϯ2, Ϯ4

69. Ϯ , Ϯ , Ϯ1, Ϯ3

Lesson 7-6 Rational Zero Theorem
Pages 380–382
1. Sample answer: You limit the number of possible solutions.

2. Sample answer: 2x 2 ϩ x ϩ 3

3. Luis; Lauren found numbers q p in the form , not as p q
Luis did according to the
Rational Zero Theorem.

4. Ϯ1, Ϯ2, Ϯ5, Ϯ10

1
2

1
3

1
6

2
3

2 Ϫ3 Ϯ 217
,
3
4

5. Ϯ1, Ϯ2, Ϯ , Ϯ , Ϯ , Ϯ

6. Ϫ4, 2, 7

7. Ϫ2, Ϫ4, 7

8. 2, Ϫ2, 3, Ϫ3

9. Ϫ2, 2,

7
2

10.

11. 10 cm ϫ 11 cm ϫ 13 cm

12. Ϯ1, Ϯ2

13. Ϯ1, Ϯ2, Ϯ3, Ϯ6

14. Ϯ1, Ϯ3, Ϯ5, Ϯ15, Ϯ , Ϯ

15. Ϯ1, Ϯ2, Ϯ3, Ϯ6, Ϯ9, Ϯ18

16. Ϯ1, Ϯ , Ϯ3

1
3

1
3

5
3

1
3

1
9

17. Ϯ1, Ϯ , Ϯ , Ϯ3, Ϯ9, Ϯ27

18. Ϫ6, Ϫ5, 10

19. Ϫ1, Ϫ1, 2

20. 1, Ϫ1

21. 0, 9

22.

23. 0, 2, Ϫ2

24. 0, 3

25. Ϫ2, Ϫ4

26. Ϫ7, 1, 3

188

1
,
2

Ϫ1, 1

Algebra 2

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27.

1
,
2

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1
3

1 1
2 5

Ϫ , Ϫ2

28. Ϫ , , 2

1 1 1 3
2 3 2 4

30. Ϫ2, ,

5 Ϯ i 13
2

35. 2, Ϫ2 Ϯ i 13; 2

31.

4
,
5

0,

2 Ϫ3 Ϯ 213
3
2

4 Ϫ3 Ϯ i
3
2

29. Ϫ , , ,

2
3

32. 3, , Ϫ ,
1
3

4
3

34. V ϭ ␲r 3 ϩ ␲r 2

33. Ϫ1, Ϫ2, 5, i, Ϫi

˛

˛

36. r ϭ 2 in., h ϭ 6 in.

37. V ϭ 2h3 Ϫ 8h 2 Ϫ 64h

38. / ϭ 36 in., w ϭ 48 in., h ϭ 32 in.
1
3

1
3

39. V ϭ / 3 Ϫ 3/ 2

40. 6300 ϭ /3 Ϫ 3/2

41. / ϭ 30 in., w ϭ 30 in., h ϭ 21 in.

42. k ϭ Ϫ3; Ϫ3, Ϫ6, 5

43. The Rational Zero Theorem helps factor large numbers by eliminating some possible zeros because it is not practical to test all of them using synthetic substitution.
Answers should include the following. • The polynomial equation that represents the volume of the compartment is
V ϭ w 3 ϩ 3w 2 Ϫ 40w.
• Reasonable measures of the width of the compartment are, in inches, 1, 2, 3, 4, 6, 7, 9,
12, 14, 18, 21, 22, 28, 33,
36, 42, 44, 63, 66, 77, and
84. The solution shows that w ϭ 14 in., / ϭ 22 in., and d ϭ 9 in.

44. D

45. Sample answer: x 5 Ϫ x 4 Ϫ 27x 3 ϩ 41x 2 ϩ
106x Ϫ 120

46. Ϫ6, Ϫ3, 5

189

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53. Ϯ3xy 22x

47. Ϫ4, 2 ϩ i, 2 Ϫ i

48. Ϫ5, 3i, Ϫ3i

49. Ϫ7, 5 ϩ 2i, 5 Ϫ 2i

50. 4x ϩ 3, 5x Ϫ 1

51. x Ϫ 4, 3x 2 ϩ 2

52. 725

55. 6 cm, 8 cm, 10 cm

54. 0 4x Ϫ 5 0

56. x 3 ϩ 4x 2 Ϫ 6

57. 4x 2 Ϫ 8x ϩ 3

58. x 3 ϩ 5x 2 ϩ x Ϫ 10

59. x 5 Ϫ 7x 4 Ϫ 8x 3 ϩ 106x 2 Ϫ
85x ϩ 25

60. x Ϫ 9 ϩ

61. x 2 ϩ x Ϫ 4 ϩ

33 xϩ7 5 xϩ1 Chapter 7
Practice Quiz 2
Page 382
1. Ϫ930, Ϫ145

2. 0, Ϫ180

3. x 4 Ϫ 4x 3 Ϫ 7x 2 ϩ
22x ϩ 24 ϭ 0

4.

4
5

3
2

5. Ϫ

Lesson 7-7

Operations on Functions
Pages 386–389

1. Sometimes; sample answer:
If f(x ) ϭ x Ϫ 2, g(x ) ϭ x ϩ 8, then f ‫ ؠ‬g ϭ x ϩ 6 and g ‫ ؠ‬f ϭ x ϩ 6.

2. Sample answer: g(x ) ϭ
{(Ϫ2, 1), (Ϫ1, 2), (4, 3)}, f (x ) ϭ {(1, 7),(2, 9), (3, 3)}

3. Danette; [g ‫ ؠ‬f ](x ) ϭ g [f (x )] means to evaluate the f function first and then the g function. Marquan evaluated the functions in the wrong order.

4. 4x ϩ 9; 2x Ϫ 1;
3x 2 ϩ 19x ϩ 20; 3x ϩ 4,

xϩ5

x

190

Ϫ5

Algebra 2

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5. x 2 ϩ x Ϫ 1; x 2 Ϫ x ϩ 7; x 3 Ϫ 4x 2 ϩ 3x Ϫ 12; x2 ϩ 3
,
xϪ4

x

6. {(Ϫ5, 7), (4, 9)}; {(4, 12)}

4

7. {(2, Ϫ7)}; {(1, 0), (2, 10)}

8. 6x Ϫ 8; 6x Ϫ 4

9. x 2 ϩ 11; x 2 ϩ 10x ϩ 31

10. 30

11. 11

12. 1
3
4

13. p(x ) ϭ x; c (x ) ϭ x Ϫ 5

14. \$32.49; price of CD when
25% discount is taken and then the coupon is subtracted 15. \$33.74; price of CD when coupon is subtracted and then 25% discount is taken

16. Discount first, then coupon; sample answer: 25% of
49.99 is greater than 25% of
44.99.
18. 6x ϩ 6; Ϫ2x Ϫ 12;
8x 2 ϩ 6x Ϫ 27;

17. 2x ; 18; x 2 Ϫ 81;

xϩ9
,
xϪ9

x

9

2x Ϫ 3
,
4x ϩ 9

19. 2x 2 Ϫ x ϩ 8; 2x 2 ϩ x Ϫ 8;
Ϫ2x 3 ϩ 16x 2;

21.

2x 2
,
8Ϫx

x

8

xϩ3
,
2

22.

Ϫ1; x 2 Ϫ x ; x

x3 ϩ x2 Ϫ x ϩ 1
,
x

x

9
4

Ϫ

20. x 2 ϩ 8x ϩ 15; x 2 ϩ 4x ϩ 3;
2x 3 ϩ 18x 2 ϩ 54x ϩ 54;

x3 ϩ x2 Ϫ 1
, x Ϫ1 xϩ1 x 3 ϩ x 2 Ϫ 2x Ϫ 1
,
xϩ1

x

x

x

Ϫ3

x 3 ϩ x 2 Ϫ 7x Ϫ 15
,x
xϩ2 x 3 ϩ x 2 Ϫ 9x Ϫ 9
,x
xϩ2

x 2 Ϫ 6x ϩ 9, x x 2 ϩ 4x ϩ 4, x

Ϫ1
0

Ϫ2;
Ϫ2;

2;
Ϫ2, 3

23. {(1, Ϫ3), (Ϫ3, 1), (2, 1)};
{(1, 0), (0, 1)}

24. {(2, 4), (4, 4)}; {(1, 5), (3, 3),
(5, 3)}

25. {(0, 0), (8, 3), (3, 3)};
{(3, 6), (4, 4), (6, 6), (7, 8)}

26. {(4, 5), (2, 5), (6, 12), (8, 12)}; does not exist

27. {(5, 1), (8, 9)}; {(2, Ϫ4)}

28. {(2, 3), (2, 2)}; {(Ϫ5, 6), (8, 6),
(Ϫ9, Ϫ5)}

191

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29. 8x Ϫ 4; 8x Ϫ 1

30. 15x Ϫ 5; 15x ϩ 1

31. x 2 ϩ 2; x 2 ϩ 4x ϩ 4

32. 3x 2 Ϫ 4; 3x 2 Ϫ 24x ϩ 48

33. 2x 3 ϩ 2x 2 ϩ 2x ϩ 2;
8x 3 ϩ 4x 2 ϩ 2x ϩ 1

34. 2x 2 Ϫ 5x ϩ 9; 2x 2 Ϫ x ϩ 5

35. Ϫ12

36. 50

37. 39

38. 68

39. 25

40. Ϫ48

41. 2

42. 1

43. 79

44. 104

45. 226

46. 36

47. P(x ) ϭ Ϫ50x ϩ 1939

48. 939,000

49. p(x ) ϭ 0.70x; s(x ) ϭ 1.0575x

50. s[p(x )]; The 30% would be taken off first, and then the sales tax would be calculated on this price.

51. \$110.30

52. [K ‫ ؠ‬C](F ) ϭ

53. 373 K; 273 K

54. 309.67 K

55. \$700, \$661.20, \$621.78,
\$581.73, \$541.04

56. 244

57. Answers should include the following. • Using the revenue and cost functions, a new function that represents the profit is p(x ) ϭ r(c(x )).
• The benefit of combining two functions into one function is that there are fewer steps to compute and it is less confusing to the general population of people reading the formulas. 58. A

59. C

60. Ϯ1, Ϯ2, Ϯ4, Ϯ8

1
2

192

5
9

(F Ϫ 32) ϩ 273

Algebra 2

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2

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1
4

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3
2

1
3

61. Ϯ1, Ϯ , Ϯ , Ϯ2, Ϯ3, Ϯ ,

1
9

62. Ϯ1, Ϯ , Ϯ

3
Ϯ , Ϯ6
4

63. x 3 Ϫ 4x 2 Ϫ 17x ϩ 60

64. x 3 Ϫ 3x 2 Ϫ 34x Ϫ 48

65. 6x 3 Ϫ 13x 2 ϩ 9x Ϫ 2

66. x 3 Ϫ 6x 2 ϩ 4x Ϫ 24

67. x 3 Ϫ 9x 2 ϩ 31x Ϫ 39

68. x 4 ϩ x 3 Ϫ 14x 2 ϩ 26x Ϫ 20
5 Ϫ6
1
70. Ϫ B
R
2 Ϫ7
8
72. does not exist

69. 10 ϩ 2j
3 Ϫ2
71. c d Ϫ1
1
73. Ϫ B

Ϫ1 Ϫ2
R
Ϫ3 Ϫ4

1
2

1
16

75. Ϫ

74. does not exist

B

Ϫ5 Ϫ2
R
Ϫ3
2

76. x ϭ

6 ϩ 3y
2
Ϫ2
3 ϩ 7y

77. y ϭ

1 Ϫ 4x 2
Ϫ5x

78. x ϭ

79. t ϭ

I pr 80. F ϭ C ϩ 32

81. m ϭ

9
5

Fr 2
GM

Lesson 7-8

Inverse Functions and Relations
Pages 393–394

1. no

2. Switch x and y in the equation and solve for y.

3. Sample answer: f (x) ϭ 2x, f Ϫ1(x) ϭ 0.5x; f [f Ϫ1(x )] ϭ f Ϫ1[f (x )] ϭ x

4. n is an odd whole number.

5. {(4, 2), (1, Ϫ3), (8, 2)}

6. {(3, 1), (Ϫ1, 1), (Ϫ3, 1), (1, 1)}

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1
3

8. g Ϫ1(x ) ϭ x Ϫ

7. f Ϫ1(x ) ϭ Ϫx
4

f (x )

Ϫ4

O

Ϫ2

g (x )

f (x) ϭ Ϫx f Ϫ1(x ) ϭ Ϫx

2

g (x ) ϭ 3x ϩ 1
2

4x

2

Ϫ4

O

Ϫ2

Ϫ2

2

4x

Ϫ2

gϪ1(x) ϭ 1 x Ϫ 1
3
3

Ϫ4

10. yes

9. y ϭ 2x Ϫ 10
12

1
3

y y ϭ 1x ϩ 5
2

8
4
O

4

Ϫ4

8

y

Ϫ1

12

x

ϭ 2x Ϫ10

11. no

12. 32.2 ft/s2

13. 15.24 m/s2

14. {(6, 2), (5, 4), (Ϫ1, Ϫ3)}

15. {(8, 3), (Ϫ2, 4), (Ϫ3, 5)}

16. {(Ϫ4, 7), (5, 3), (4, Ϫ1), (5, 7)}

17. {(Ϫ2, Ϫ1), (Ϫ2, Ϫ3),
(Ϫ4, Ϫ1), (6, 0)}

18. {(11, 6), (7, Ϫ2), (3, 0),
(3, Ϫ5)}

19. {(8, 2), (5, Ϫ6), (2, 8), (Ϫ6, 5)}

20. x ϭ Ϫ3 y 4 x ϭ Ϫ3
2
Ϫ4

Ϫ2 O

2

4x

Ϫ2
Ϫ4

194

y ϭ Ϫ3

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1
2

22. f Ϫ1(x ) ϭ x ϩ 5

21. gϪ1(x ) ϭ Ϫ x g (x )

4

g (x ) ϭ Ϫ 1 x
2
2
Ϫ1

2

O

Ϫ2

2

Ϫ4

4

Ϫ2 O

g (x ) ϭ Ϫ2x

24. f Ϫ1(x ) ϭ x Ϫ 1

g (x )
4

f (x )
4

2 g (x ) ϭ x ϩ 4
O

2

2 f (x) ϭ 3x ϩ 3
4x

Ϫ4

gϪ1(x) ϭ xϪ2 4
Ϫ

Ϫ2 O

Ϫ4

1
2

1
2

25. y ϭ Ϫ x Ϫ

Ϫ1

Ϫ2

y Ϫ1 ϭ 3x

ϭ Ϫ 1x Ϫ 1
2
2

O

f Ϫ1(x) ϭ 1 x Ϫ1
3

2

4

y

2
Ϫ4

4x

Ϫ2 y ϭ Ϫ2x Ϫ 1
Ϫ4

4x

26. y ϭ 3x

y

4

Ϫ4

2

Ϫ2

Ϫ4

y

4

1
3

23. gϪ1(x ) ϭ x Ϫ 4

Ϫ2

2

Ϫ2 f (x) ϭ x Ϫ 5
Ϫ4

Ϫ2

Ϫ4

f Ϫ1(x) ϭ x ϩ 5 x x
Ϫ4

f (x )

4

y ϭ 1x
3

195

O

Ϫ2

2

4x

Ϫ2
Ϫ4

Algebra 2

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8
5

27. f Ϫ1(x ) ϭ x

28. f Ϫ1(x ) ϭ 3x Ϫ 12 f (x )
8
7 f (x) ϭ 1 x ϩ 4
3
6
5

f (x )

4
2
Ϫ4

O

Ϫ2

4x

2

3
2
1

f (x) ϭ 5 x
8

Ϫ2

O

Ϫ4

1 2 3 4 5 6 7 8x

f Ϫ1(x ) ϭ 3x Ϫ 12

f Ϫ1(x) ϭ 8 x
5

5
4

29. f Ϫ1(x ) ϭ x ϩ

35
4

30. gϪ1(x ) ϭ 3x Ϫ

f (x)Ϫ1 ϭ 5 x ϩ 35
4
4
Ϫ40

4

f (x )
O x
Ϫ30 Ϫ20 Ϫ10

Ϫ4

Ϫ2

O

Ϫ20
Ϫ30

4

2

4x

Ϫ2
Ϫ4

f (x) ϭ 4 x Ϫ 7
5
Ϫ40

8
7

g (x )

ϩ2 g (x) ϭ 2x 6 3

Ϫ10

31. f Ϫ1(x ) ϭ x ϩ

3
2

gϪ1(x) ϭ 3x Ϫ 3
2

4
7

32. yes

f (x )

f Ϫ1(x) ϭ 8 x ϩ2 4
7
7
Ϫ4

O

Ϫ2

Ϫ2

2

4x

Ϫ f (x) ϭ 7x 8 4

Ϫ4

33. no

34. no

35. yes

36. yes

37. yes

38. y ϭ
1
2

39. y ϭ x Ϫ

11
2

4(x ϩ 7) Ϫ 6
2

40. 12

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9
5

41. I(m) ϭ 320 ϩ 0.04m; \$4500

42. C Ϫ1(x ) ϭ x ϩ 32;
C [CϪ1(x )] ϭ C Ϫ1[C (x )] ϭ x

43. It can be used to convert
Celsius to Fahrenheit.

44. Sample answer: f(x ) ϭ x and f Ϫ1(x ) ϭ x or f(x ) ϭ Ϫx and f Ϫ1(x ) ϭ Ϫx

45. Inverses are used to convert between two units of measurement. Answers should include the following.
• Even if it is not necessary, it is helpful to know the imperial units when given the metric units because most measurements in the
U.S. are given in imperial units so it is easier to understand the quantities using our system.
• To convert the speed of light from meters per second to miles per hour,
3.0 ϫ 108 meters
ؒ
1 second
3600 seconds
1mile
ؒ
1 hour
1600 meters

46. A

f (x) Ϸ

Ϸ

675,000,000 mi/hr

47. B

48. g [h(x)] ϭ 4x ϩ 20; h[g(x)] ϭ 4x ϩ 5

49. g[h(x )] ϭ 6x Ϫ 10; h[g(x )] ϭ 6x

50. g [h(x)] ϭ x 2 Ϫ 3x Ϫ 24; h[g(x)] ϭ x 2 ϩ 5x Ϫ 24

51. Ϫ7, Ϫ2, 3

52. Ϫ , ,

53. 64

54. 32

55. 3

56. 4

57. 117

58. 196

1 4 5
4 3 2

197

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60. ٠

59. Ϫ7
61.

25
4

Lesson 7-9

2. Both have the shape of the graph of y ϭ 2x, but y ϭ 2x Ϫ 4 is shifted down
4 units, and y ϭ 2x Ϫ 4 is shifted to the right 4 units.

Square Root Functions and Inequalities
Pages 397–399

1. In order for it to be a square root function, only the nonnegative range can be considered. 3. Sample answer: y ϭ 22x Ϫ 4

4.

8
7
6
5
4
3
2
1

y

y ϭ ͙x ϩ 2
1 2 3 4 5 6 7 8x

O

D: x Ն 0, R: y Ն 2
5.

8
7
6
5
4
3
2
1

6.

y

4

x
O

y ϭ ͙4x

O

4

8

12

Ϫ2
1 2 3 4 5 6 7 8x

D: x Ն 0; R: y Ն 0
8
7
6
5
4
3
2
1

y ϭ 3 Ϫ ͙x

2

O

7.

y

D: x Ն 0; R: y Յ 3
8.

y

y ϭ ͙x Ϫ 1 ϩ 3

1 2 3 4 5 6 7 8x

8
7
6
5
4
3
2
1
O

y

y ϭ͙x Ϫ 4 ϩ 1

1 2 3 4 5 6 7 8x

D: x Ն 1; R: y Ն 3

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Algebra 2

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10.

y

6

y ϭ ͙2x ϩ 4

4

O

11.

4
3
2
1
O

Ϫ2

2

y

4

y y ϭ 3 Ϫ ͙5x ϩ 1

12. v ϭ 22gh
Ϫ1 O

2
Ϫ2

4
3
2
1

1 2 3 4 5 6 7x

Ϫ2
Ϫ3
Ϫ4

6x

y ϭ ͙x ϩ 2 Ϫ 1

1 2 3 4 5 6x

Ϫ2
Ϫ3
Ϫ4

14.

13. Yes; sample answer: The advertised pump will reach a maximum height of 87.9 ft.

8
7
6
5
4
3
2
1

y

y ϭ ͙3x
1 2 3 4 5 6 7 8x

O

D: x Ն 0, R: y Ն 0
15.

16.

y
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8

1 2 3 4 5 6 7 8x

y ϭ Ϫ͙5x

1 2 3 4 5 6 7 8x

y ϭ Ϫ4͙x

D: x Ն 0, R: y Յ 0

D: x Ն 0, R: y Յ 0

y
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8

199

Algebra 2

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18.

y

8
7
6
5
4
3
2
1

2

2

1 2 3 4 5 6 7 8x

20.

1 2 3 4 5 6 7 8x

y ϭ Ϫ͙2x ϩ 1

22.

4

y

2

x

O
Ϫ4

Ϫ6

y ϭ ͙5x Ϫ 3

y ϭ ͙x ϩ 6 Ϫ 3 Ϫ4

1 2 3 4 5 6 7 8x

23.

8

D: x Ն Ϫ6, R: y Ն Ϫ3
24.

y

6
4

y ϭ 5 Ϫ͙x ϩ 4

2
O
Ϫ2

2

4x

8
7
6
5
4
3
2
1
O

D: x Ն Ϫ4, R: y Յ 5

2

Ϫ2
Ϫ2

D: x Ն 0.6, R: y Ն 0

Ϫ4

1 2 3 4 5 6 7 8x

D: x Ն Ϫ0.5, R: y Յ 0

y

O

6x

4

y
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8

D: x Ն 7, R: y Ն 0
8
7
6
5
4
3
2
1

2

D: x Ն Ϫ2, R: y Ն 0

y ϭ ͙x Ϫ 7

O

O

Ϫ2

y

8
7
6
5
4
3
2
1

y ϭ ͙x ϩ 2

4

y ϭ 1 ͙x

D: x Ն 0, R: y Ն 0

21.

y

6

O

19.

8

y

y ϭ ͙3x Ϫ 6 ϩ 4

1 2 3 4 5 6 7 8x

D: x Ն 2, R: y Ն 4

200

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25.

8

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26.

y

y
O

y ϭ 2͙3 Ϫ 4x ϩ 3 6
4
2
Ϫ3

Ϫ2

x

O

Ϫ1

1 2 3 4 5 6 7 8x
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12 y ϭ Ϫ6͙x
Ϫ14
Ϫ16

D: x Յ 0.75, R: y Ն 3
27.

8

28.

y

8

6
4

4

2

y ϭ ͙x ϩ 5

6

2
O

Ϫ4

29.

8
7
6
5
4
3
2
1
O

31.

8
7
6
5
4
3
2
1
O

Ϫ2

y

2

y ϭ ͙2x ϩ 8

O

x

Ϫ4

30.

y

y ϭ ͙5x Ϫ 8
1 2 3 4 5 6 7 8x

8
7
6
5
4
3
2
1
O

Ϫ2

2

4x

y

y ϭ ͙x Ϫ 3 ϩ 4

1 2 3 4 5 6 7 8x

32. 125 ft

y

y ϭ ͙6x Ϫ 2 ϩ 1

1 2 3 4 5 6 7 8x

33. 317.29 mi

34. 119 lb

35. See students’ work.

36. If a is negative, the graph is reflected over the x-axis. The larger the value of a, the less steep the graph. If h is positive, the origin is

201

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translated to the right, and if h is negative, the origin is translated to the left. When k is positive, the origin is translated up, and when k is negative, the origin is translated down.
37. Square root functions are used in bridge design because the engineers must determine what diameter of steel cable needs to be used to support a bridge based on its weight. Answers should include the following.
• Sample answer: When the weight to be supported is less than 8 tons.
• 13,608 tons

38. C

39. D

40. yes

41. no

42. yes

43. 2x ϩ 2; 8; x 2 ϩ 2x Ϫ 15;

44. 11x Ϫ 22; 9xϪ18;
10x 2 Ϫ 40x ϩ 40; 10; x

45.

xϩ5
,x 3 xϪ3 8x 3 ϩ 12x 2 Ϫ 18x Ϫ 26
,
2x ϩ 3
3
x Ϫ ;
2
3 ϩ 12x 2 Ϫ 18x Ϫ 28
8x
,
2x ϩ 3
3
3 x Ϫ ; 2x Ϫ 3, x Ϫ ;
2
2

2

46. 4; If x is your number, you can write the expression
3x ϩ x ϩ 8
,
xϩ2

which equals 4

after dividing the numerator and denominator by the GCF, x ϩ 2.

8x 3 ϩ 12x 3 Ϫ 18x Ϫ 27, x 3
2

Ϫ

47. 2x 2 Ϫ 4x Ϫ 16

48. 6p 2 Ϫ 2p Ϫ 20

49. a3 Ϫ 1

202

Algebra 2

Chapter 7

Chapter 8 Conic Sections
Lesson 8-1 Midpoint and Distance Formulas
Pages 414–416
1. Since the sum of the x-coordinates of the given points is negative, the x-coordinate of the midpoint is negative. Since the sum of the y-coordinates of the given points is positive, the y-coordinate of the midpoint is positive. Therefore, the midpoint is in Quadrant II.

2. all of the points on the perpendicular bisector of the segment 3. Sample answer: (0, 0) and
(5, 2)

4. aϪ2,

5. (2.5, 2.25)

6. 10 units

8. 12.61 units

7. 1122 units

13 b 2

10. (12, 5)

9. D

12. (2, Ϫ6)

11. (Ϫ4, Ϫ2)
17 27 b 2 2

13. a ,

14. (0.075, 3.2)

15. (3.1, 2.7)

16. a , Ϫ b

1
24

17. a ,

5
12

5
24

1 13
5
b, a ,
2 2
2

5 b 8

18. a ,

1
2

2b, a5, b

19. (7, 11)

20. around 8th St. and 10th Ave.

21. Sample answer: Draw several line segments across the U.S. One should go from the NE corner to the SW corner; another should go from the SW corner to the
NW corner; another should go across the middle (east to west); and so on. Find the midpoints of these segments.
Locate a point to represent all of these midpoints.

22. near Lebanon, Kansas

203

Algebra 2

Chapter 8

25. 25 units

26. 12 units

27. 3117 units

28. 0.75 unit

30. 165 units

29. 170.25 units

24. 13 units

23. See students’ work.
1813
12

32. 1271 units

36. 165 ϩ 2 12 ϩ 1122 ϩ
1277 units

35. 712 ϩ 158 units, 10 units2
31. 1 unit

37. 1130 units
33.

34. 6110␲ units, 90␲ units2

units

38. about 85 mi

39. about 0.9 h

40. 14 in.

41. The slope of the line through

42. The formulas can be used to decide from which location an emergency squad should be dispatched. Answers should include the following.
• Most maps have a superimposed grid. Think of the grid as a coordinate system and assign approximate coordinates to the two cities. Then use the Distance Formula to find the distance between the points with those coordinates. • Suppose the bottom left of the grid is the origin. Then the coordinates of Lincoln are about (0.7, 0.3); the coordinates of Omaha are about (4.6, 3.3); and the coordinates of Fremont are about (1.5, 4.6). The distance from Omaha to

y Ϫy

1
(x1, y1) and (x2, y2) is 2 x2 Ϫ x1 and the point-slope form of the equation of the line is

y2 Ϫ y1
(x Ϫ x1). x2 Ϫ x1 x ϩ x2 y1 ϩ y2
Substitute ¢ 1
,

2
2

y Ϫ y1 ϭ

into this equation. The left

y2 Ϫ y1
.
2 y Ϫ y1
The right side is 2 x2 Ϫ x1 y2 Ϫ y1 x2 Ϫ x1 x1 ϩ x2
¢
Ϫ x1≤ ϭ

¢ x2 Ϫ x1
2
2 y Ϫ y1 or 2
. Therefore, the
2

side is

y1 ϩ y2
2

Ϫ y1 or

point with coordinates
¢

x1 ϩ x2 y1 ϩ y2
,

2
2

lies on the

line through (x1, y1) and
(x2, y2). The distance from
B
¢

x1 ϩ x2 y1 ϩ y2
,

2
2

y1 Ϫ y2 2 x1 Ϫ x2 2
≤ ϩ¢
≤ . The
B
2
2

¢x1 Ϫ

or

to (x1, y1) is

¢

y1 ϩ y2 2 x1 ϩ x2 2
≤ ϩ ¢y1 Ϫ

2
2

102(1.5 Ϫ 4.6)2 ϩ (4.6 Ϫ 3.3)2

or 34 miles. The distance from Lincoln to Fremont is
204

Algebra 2

Chapter 8

distance from ¢

B

to (x2, y2) is

x1 ϩ x2 y1 ϩ y2
,

2
2

y2 Ϫ y1 2 x2 Ϫ x1 2
≤ ϩ¢

B
2
2
¢x1 Ϫ

102(1.5 Ϫ 0.7)2 ϩ (4.6 Ϫ 0.3)2

or 44 miles. Since Omaha is closer than Lincoln, the helicopter should be dispatched from Omaha.

y1 ϩy2 x1 ϩ x2 b ϩ ¢y2 Ϫ
≤ ϭ
2
2
2

y1 Ϫ y2 2 x1 Ϫ x2 2
≤ ϩ¢
≤.
B
2
2
¢

¢

2

or

Therefore, the point with coordinates ¢

x1 ϩ x2 y1 ϩ y2
,

2
2

is equidistant from (x1, y1) and (x2, y2).
43. C

44. B

45. on the line with equation y ϭ x

ឈជ
46. Ϫ1; AA ¿ is perpendicular to the line with equation y ϭ x, which has slope 1.
48. D ϭ 5x 0 x Ն 06,
R ϭ 5y 0 Ն Ϫ16

47. D ϭ 5x 0 x Ն 26, R ϭ 5y 0 y Ն 06 y y

y ϭ ͙x Ϫ 1

y ϭ ͙x Ϫ 2

O

x

x

O

49. D ϭ 5x 0 x Ն 06, R ϭ 5y 0 y Ն 16

50. no

y

y ϭ 2͙x ϩ 1

O

x

51. Ϫ1 ϩ 13i

52. 6 Ϫ 2i

53. 4 Ϫ 3i

54. y ϭ (x ϩ 3)2

205

Algebra 2

Chapter 8

55. y ϭ (x Ϫ 2)2 Ϫ 3

56. y ϭ 2(x ϩ 5)2

57. y ϭ 3(x Ϫ 1)2 ϩ 2

58. y ϭ Ϫ(x ϩ 2)2 ϩ 10

59. y ϭ Ϫ3(x ϩ 3)2 ϩ 17

Lesson 8-2 Parabolas
Pages 423–425
15
16

1. (3, Ϫ7), a3, Ϫ6 b, x ϭ 3, y ϭϪ7

2. Sample answer: x ϭϪy 2

1
16

4. y ϭ 2(x Ϫ 3)2 Ϫ 12

3. When she added 9 to complete the square, she forgot to also subtract 9. The standard form is y ϭ (x ϩ 3)2 Ϫ
9 ϩ 4 or y ϭ (x ϩ 3)2 Ϫ 5.
3
4

5. (3, Ϫ4), a3, Ϫ3 b, x ϭ 3,
1
4

7
8

y ϭ Ϫ4 , upward, 1 unit

y ϭ 2 , upward,

y

16
14
12
10
8
6
4 y ϭ 2(x ϩ 7)2 ϩ 3 2

x

O

Ϫ Ϫ12Ϫ Ϫ8 Ϫ6 Ϫ4 Ϫ2
14
10

y ϭ (x Ϫ 3) 2 Ϫ 4

1
8

6. (Ϫ7, 3), aϪ7, 3 b, x ϭ Ϫ7,

206

1
2

unit

y

2x

Algebra 2

Chapter 8

4
3

2
3
7 ϭϪ ,
12

4
3

3
4

4
3

3 9
2 2

7. aϪ , Ϫ b, aϪ , Ϫ b, x ϭ Ϫ , y downward,

y ϭ Ϫ3x 2 Ϫ 8x Ϫ 6

1
3

9 9
8 2

9
2

8. aϪ , b, aϪ , b, y ϭ ,
15
8

x ϭ Ϫ , right,

unit

3
2

units

y

y x O

x ϭ 2 y 2 Ϫ 6y ϩ 12
3

x

O

1
8

1
8

9. y ϭ (x Ϫ 3)2 ϩ 6

10. x ϭ Ϫ (y ϩ 1)2 ϩ 5 y y

8

8

x

O

11. x ϭ

x

O x ϭ Ϫ 1 (y ϩ 1) 2 ϩ 5

y ϭ 1 (x Ϫ 3) 2 ϩ 6

1 2 y 24

12. y ϭ (x Ϫ 3)2 ϩ 2

Ϫ6

1
2

14. y ϭ (x ϩ 12)2 Ϫ 80

13. x ϭ (y ϩ 7)2 Ϫ 29
5 2

3
2

1
12

15. x ϭ 3 ay ϩ b Ϫ 11
6

3
2

16. (0, 0), a0, Ϫ b, x ϭ 0, y ϭ , downward, 6 units y Ϫ6y ϭ x 2
O

207

x

Algebra 2

Chapter 8

1
2

1
2

17. (0, 0), a , 0b, y ϭ 0, x ϭ Ϫ , right, 2 units

3
4

18. (Ϫ6, 3), aϪ6, 3 b, x ϭ Ϫ6,
1
4

y ϭ 2 , upward, 3 units

y

y y 2 ϭ 2x x O

3(y Ϫ 3) ϭ (x ϩ 6)2
O x

1
2

1
2

19. (1, 4), a1, 3 b, x ϭ 1, y ϭ 4 , downward, 2 units

20. (2, Ϫ3), (3, Ϫ3), y ϭ Ϫ3, x ϭ 1, right, 4 units y y

x

O

4(x Ϫ 2) ϭ (y ϩ 3)2
O
Ϫ2(y Ϫ 4) ϭ (x Ϫ 1)2

x

16
14
12
10
8
6
4
2
Ϫ4 Ϫ3Ϫ2Ϫ1

3
4

22. (6, Ϫ16), a6, Ϫ15 b, x ϭ 6,

21. (4, 8), (3, 8), y ϭ 8, x ϭ 5, left, 4 units

1
4

y ϭ Ϫ16 , upward, 1 unit

y

Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
Ϫ14
Ϫ16

(y Ϫ 8)2 ϭ Ϫ4(x Ϫ 4)
O
1 2 3 4x

208

y
O2 4 6 8 10 12 14 x

y ϭ x 2 Ϫ 12x ϩ 20

Algebra 2

Chapter 8

5
115
144
, Ϫ b, a ,
4
2
5
287
1

, right,
10
5

3
4

24. a

23. (Ϫ24, 7), aϪ23 , 7b, y ϭ 7,
1
4

x ϭ Ϫ24 , right, 1 unit
24

y

y

x ϭ y 2 Ϫ 14y ϩ 25 16

Ϫ8

25. (4, 2), a4, 2
11
12

1 b, 12

unit

x ϭ 5y 2 ϩ 25y ϩ 60

2
1

Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6

8x

O

Ϫ8

5
2

O 10 20 30 40 50 60 70 80 x

8
Ϫ24 Ϫ16

5
2

Ϫ b, y ϭ Ϫ ,

5
4

y ϭ 1 , upward,

1
3

55
5
5
8
4
4
27
1
Ϫ , downward, unit
4
2

26. a , Ϫ b, a , Ϫ7b, x ϭ ,

x ϭ 4,

unit

y
O

y
Ϫ2

2

x

4

Ϫ4 x ϭ Ϫ2x 2 ϩ 5x Ϫ 10
Ϫ8
Ϫ12

y ϭ 3x 2 Ϫ 24x ϩ 50

Ϫ16

x

O

17 3
67 3
4 4
16 4
69
1 x ϭ , left,
4
16

3
4

1
2

27. a , b, a , b, y ϭ ,

1
2

28. (3, 5), a3, 5 b, x ϭ 3, y ϭ 4 , upward, 2 units

unit

y

y x ϭ Ϫ4y 2 ϩ 6y ϩ 2

O

y ϭ 1 x 2 Ϫ 3x ϩ 19
2

x
O

209

2

x

Algebra 2

Chapter 8

1
4
3 x ϭ 123 , left,
4

29. (123, Ϫ18), a122 , Ϫ18b, y ϭ Ϫ18,

20
Ϫ120 Ϫ60

O

30.

y x ϭ 3y 2 ϩ 4y ϩ 1

3 units

y

O

x

120x

60

Ϫ20
Ϫ40
Ϫ60

x ϭ Ϫ 1 y 2 Ϫ 12y ϩ 15
3

1
3

31. 1

32. Ϫ1 and Ϫ
1
3

2
3

2
3

33. y ϭ Ϫ

34. aϪ , Ϫ b

35. 0.75 cm

36. y ϭ

1 2 x 16

ϩ1

y ϭ 1 x2 ϩ 1
16

y

x

O

1
(y
24

37. x ϭ Ϫ y O

1
8

38. x ϭ (y ϩ 2)2 Ϫ 6

Ϫ 6)2 ϩ 8

y

14
12
10
8
6 x ϭ Ϫ 1 (y Ϫ 6)2ϩ 8
24
4
2

x

O

x ϭ 1 (y ϩ 2)2Ϫ 6
8

1 2 3 4 5 6 7 8x
Ϫ2

210

Algebra 2

Chapter 8

39. y ϭ

1
(x
16

1
6

40. y ϭ Ϫ (x ϩ 7)2 ϩ 4

Ϫ 1)2 ϩ 7
8
6
4
2

Ϫ4 Ϫ3Ϫ2Ϫ1
Ϫ2
Ϫ4
Ϫ6
Ϫ8

y

8

y ϭ Ϫ 1 (x ϩ 7)2ϩ 4
6

y

4

y ϭ 1 (x Ϫ 1)2ϩ 7
16

O

O1 2 3 4 5 6 x

Ϫ12

Ϫ8

Ϫ4

x

Ϫ4

XBox.

Ϫ8

2
9

1
4

42. y ϭ x 2 Ϫ 2

41. x ϭ (y Ϫ 3)2 ϩ 4 y x ϭ 1 (y Ϫ 3)2ϩ 4
4

x

O

1
(x
100

43. about y ϭ Ϫ0.00046x 2 ϩ 325
1
x2
26,200

45. y ϭ Ϫ

44. y ϭ Ϫ

Ϫ 50)2 ϩ 25

46. x ϭ (y Ϫ 3)2 ϩ 4

ϩ 6550

47. A parabolic reflector can be used to make a car headlight more effective. Answers should include the following.
• Reflected rays are focused at that point.
• The light from an unreflected bulb would shine in all directions. With a parabolic reflector, most of the light can be directed forward toward the road.

48. B

49. A

50. 13 units

52. 234 units

51. 10 units

211

Algebra 2

Chapter 8

53.

54. 2.016 ϫ 105

y

O

y ϭ͙x ϩ 1

x

55. 4

56. 5

57. 9

58. 12

59. 223

60. 322

61. 423

62. 622

Lesson 8-3 Circles
Pages 428–431
1. Sample answer: (x Ϫ 6)2 ϩ
(y ϩ 2)2 ϭ 16

2. (x ϩ 3)2 ϩ (y Ϫ 1)2 ϭ 64; left
3 units, up 1 unit

3. Lucy; 36 is the square of the radius, so the radius is 6 units. 4. (x Ϫ 3)2 ϩ (y ϩ 1)2 ϭ 9

5. (x ϩ 1)2 ϩ (y ϩ 5)2 ϭ 4

6. x 2 ϩ (y ϩ 2)2 ϭ 25

7. (x Ϫ 3)2 ϩ (y ϩ 7)2 ϭ 9

8. (4, 1), 3 units y O

x
(x Ϫ 4)2 ϩ (y Ϫ 1)2 ϭ 9

212

Algebra 2

Chapter 8

9. (0, 14), 234 units
24

y

10. (4, 0),

(x Ϫ 4)2 ϩ y 2 ϭ 16
25

8

x

O

O

Ϫ8

unit

y

x 2 ϩ (y Ϫ 14)2 ϭ 34

16

Ϫ16

4
5

16x

8

Ϫ8

2 1
3 2

212
3

11. aϪ , b

12. (Ϫ4, 3), 5 units

unit

y

y

(x ϩ 4)2 ϩ (y Ϫ 3)2 ϭ 25

x

O

(x ϩ 2 ) ϩ (y Ϫ 1 )
3
2
2

2

ϭ 8

O

9

x

14. x 2 ϩ y 2 ϭ 42,2002

13. (Ϫ2, 0), 223 units y x

O

(x ϩ 2)2 ϩ y 2 ϭ 12

16. (x ϩ 1)2 ϩ (y Ϫ 1)2 ϭ 16

15.
Earth

y
Satellite
35,800 km x

42,200 km 213

Algebra 2

Chapter 8

17. (x Ϫ 2)2 ϩ (y ϩ 1)2 ϭ 4

18. x 2 ϩ (y Ϫ 3)2 ϭ 49

1
4

23. (x ϩ 213)2 ϩ (y Ϫ 42)2 ϭ
1777

20. (x ϩ 1)2 ϩ (y Ϫ 4)2 ϭ 20

1945
4

22. (x Ϫ 8)2 ϩ (y ϩ 9)2 ϭ 1130

25. (x Ϫ 4)2 ϩ (y Ϫ 2)2 ϭ 4

26. (x Ϫ 1)2 ϩ (y Ϫ 4)2 ϭ 16

27. (x ϩ 5)2 ϩ (y Ϫ 4)2 ϭ 25

28. x 2 ϩ y 2 ϭ 18

29. (x ϩ 2.5)2 ϩ (y ϩ 2.8)2 ϭ 1600

30. (0, Ϫ2), 2 units

19. (x ϩ 8)2 ϩ (y Ϫ 7)2 ϭ
1 2
2

21. (x ϩ 1)2 ϩ ay ϩ b ϭ

24. (x ϩ 8)2 ϩ (y ϩ 7)2 ϭ 64

y x 2 ϩ (y ϩ 2)2 ϭ 4
O

32. (3, 1), 5 units

31. (0, 0), 12 units
16

y

x

y

x 2 ϩ y 2 ϭ 144

8
O

Ϫ16 Ϫ8

8

16x

(x Ϫ 3)2 ϩ (y Ϫ 1)2 ϭ 25

Ϫ8

x

O

Ϫ16

214

Algebra 2

Chapter 8

34. (3, 0), 4 units

33. (Ϫ3, Ϫ7), 9 units
(x ϩ 3)2 ϩ (y ϩ 7)2 ϭ 81 4
2

y

y

Ϫ12Ϫ10Ϫ8Ϫ6Ϫ4Ϫ2 O2 4 6 8 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
Ϫ14
Ϫ16

(x Ϫ 3)2 ϩ y 2 ϭ 16

36. (Ϫ25, 4), 5 units

35. (3, Ϫ7), 522 units
2
Ϫ6Ϫ4 Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
Ϫ14

x

O

y

y (x Ϫ 3)2 ϩ (y ϩ 7)2 ϭ 50
O
2 4 6 8 10 x

37. (Ϫ2, 23), 229 units

O

x

38. (Ϫ7, Ϫ3), 222 units

y

y
O x

O

x

215

Algebra 2

Chapter 8

40. (Ϫ1, 0), 211 units

39. (0, 3), 5 units y y

O

41. (9, 9), 2109 units
O

18
16
14
12
10
8
6
4
2
Ϫ2O
Ϫ2

x

9
2

42. aϪ , 4b,

y

2129
2

x

units y 3 217
2

2 4 6 8 10 12 14 16 18 x

3
2

43. a , Ϫ4b,
4
2
Ϫ6 Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12

O x

44. (6, 8), 4 units

units

y
O

16
14
12
10
8
6
4
2

2 4 6 8 10 x

Ϫ2
Ϫ2

216

y

O

2 4 6 8 10 12 14 x

Algebra 2

Chapter 8

45. (Ϫ1, Ϫ2), 214 units

46. (Ϫ2, 1), 22 units

y

O

y

x
O

47. a0, Ϫ b, 219 units
9
2

x

48. about 109 mi

y
O

x

49. (x ϩ 1)2 ϩ (y ϩ 2)2 ϭ 5

50. A circle can be used to represent the limit at which planes can be detected by radar. Answers should include the following.
• x 2 ϩ y 2 ϭ 2500
• The region whose boundary is modeled by x 2 ϩ y 2 ϭ 4900 is larger, so there would be more planes to track.

51. A

52. D

54. y ϭ 216 Ϫ (x ϩ 3)2,

53. y ϭ Ϯ216 Ϫ (x ϩ 3)2

y ϭ Ϫ216 Ϫ (x ϩ 3)2

217

Algebra 2

Chapter 8

56. x ϭ Ϫ3 Ϯ 216 Ϫ y 2;
The equations with the ϩ symbol and Ϫ symbol represent the right and left halves of the circle, respectively. 55.

[Ϫ10, 10] scl:1 by [Ϫ10, 10] scl:1

1
12

11
12

left,

1
3

1
4

58. (3, Ϫ2), a3, Ϫ2 b, x ϭ 3,

57. (1, 0), a , 0b, y ϭ 0, x ϭ 1 ,

3
4

y ϭ Ϫ1 , downward, 1 unit

unit y y
O

x ϭ Ϫ3y 2 ϩ 1

y ϩ 2 ϭ Ϫ(x Ϫ 3)2

x

x

O

3
4

59. (Ϫ2, Ϫ4), aϪ2, Ϫ3 b, x ϭϪ2,
1
4

60. (4, Ϫ4)

y ϭ Ϫ4 , upward, 1 unit y 2

y ϭ x ϩ 4x

O

x

3
2

61. (Ϫ1, Ϫ2)

62. a , 6b

63. Ϫ4, Ϫ2, 1

64. Ϫ , 2, 3

65. 28 in. by 15 in.

66. 12

67. 6

68. 4

69. 25

70. 225

1
2

71. 222

218

Algebra 2

Chapter 8

Chapter 8
Practice Quiz 1
Page 431
2. 2226 units

1. 13 units
1
2
1
Ϫ1 ,
2

1
4

4. (Ϫ4, 4), aϪ4, 4 b, x ϭ Ϫ4,

3. (0, 0), a1 , 0b, y ϭ 0,

3
4

y ϭ 3 , upward, 1 unit

right, 6 units

y

y y 2 ϭ 6x x O

y ϭ x 2 ϩ 8x ϩ 20
O x

5. (0, 4), 7 units
12
10
8
6
4
2
Ϫ8Ϫ6Ϫ4Ϫ2O
Ϫ2
Ϫ4

y

2 4 6 8x

x 2 ϩ (y Ϫ 4)2 ϭ 49

219

Algebra 2

Chapter 8

Lesson 8-4 Ellipses
Pages 437–440
2. Let the equation of a circle be (x Ϫ h)2 ϩ (y Ϫ k)2 ϭ r 2.
Divide each side by r 2 to get

1. x ϭ Ϫ1, y ϭ 2

(y Ϫ k)2
(x Ϫ h)2 ϩ r2 r2 ϭ1. This

is the equation of an ellipse with a and b both equal to r. In other words, a circle is an ellipse whose major and minor axes are both diameters.
4.

(x Ϫ
4

5.

2)2

(y ϩ 4)2
36

5)2

ϩ

(y ϩ
1

ϩ

(x Ϫ 2)2
4

x2
36

6.

y2
100

ϭ1 ϭ1 7. (0, 0): (0, Ϯ3); 612; 6

ϩ

y2
20

ϩ

ϭ1

x2
36

ϭ1

8. (1, Ϫ2); (5, Ϫ2), (Ϫ3, Ϫ2);
415; 4

y

y

O

x

O

y2 x2 ϩ ϭ1 18
9

x

( y ϩ 2)2
(x Ϫ 1)2 ϩ ϭ1
4
20

9. (0, 0); (Ϯ2, 0); 412; 4

10. (4, Ϫ2); (4 Ϯ 216, Ϫ2); 10; 2 y y

x

O

x

O
2

2

4x ϩ 8y ϭ 32

220

Algebra 2

Chapter 8

x2
1.32 ϫ 1015

12.

y2
64

14.

(x Ϫ 5)2
64

ϩ

(y Ϫ 4)2
9

ϭ1

16.

(x ϩ 2)2
81

ϩ

(y Ϫ 5)2
16

ϭ1

ϭ1

18.

(y Ϫ 2)2
100

ϩ

(x Ϫ 42 2
9

ϭ1

ϭ1

20.

(x Ϫ 1)2
81

ϩ

(y Ϫ 2)2
56

ϭ1

22.

x2
324

24.

x2
193,600

ϩ

y2
279,312.25

26.

(x Ϫ 1)2
30

ϩ

(y ϩ 1)2
5

ϩ

y2
1.27 ϫ 1015

ϩ

x2
39

ϭ1

ϭ1
13.

x2
16

ϩ

y2
7

15.

y2
16

ϩ

(x ϩ 2)2
4

17.

(y Ϫ 4)2
64

ϩ

19.

(x Ϫ 5)2
64

ϩ

21.

x2
169

ϩ

ϭ1

y2
25

ϭ1

(x Ϫ 2)2
4
(y Ϫ 4)2
81
4

ϭ1

x2
2.02 ϫ 1016

ϩ

y2
2.00 ϫ 1016

ϩ

y2
196

ϭ1 ϭ1 ϭ1
25.

y2
20

ϩ

x2
4

ϭ1

27. (0, 0); (0, Ϯ15); 2110; 215

28. (0, 0); (Ϯ4, 0); 10; 6

y

O

y

x

x

O

y2 x2 ϩ ϭ1 10
5

ϭ1

x2 y2 ϩ ϭ1 25
9

221

Algebra 2

Chapter 8

29. (Ϫ8, 2); (Ϫ8 Ϯ 3 17, 2); 24; 18
16

30. (5, Ϫ11); (5, Ϫ11 Ϯ 123);
24; 22

y

4

8

Ϫ12Ϫ8Ϫ4 O4 8 12 16 20 x
Ϫ4
Ϫ8
Ϫ12
Ϫ16
Ϫ20
Ϫ24
Ϫ28
(y ϩ 11)2
(x Ϫ 5)2 ϩ ϭ1

O
Ϫ24 Ϫ16

8x

Ϫ8

y

32. (0, 0); (0, Ϯ 16); 6; 213

Ϫ8
(x ϩ 8) 2
(y Ϫ 2)2 ϩ ϭ1
Ϫ16
144
81

144

31. (0, 0); (Ϯ16, 0); 6; 213

121

y

y

x

O

3x 2 ϩ 9y 2 ϭ 27

27x 2 ϩ 9y 2 ϭ 81

34. (0, 0); (Ϯ315, 0); 18; 12

33. (0, 0); (0, Ϯ17); 8; 6 y 8
6
4
2

x

O

2

x

O

Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8

2

16x ϩ 9y ϭ 144

y

O
2 4 6 8x

36x 2 ϩ 81y 2 ϭ 2916

36. (Ϫ2, 7); (Ϫ2 Ϯ 412, 7);
4110 ; 412

35. (Ϫ3, 1); (Ϫ3, 5), (Ϫ3, Ϫ3);
416; 412 y 12

y

8
4
O
O

x

Ϫ8

4

Ϫ4

x

Ϫ4

222

Algebra 2

Chapter 8

37. (2, 2); (2, 4), (2, 0); 217; 213

38. (Ϫ1, 3); (2, 3), (Ϫ4, 3); 10; 8

y

y

x

O

x

O

39.

x2
12

ϩ

y2
9

ϭ1

40. Knowledge of the orbit of
Earth can be used in predicting the seasons and in space exploration. Answers should include the following.
• Knowledge of the path of another planet would be needed if we wanted to send a spacecraft to that planet. • 1.55 million miles
42. B

41. C

x2
1.35ϫ1019

ϩ

44. (x Ϫ 3)2 ϩ (y ϩ 2)2 ϭ 25

y2
1.26 ϫ 1019

ϭ1
45. (x Ϫ 4)2 ϩ (y Ϫ 1)2 ϭ 101

46. (x ϩ 1)2 ϩ y 2 ϭ 45

47. (x Ϫ 4)2 ϩ (y ϩ 1)2 ϭ 16

48. y ϭ (x Ϫ 3)2 ϩ 1

1
2

y

O

223

1
2
y ϭ 2 (x Ϫ 3) ϩ 1 x

Algebra 2

Chapter 8

49.

50. Sample answer using
(0, 104.6) and (10, 112.6): y ϭ 0.8x ϩ 104.6

People (millions)

Married Americans
120
118
116
114
112
110
108
106
104
0

0 2 4 6 8 10 12 14 16 18 20

51. Sample answer: 128,600,000

52.

y

O

y ϭ 2x

53.

y y ϭ Ϫ 1x
2

x

O

x

O

55.

54.

y

x

y ϭ Ϫ2x

56.

y

y

y ϭ 1x
2

O

x

O

x y ϩ 2 ϭ 2(x Ϫ1)

224

Algebra 2

Chapter 8

57.

y

O

x y ϩ 2 ϭ Ϫ2(x Ϫ 1)

Lesson 8-5 Hyperbolas
Pages 445–448
1. sometimes

5.

x2
1

Ϫ

y2
15

2. As k increases, the branches of the hyperbola become wider. x2
4

Ϫ

y2
9

6. (0, Ϯ3 22); (0, Ϯ238);
4.

ϭ1

ϭ1

y2
4

Ϫ

x2
21

yϭϮ

ϭ1

3110 x 10
8
6
4
2

Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8

225

y

y2

x2
Ϫ 20 ϭ 1
18
O2 4 6 8 x

Algebra 2

Chapter 8

7. (1, Ϫ6 Ϯ 2 25);
(1, Ϫ6 Ϯ 3 25); y ϩ 6 ϭϮ

2 25
(x
5

8. (Ϯ6, 0); (Ϯ237, 0);
1
6

yϭϮ x
Ϫ 1)

16

y

2
2
8 x Ϫ 36y ϭ 36

x

O

y

Ϫ16

O

Ϫ8

8

16x

Ϫ8
2

(y ϩ 6)

2

Ϫ

20

(x Ϫ 1)
25

ϭ1

Ϫ16

9. (4 Ϯ 2 25, Ϫ2);
(4 Ϯ 3 25, Ϫ2); yϩ2ϭϮ 25
(x
2

3
4

10. (0, Ϯ15); (0, Ϯ25); y ϭ Ϯ x
20
15
10
5

Ϫ 4)

y

16
12
8
4

13.

x2
4

Ϫ

y2
12

ay

11 ϩ 2b

12.

Ϫ

15.

x2
25

17.

(x Ϫ 2)2
49

Ϫ

19.

x2
16

y2
36

(x ϩ 2)2
4

ϭ1

(x Ϫ 3)2
4

Ϫ

(y ϩ 5)2
9

ϭ1

16.

y2
16

ϭ1

18.

(y Ϫ 5)2
16

Ϫ

20.

ϭ1

ϭ1

Ϫ

Ϫ

y2
36

ϭ1

2

25
4

Ϫ

(y Ϫ 3)2
1

14.

ϭ1 x2 6

y2
9

ϭ1

x2
Ϫ 400 ϭ 1
225
y2

Ϫ Ϫ Ϫ Ϫ5 O5 10 15 20x
20 15 10
Ϫ5
Ϫ10
Ϫ15
Ϫ20

Ϫ12Ϫ8Ϫ4 O4 8 12 16 20 x
Ϫ4
Ϫ8
Ϫ12
Ϫ16

11.

y

(y ϩ 3)2
4

ϭ1

226

Ϫ

Ϫ

x2
49

x2
4

(x ϩ 4)2
81

ϭ1

Algebra 2

Chapter 8

21. (Ϯ9, 0); (Ϯ 2130, 0);

22. (0, Ϯ6); (0, Ϯ2210); y ϭϮ3x

7
9

yϭϮ x

8
6
4
2

2
2
y x Ϫ y ϭ1
81
49
16
12
8
4

4
5

24. (Ϯ3, 0); ( Ϯ234, 0); y ϭ Ϯ x

y

y x2 y2
Ϫ 25 ϭ 1
9

Ϫ8 Ϫ6 Ϫ4Ϫ2 O2 4 6 8 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8 y2 x2
Ϫ 25 ϭ 1
16

22
Ϯ x
2

26. (Ϯ2, 0); (Ϯ222, 0); y ϭ Ϯx y x 2 Ϫ y 2ϭ 4

y x 2 Ϫ 2y 2 ϭ 2

x

O

25. (Ϯ 22, 0); (Ϯ23, 0);

O

x2
Ϫ 4 ϭ1

5
3

23. (0, Ϯ4); (0, Ϯ241); y ϭ Ϯ x

y2

36

Ϫ4 Ϫ3 Ϫ2Ϫ1 O1 2 3 4 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8

Ϫ16Ϫ12Ϫ8Ϫ4 O4 8 12 16x
Ϫ4
Ϫ8
Ϫ12
Ϫ16

8
6
4
2

y

O

x

x

227

Algebra 2

Chapter 8

27. (0, Ϯ6); (0, Ϯ3 25); y ϭ Ϯ2x
16

y

8
Ϫ8

Ϫ16

Ϫ4

yϭϮ

y

y 2 ϭ 36 ϩ 4x 2
O

16x

8

6y 2 ϭ 2x 2 ϩ 12

30. (2, Ϫ2), (2, 8); (2, 3 Ϯ 241);

Ϫ12

5
4

4
3

y Ϫ 3 ϭ Ϯ (x Ϫ 2)

(Ϫ2, 9); y Ϫ 4 ϭ Ϯ (x ϩ 2)
12

y

y
10
8
(x Ϫ 2)2
)
(y Ϫ 362 ϭ1 4 Ϫ
16
25
2
O
2 4 6 8 10 x
Ϫ6 Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6

8
(y Ϫ 4
16

(x ϩ 2)2

Ϫ 49
O
Ϫ4

ϭ1

(Ϫ1 Ϯ 213, Ϫ3);
Ϫ8

4

x

(Ϫ6 Ϯ 3 25, Ϫ3);

Ϫ4

31. (Ϫ3, Ϫ3), (1, Ϫ3); yϩ3ϭ x

O

29. (Ϫ2, 0), (Ϫ2, 8); (Ϫ2, Ϫ1),

)2

23 x 3

28. (0, Ϯ22); (0, Ϯ222);

3
Ϯ (x
2

32. (Ϫ12, Ϫ3), (0, Ϫ3);
1
2

ϩ 1)

y ϩ 3 ϭ Ϯ (x ϩ 6)

y
O

6
4
2

x

y

O
2x
Ϫ Ϫ Ϫ Ϫ8 Ϫ6 Ϫ4Ϫ2
14 12 10
Ϫ2
Ϫ4
Ϫ6
2
(y ϩ 3)Ϫ8
(x ϩ 6)2 ϭ Ϫ
9 Ϫ10 1
36
(y ϩ 3)2
(x ϩ 1)2 ϭ1 Ϫ
9
4

228

Algebra 2

Chapter 8

34. (Ϫ4, 0), (6, 0); (1 Ϯ 229, 0);

33. (1, Ϫ3 Ϯ 2 26);
(1, Ϫ3 Ϯ 4 22);

2
5

y ϭ Ϯ (x Ϫ 1)

y ϩ 3 ϭ Ϯ23(x Ϫ 1)
6
4
2

8
6
4
2

y

Ϫ8 Ϫ6 Ϫ4Ϫ2 O2 4 6 8 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10

y

Ϫ8 Ϫ6 Ϫ4Ϫ2 O2 4 6 8 x
Ϫ2
Ϫ4
4x 2 Ϫ 25y 2 Ϫ 8x Ϫ 96ϭ 0
Ϫ6
Ϫ8

y 2 Ϫ 3x 2 ϩ 6y ϩ 6x Ϫ 18 ϭ 0

35.

x2
1.1025

Ϫ

y2
7.8975

36.

ϭ1

y

Station

38.

37. 120 cm, 100 cm

229

(x Ϫ 2)2
4

O

Ϫ

Station x

(y Ϫ 3)2
4

ϭ1

Algebra 2

Chapter 8

39. about 47.32 ft

40. Hyperbolas and parabolas have different graphs and different reflective properties.
Answers should include the following. • Hyperbolas have two branches, two foci, and two vertices. Parabolas have only one branch, one focus, and one vertex.
Hyperbolas have asymptotes, but parabolas do not.
• Hyperbolas reflect rays directed at one focus toward the other focus.
Parabolas reflect parallel incoming rays toward the only focus.

41. C

42. B

43.

44. (22, 22), (Ϫ22, Ϫ22)

y xy ϭ 2

x

O

45.

46. The graph of xy ϭ Ϫ2 can be obtained by reflecting the graph of xy ϭ 2 over the x-axis or over the y-axis. The graph of xy ϭ Ϫ2 can also be obtained by rotating the graph of xy ϭ 2 by 90Њ.

y xy ϭ Ϫ2 x O

47.

(x Ϫ 5)2
16

ϩ

(y Ϫ 2)2
1

48.

ϭ1

230

(y Ϫ 1)2
16

ϩ

(x ϩ 3)2
9

ϭ1

Algebra 2

Chapter 8

49.

(x Ϫ 1)2
25

ϩ

(y Ϫ 4)2
9

ϭ1

50. (5, Ϫ1), 2 units y x

O

3
2

51. Ϫ4, Ϫ2
Ϫ7 0
S
53. C
5 20
55. about 5,330,000 subscribers per year

52. Ϫ7,

57. 2x ϩ 17y

58. 2, 3, Ϫ5

59. 1, Ϫ2, 9

60. Ϫ3, 1, 2

61. 5, 0, Ϫ2

62. 1, 0, 0

54. [13 Ϫ8 1]
56. Ϫ5, 4

63. 0, 1, 0
2. (4, Ϫ2); (4 Ϯ 2 22, Ϫ2); 6; 2

Chapter 8
Practice Quiz 2
Page 448
1.

(y Ϫ 1)2
81

ϩ

(x Ϫ 3)2
32

ϭ1

y

O

x

( y ϩ 2) 2
(x Ϫ 4)2 ϭ1 ϩ
1
9

231

Algebra 2

Chapter 8

3. (Ϫ1, 1); (Ϫ1, 1 Ϯ 211); 8;

4.

x2
9

Ϫ

y2
16

ϭ1

225 y x

O

5.

(x Ϫ 2)2
16

Ϫ

(y Ϫ 2)2
5

ϭ1

Lesson 8-6 Conic Sections
Pages 450–452
2x 2 ϩ 2y 2 Ϫ 1 ϭ 0

2. 2x 2 Ϫ 4x ϩ 7y ϩ 1 ϭ 0

3. The standard form of the equation is (x Ϫ 2)2 ϩ
(y ϩ 1)2 ϭ 0. This is an equation of a circle centered at (2, Ϫ1) with radius 0. In other words, (2, Ϫ1) is the only point that satisfies the equation. 4. y ϭ ax ϩ b Ϫ , parabola

5.

y2
16

Ϫ

x2
8

3 2
2
y

Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8

x

O

1 2
2

9
4

6. ax Ϫ b ϩ y 2 ϭ , circle

ϭ 1, hyperbola
8
6
4
2

5
4

y

y

O
2 4 6 8x

O

232

x

Algebra 2

Chapter 8

7.

(x ϩ 1)2
4

ϩ

(y Ϫ 3)2
1

8. parabola

ϭ 1,

ellipse y x

O

9. ellipse
11.

10
8
6
4
2
Ϫ2
Ϫ4
Ϫ6

10. hyperbola
12. x 2 ϩ y 2 ϭ 27, circle

y

8

y

4
O2 4 6 8 10 12 14

x

Ϫ8

Ϫ4

O

8x

4

Ϫ4
Ϫ8

13.

y2
4

ϩ

x2
2

1
8

14. y ϭ x 2, parabola

ϭ 1, ellipse

y

y

O

O

x

233

x

Algebra 2

Chapter 8

15.

x2
4

Ϫ

y2
1

16.

ϭ 1, hyperbola

(x Ϫ 1)2
36

Ϫ

(y Ϫ 4)2
4

ϭ 1,

hyperbola

y

12

y

8

x

O

4
O
Ϫ8

4

Ϫ4

8

12x

Ϫ4

1
9

18. x ϭ (y Ϫ 4)2 ϩ 4, parabola

17. y ϭ (x Ϫ 2)2 Ϫ 4, parabola y y

x

O

x

O

20. x 2 ϩ (y ϩ 3)2 ϭ 36, circle

19. (x ϩ 2)2 ϩ (y Ϫ 3)2 ϭ 9, circle y

y

4
O
Ϫ8

4

Ϫ4

8x

Ϫ4
Ϫ8
O

x

234

Algebra 2

Chapter 8

21.

(x ϩ 4)2
32

Ϫ

y2
32

22.

ϭ 1, hyperbola
8
6
4
2

(x Ϫ 1)2
9

ϩ

y

24.

y

(y ϩ 1)2
25

Ϫ
8
6
4
2

x2
4

ϩ

Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8

x

(y ϩ 1)2
3

x

O

23. x 2 ϩ (y Ϫ 4)2 ϭ 5, circle

25.

26.

ϭ 1, ellipse

(x ϩ 1)2
16

ϩ

x2
9

O
2 4 6 8x

(y Ϫ 1)2
4

ϭ 1, ellipse

y

x
O

ϭ 1, hyperbola

y

y

O

ϭ 1, ellipse

9
2

y

Ϫ12Ϫ10Ϫ8Ϫ6 Ϫ4Ϫ2 O 2 4 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8

O

y2

235

x

Algebra 2

Chapter 8

27. y ϭ Ϫ(x ϩ 4)2 Ϫ 7, parabola
Ϫ12

Ϫ8

y
O

Ϫ4

28.

(x Ϫ 2)2
5

Ϫ

(y ϩ 1)2
6

ϭ 1,

hyperbola

4x

y

Ϫ4
Ϫ8

x

O

Ϫ12
Ϫ16

29.

(x Ϫ 3)2
25

ϩ

(y Ϫ 1)2
9

30. parabolas and hyperbolas

ϭ 1, ellipse

y

O

x

32.

31. hyperbola

y

x

O

33. circle

34. hyperbola

35. parabola

36. ellipse

37. ellipse

38. circle

39. parabola

40. hyperbola

41. b

42. a

43. c

44. 2 intersecting lines

236

Algebra 2

Chapter 8

45. The plane should be vertical and contain the axis of the double cone.

46. If you point a flashlight at a flat surface, you can make different conic sections by varying the angle at which you point the flashlight.
Answers should include the following. • Point the flashlight directly at a ceiling or wall. The light from the flashlight is in the shape of a cone and the ceiling or wall acts as a plane perpendicular to the axis of the cone.
• Hold the flashlight close to a wall and point it directly vertically toward the ceiling.
A branch of a hyperbola will appear on the wall. In this case, the wall acts as a plane parallel to the axis of the cone.

47. D

48. C

49. 0 Ͻ e Ͻ 1, e Ͼ 1

50.

51.

(x Ϫ 3)2
9

Ϫ

(y ϩ 6)2
4

52. (3, Ϫ4); (3 Ϯ 25, Ϫ4); 6; 4

ϭ1

(y Ϫ 4)2
36

Ϫ

(x Ϫ 5)2
16

ϭ1

y x O

54. m12n

53. x12
55.

x7 y4 56. 196 beats per min

57. (2, 6)

58. (3, 2)

59. (0, 2)

237

Algebra 2

Chapter 8

Lesson 8-7

Pages 458–460
2. The vertex of the parabola is on the ellipse. The parabola opens toward the interior of the ellipse and is narrow enough to intersect the ellipse in two other points.
Thus, there are exactly three points of intersection.

1a. (Ϫ3, Ϫ4), (3, 4) y 4x Ϫ 3y ϭ 0
O

x

x 2 ϩ y 2 ϭ 25

1b. (Ϯ1, 4) y y ϭ 2x 2 ϩ 2

y ϭ 5 Ϫ x2
O

x

3. Sample answer: x 2 ϩ y 2 ϭ 40, y ϭ x 2 ϩ x

4. (Ϯ4, 5)

5. (Ϫ4, Ϫ3), (3, 4)

6. no solution

7. (1, Ϯ5), (Ϫ1, Ϯ5)

8.

14
12
10
8
6
4
2
Ϫ10Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6

238

y

O

2 4 6 8 10 x

Algebra 2

Chapter 8

9.

10. (40, 30)

y

O

x

13. (Ϫ1 ϩ 217, 1 ϩ 217),
(Ϫ1 Ϫ 217, 1 Ϫ 217)

3
2

12. a ,

15. ( 25, 25), (Ϫ25, Ϫ25)

11. (2, 4), (Ϫ1, 1)

9 b, 2

(Ϫ1, 2)
123
11
,Ϫ b
2
4

14. no solution
16. no solution

17. (5, 0), (Ϫ4, Ϯ6)

18. (0, 3), aϮ

19. (Ϯ8, 0)

20. (0, Ϯ5)

21. no solution

22. (4, Ϯ3), (Ϫ4, Ϯ3)

23. (Ϫ5, 5), (Ϫ5, 1), (3, 3)

24. (6, 3), (6, 1), (Ϫ4, 4), (Ϫ4, 0)

5
3

7
3

25. aϪ , Ϫ b, (1, 3)

26. (3, Ϯ4), (Ϫ3, Ϯ4)

27. 0.5 s

29. a

x2 y2 ϩ ϭ 1,
36
16 x2 (y Ϫ 2)2 ϩ ϭ
16
4 x2 y2 ϩ ϭ1
2
16

40 Ϫ 24 25 45 Ϫ 12 25 b ,
5
5

1,

30. (39.2, Ϯ4.4)

239

Algebra 2

Chapter 8

32.

31. No; the comet and Pluto may not be at either point of intersection at the same time. y

x

O

33.

34.

y

y

x

O

x

O

35.

36.

y

37.

O
Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8

x

O

y

2 4 6 8x

38. k Ͻ Ϫ3, Ϫ2 Ͻ k Ͻ 2, or k Ͼ 3

y

O

8
6
4
2

x

39. none

40. k ϭ Ϯ2 or k ϭ Ϯ3

41. none

42. Ϫ3 Ͻ k Ͻ Ϫ2 or 2 Ͻ k Ͻ 3

240

Algebra 2

Chapter 8

43. Systems of equations can be used to represent the locations and/or paths of objects on the screen.
Answers should include the following. • y ϭ 3x, x 2 ϩ y 2 ϭ 2500
• The y-intercept of the graph of the equation y ϭ 3x is 0, so the path of the spaceship contains the origin. • (Ϫ5 210, Ϫ15 210) or about (Ϫ15.81, Ϫ47.43)
45. B

44. A

46. Sample answer: y ϭ x 2, x ϭ (y Ϫ 2)2

x 2 ϩ y 2 ϭ 36,
(x ϩ 2)2
16

Ϫ

y2
4

ϭ1

49. Sample answer: x 2 ϩ y 2 ϭ 81, x2 4

ϩ

y2

100

x2
16

x 2 ϩ y 2 ϭ 100,

ϭ1

y2 x2 Ϫ
64
16

ϩ

x2
64

y2
4

ϩ

ϭ1

y2
16

ϭ 1,

ϭ1

52. (x ϩ 2)2 ϩ (y ϩ 1)2 ϭ 11, circle 51. impossible

y

O

241

x

Algebra 2

Chapter 8

53.

(y Ϫ 3)2
9

ϩ

x2
4

54. (0, Ϯ2); (0, Ϯ4); y ϭ Ϯ

ϭ 1, ellipse

y

23 x 3

y

6y 2 Ϫ 2x 2 ϭ 24

x

O
O

x

55. Ϫ7, 0

56. 0, 3

57. Ϫ7, 3

58. 7, Ϫ5

4
3

59. Ϫ

3
4

60. Ϫ ,

210
5

1
2

2i 23
3

61a. 40
61b. two real, irrational

62a. Ϫ48
62b. two imaginary

61c. Ϯ

62c. 1 Ϯ

63. 2 ϩ 9i
65.

8
5

64. 29 Ϫ 28i

1
5

66. about 1830 times

Ϫ i

67. 6

68. Ϫ2

69. Ϫ51

70. (5, 3, 7)

71. y ϭ 3x Ϫ 2

72. y ϭ Ϫ x Ϫ

5
3

242

4
3

Algebra 2

Chapter 8

PQ245-6457F-P09[243-266].qxd 7/31/02 9:54 AM Page 243 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-09:

Chapter 9 Rational Expressions and Equations
Lesson 9-1 Multiplying and Dividing Rational Expressions
Pages 476–478
4
6

1. Sample answer: ,

4(x ϩ 2)
6(x ϩ 2)

2. To multiply rational numbers or rational expressions, you multiply the numerators and multiply the denominators. To divide rational numbers or rational expressions, you multiply by the reciprocal of the divisor. In either case, you can reduce your answer by dividing the numerator and the denominator of the results by any common factors.
4.

3. Never; solving the equation using cross products leads to
15 ϭ 10, which is never true.

9m
4n4

5.

1 aϪb 6.

3y 2 yϩ4 7.

3c
20b

8.

5
12x

9.

6
5

10.

pϩ5 pϩ1 11. cd 2x

12.

2y(y Ϫ 2)
3(y ϩ 2)

13. D

14.

5c
2b

n2
7m

16. Ϫ3x 4y

15. Ϫ
17.

s
3

18.

5 tϩ1 19.

1
2

20.

yϩ2
3y Ϫ 1

21.

aϩ1
2a ϩ 1

22.

3x 2
2y

Glencoe/McGraw-Hill

243

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4bc
27a

23. Ϫ

24. Ϫf

25. Ϫ2p 2

26.

xz
8y

27.

b3 x 2y 2

28. 3

29.

4
3

30.

2
3

31. 1

32.

5(x Ϫ 3)
2(x ϩ 1)
3(r ϩ 4) rϩ3 33.

wϪ3 wϪ4 34.

35.

2(a ϩ 5)
(a Ϫ 2)(a ϩ 2)

36. Ϫ

3n m 37. Ϫ2p

38.

mϩn m2 ϩ n2

39.

2x ϩ y
2x Ϫ y

40. y ϩ 1

41.

4
3

42. d ϭ Ϫ2, Ϫ1 or 2

43. a ϭ Ϫb or b
45.

44.

6827 ϩ m
13,129 ϩ a

6827
13,129

46. 2x ϩ 1 units
1
aϪ2

47. (2x 2 ϩ x Ϫ 15)m 2

48.

49. A rational expression can be used to express the fraction of a nut mixture that is peanuts. Answers should include the following.
• The rational expression
8ϩx
is in simplest form

50. C

13 ϩ x

because the numerator and the denominator have no common factors.

244

Algebra 2

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8ϩx

13 ϩ x ϩ y could be used to represent the fraction that is peanuts if x pounds of peanuts and y pounds of cashews were added to the original mixture. 51. A

52. (Ϫ1, Ϯ4), (5, Ϯ2)

53. (Ϯ217, Ϯ222)

54. x ϭ

1
3

(y ϩ 3)2 ϩ 1; parabola

y
O

x x ϭ 1 (y ϩ 3)2 ϩ 1
3

55.

(x Ϫ 7)2 (y Ϫ 2)2
9
1

56. even; 2

ϭ 1;

hyperbola
8

y

4

x

O
4
Ϫ4

8

(x Ϫ 7)
9

2

12
2

Ϫ

(y Ϫ 2) ϭ1 1

Ϫ8

AA

57. odd; 3

C A BLACK

58. even; 0
1 1
6 3

59. Ϫ1, 4

60. Ϫ ,

61. 0, 5

62. 4.99 ϫ 102 s or about 8 min
19 s

63. л

64.
1
9

11
24

65. Ϫ1

3 19
,
2 16

66. Ϫ1
245

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4
15

11
16

67. 1

68. Ϫ

11
18

70.

69. Ϫ

Lesson 9-2

1
6

Adding and Subtracting Rational Epressions
Pages 481–484

1. Catalina; you need a common denominator, not a common numerator, to subtract two rational expressions. 2. Sample answer: d 2 Ϫ d, d ϩ 1

3a. Always; since a, b, and c are factors of abc, abc is always a common denominator of

4. 12x 2y 2

1
1
1 ϩ ϩ . a c b 3b. Sometimes; if a, b, and c have no common factors, then abc is the LCD of
1
1
1
ϩ ϩ . a c b 3c. Sometimes; if a and b have no common factors and c is a factor of ab, then ab is the
1
1
1
LCD of ϩ ϩ . a b

c

b

c

3d. Sometimes; if a and c are factors of b, then b is the
1
1
1
LCD of ϩ ϩ . a 3e. Always; since

1
1
1 ϩ ϩ ϭ a c b ac ab bc ϩ ϩ
, the sum abc abc abc bc ϩ ac ϩ ab
.
always abc is

5. 80ab3c
7.

6. x(x Ϫ 2)(x ϩ 2)

2 Ϫ x3 x 2y

8.
246

42a 2 ϩ 5b 2
90ab 2

Algebra 2

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37
42m

10.

5d ϩ 16
(d ϩ 2) 2

11.

3a Ϫ 10
(a Ϫ 5)(a ϩ 4)

12.

8
5

13.

13x 2 ϩ 4x Ϫ 9
2x (x Ϫ 1)(x ϩ 1)

9.

14. 70s 2t 2

units

15. 180x 2yz

16. 420a3b3c 3

17. 36p 3q 4

18. 4(w Ϫ 3)

19. x 2(x Ϫ y)(x ϩ y)

20. (2t ϩ 3)(t Ϫ 1)(t ϩ 1)

21. (n Ϫ 4)(n Ϫ 3)(n ϩ 2)

22.

6 ϩ 8b ab 23.

31
12v

24.

5 ϩ 7r r 25.

2x ϩ 15y
3y

26.

9x 2 Ϫ 2y 3
12x 2y

27.

25b Ϫ 7a3
5a 2b 2

28. Ϫ

29.

110w Ϫ 423
90w

30.

13 yϪ8 31.

aϩ3 aϪ4 32.

5m Ϫ 4
3(m ϩ 2)(m Ϫ 2)

33.

y (y Ϫ 9)
(y ϩ 3)(y Ϫ 3)

34.

7x ϩ 38
2(x Ϫ 7)(x ϩ 4)

35.

Ϫ8d ϩ 20
(d Ϫ 4)(d ϩ 4)(d Ϫ 2)

36.

Ϫ4h ϩ 15
(h Ϫ 4)(h Ϫ 5)2

37.

x2 Ϫ 6
(x ϩ 2)2(x ϩ 3)

38. 0

39.

2y 2 ϩ y Ϫ 4
(y Ϫ 1)(y Ϫ 2)

40.

1 bϩ1 42.

2s Ϫ 1
2s ϩ 1

aϩ7 aϩ2 44.

3x Ϫ 4
2x (x Ϫ 2)

45. 12 ohms

46.

24 h x

3
20q

41. Ϫ1
43.

247

Algebra 2

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47.
49.

24 h xϪ4
2md
(d Ϫ L)2(d ϩ L) 2
2md
2
(d Ϫ L2 ) 2

48.

or

1
1
, xϩ1 xϪ2

51. Subtraction of rational expressions can be used to determine the distance between the lens and the film if the focal length of the lens and the distance between the lens and the object are known. Answers should include the following.
• To subtract rational expressions, first find a common denominator.
Then, write each fraction as an equivalent fraction with the common denominator. Subtract the numerators and place the difference over the common denominator. If possible, reduce the answer. •

1
1
1 ϭ Ϫ q 10
60

48(x Ϫ 2) h x(x Ϫ 4)

52. B

could be used

to determine the distance between the lens and the film if the focal length of the lens is 10 cm and the distance between the lens and the object is 60 cm.
54.

53. C

248

4
15xyz 2

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55.

a(a ϩ 2) aϩ1 56.

y
8
x 2 ϩ y 2 ϭ 16
8x

O

Ϫ8

Ϫ8
9x 2 ϩ y 2 ϭ 81

57.

58. 2.5 ft

y
2

(y Ϫ 3) ϭ x ϩ 2

x

O

x2ϭ yϩ 4

59.

60.

y

y
15

6
2
Ϫ8

10
O

5

8x

Ϫ2

Ϫ10 Ϫ5 O
2

2

x Ϫ6 y ϭ 1
Ϫ 20

5

10x

Ϫ5

16

Ϫ10

y2 x2
Ϫ15 ϭ 1
Ϫ

49

61.

10
8
6
4
2

25

y

O
Ϫ2
2 4 6x

Ϫ6

Ϫ4
2

2

(x ϩ 2)
(y Ϫ 5) ϭ1 Ϫ Ϫ8
16
25

249

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Chapter 9
Practice Quiz 1
Page 484 tϩ2 tϪ3

2.

c
6b 2

3. Ϫ

y2
32

4.

7
2

5. (w ϩ 4)(3w ϩ 4)

6. x Ϫ 1

1.

7.

4a ϩ 1 aϩb 9.

n Ϫ 29
(n ϩ 6)(n Ϫ 1)

8.

Lesson 9-3

10.

1
4

Graphing Rational Functions
Pages 488–490

1. Sample answer: f(x) ϭ

6ax ϩ 20by
15a 2b3

2. Each of the graphs is a straight line passing through
(Ϫ5, 0) and (0, 5). However,

1
(x ϩ 5)(x Ϫ 2)

the graph of f (x) ϭ

(x Ϫ 1)(x ϩ 5) xϪ1 has a hole at (1, 6), and the graph of g(x ) ϭ x ϩ 5 does not have a hole.
3. x ϭ 2 and y ϭ 0 are asymptotes of the graph. The y-intercept is 0.5 and there is no x-intercept because y ϭ 0 is an asymptote.

4. asymptote: x ϭ 2

5. asymptote: x ϭ Ϫ5; hole: xϭ1 6.

f (x )

O

f (x ) ϭ

250

x x x ϩ1

Algebra 2

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7.

8.

f (x )
4

f (x )
10

2
6
O

f (x ) ϭ

8x

4

Ϫ2

Ϫ8

Ϫ4

2

6
(x Ϫ 2)(x ϩ 3)

Ϫ4 O

10.

f (x )

4
2
Ϫ8

O

Ϫ2

10

x

4
( x Ϫ 1)2

8x

4

f (x ) ϭ

x Ϫ5 x ϩ1

Ϫ4

O

C

11.

6

f (x ) f (x ) ϭ

Ϫ4

2

Ϫ4

Ϫ4

9.

x 2 Ϫ 25

f (x ) ϭ x Ϫ 5

C

12. 100 mg

f (x)

O

x

x f(x) ϭ

13.

x ϩ2 x2 Ϫ x Ϫ 6

14. y ϭ Ϫ12, C ϭ 1; 0; 0

C
10
6

y y ϩ 12

8

16 y

2
O

Ϫ16 Ϫ8

Ϫ4

15. y Ͼ 0 and 0 Ͻ C Ͻ 1

16. asymptotes: x ϭ 2, hole: xϭ3 17. asymptotes: x ϭ Ϫ4, x ϭ 2

18. asymptotes: x ϭ Ϫ4, hole: x ϭ Ϫ3

19. asymptotes: x ϭ Ϫ1, hole: xϭ5 20. hole: x ϭ 4

251

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22.

21. hole: x ϭ 1

f (x )
1

f (x ) ϭ x

x

O

23.

24.

f (x )

f (x ) f (x ) ϭ

1 xϩ2 x

O

x

O

3

f (x ) ϭ x

25.

26.

f (x )
6

f (x ) ϭ

Ϫ5 x ϩ1

2
Ϫ8

x
O

Ϫ4

f (x )

4

x

O

8

Ϫ4

f (x ) ϭ

x x Ϫ3

Ϫ8

27.
8

C

28.

f (x ) f (x ) ϭ

f (x )

C

f (x ) ϭ

5x x ϩ1

Ϫ3
( x Ϫ 2)2

x
O

4
Ϫ8

Ϫ4

O

4

8x

Ϫ4

29.

30.

f (x ) f (x ) ϭ

1
( x ϩ 3)2

f (x ) f (x ) ϭ

x ϩ4 x Ϫ1

6
2

Ϫ8

Ϫ4 O

4

8x

Ϫ4
O

x

Ϫ8

252

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31.

32.

f (x )

4
Ϫ8

x

O

f (x ) ϭ

f(x) ϭ

f(x)

x

4

Ϫ4 O

x 2 Ϫ 36 Ϫ4 x ϩ6

x Ϫ1 x Ϫ3

Ϫ8
Ϫ12

33.

34.

f (x) f(x) ϭ

f (x )

x2 Ϫ 1 x Ϫ1

O

x

O

f (x ) ϭ

35.

f (x ) f (x ) ϭ

O

x

Ϫ1
( x ϩ 2)( x Ϫ 3)

37.

38.

f (x ) f (x ) ϭ

x x2 Ϫ 1

x

O

f (x ) ϭ

3
( x Ϫ 1)( x ϩ 5)

36.

f (x )

x

x Ϫ1 x2 Ϫ 4

f (x ) f (x ) ϭ

O

x

O

6
(x Ϫ 6)2

253

x

Algebra 2

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39.

40.

f (x ) f (x ) ϭ

f (x )

1
(x ϩ 2)2

f (x ) ϭ

64 x 2 ϩ 16

x

O

x

O

41. The graph is bell-shaped with a horizontal asymptote at f(x) ϭ 0.

Ϫ64
64
ϭ Ϫa 2 b, x ϩ 16 x ϩ 16
Ϫ64
graph of f (x) ϭ 2 x ϩ 16

42. Since the 2

would be a reflection of the graph of f (x) ϭ the x-axis.
43.
20

4

Vf ϭ

m1 Ϫ 7 m1 ϩ 7

5

O
8 m1

Ϫ16 Ϫ8 Ϫ4

45. about Ϫ0.83 m/s

f(x) ϭ f (x) ϭ f (x) ϭ
47.

8

P (x ) ϭ
Ϫ12

Ϫ8

over

44. m1 ϭ Ϫ7; 7; Ϫ5

Vf

12

64 x ϩ 16
2

xϩ2
,
(x ϩ 2)(x Ϫ 3)
2(x ϩ 2)
,
(x ϩ 2)(x Ϫ 3)
5(x ϩ 2)
(x ϩ 2)(x Ϫ 3)

48. the part in the first quadrant

P (x )

6ϩx
10 ϩ x
4

Ϫ4 O

4x

Ϫ4
Ϫ8

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254

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49. It represents her original freethrow percentage of 60%.

50. y ϭ 1; This represents 100%, which she cannot achieve because she has already missed 4 free throws.

51. A rational function can be used to determine how much each person owes if the cost of the gift is known and the number of people sharing the cost is s. Answers should include the following.

c

52. A

100
50

sϭ0

O

Ϫ100 Ϫ50

150 s 50 100 s

Ϫ50 c ϭ 0
Ϫ100

• Only the portion in the first quadrant is significant in the real world because there cannot be a negative number of people nor a negative amount of money owed for the gift.

58. (Ϫ2, 0); 113
54.

55.

3x Ϫ 16
(x ϩ 3)(x Ϫ 2)

57. (6, 2); 5 y 3m ϩ 4 mϩn 56.

53. B

5(w Ϫ 2)
(w ϩ 3) 2

y

(x Ϫ 6)2 ϩ ( y Ϫ 2)2ϭ 25
O
O

Glencoe/McGraw-Hill

x

x x 2 ϩ y 2 ϩ 4x ϭ 9

255

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Ϫ7 Ϯ 3 213
2

59. \$65,892

60. Ϫ4 Ϯ 2i

61. Ϫ12, 10

62.

63. 4.5

64. 1.4

65. 20

66. 12

Lesson 9-4

Direct, Joint, and Inverse Variation
Pages 495–498

1a. inverse
1b. direct

2. Both are examples of direct variation. For y ϭ 5x, y increases as x increases. For y ϭ Ϫ5x, y decreases as x increases. 3. Sample answers: wages and hours worked, total cost and number of pounds of apples; distances traveled and amount of gas remaining in the tank, distance of an object and the size it appears 4. inverse; 20

5. direct; Ϫ0.5

6. joint;

7. 24

8. Ϫ45

9. Ϫ8

1
2

10. P ϭ 0.43d

11. 25.8 psi

12. about 150 ft

13.

14. direct; 1.5

p

Depth(ft) Pressure(psi)
0
0
1
0.43
2
0.86
3
1.29
4
1.72

Glencoe/McGraw-Hill

256

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P
P ϭ 0.43d
O

d

15. joint; 5

16. inverse; Ϫ18

17. direct; 3

18. inverse; 12

19. direct; Ϫ7

20. joint;

21. inverse; 2.5

22. V ϭ

23. V ϭ kt

24. directly; 2␲

25. 118.5 km

26. 60

27. 20

28. 216

29. 64

30. 25

31. 4

32. 1.25

33. 9.6

34. Ϫ12.6

35. 0.83

36. 2

37.

1
3

k p 1
4

1
6

38. 30 mph

39. 100.8 cm3

40. See students’ work.

41. m ϭ 20sd

42. joint

43. 1860 lb

44. / ϭ15md

45. joint

46. See students’ work.

47. I ϭ

k d2 48.

I

I ϭ 16
2
d

O

Glencoe/McGraw-Hill

257

d

Algebra 2

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1

0.02P1P2

49. The sound will be heard as
4
intensely.

50. 0.02; C ϭ

51. about 127,572 calls

52. about 601 mi

53. no; d

54. Sample answer: If the average student spends \$2.50 for lunch in the school cafeteria, write an equation to represent the amount s students will spend for lunch in d days. How much will 30 students spend in a week? a ϭ 2.50sd; \$375

0

d2

55. A direct variation can be used to determine the total cost when the cost per unit is known. Answers should include the following.
• Since the total cost T is the cost per unit u times the number of units n or
C ϭ un, the relationship is a direct variation. In this equation u is the constant of variation.
• Sample answer: The school store sells pencils for 20¢ each. John wants to buy
5 pencils. What is the total cost of the pencils? (\$1.00)

56. D

57. C

58. asymptote: x ϭ 1; hole x ϭ Ϫ1

59. asymptotes: x ϭ Ϫ4, x ϭ 3

60. hole: x ϭ Ϫ3 t 2 Ϫ 2t Ϫ 2
(t ϩ 2)(t Ϫ 2)

61.

x yϪx 62.

63.

m (m ϩ 1) mϩ5 64. 9.3 ϫ 107

65. 0.4; 1.2

66. 3; 7

3
5

67. Ϫ ; 3

68. C

69. A

70. S

258

Algebra 2

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71. P

72. A

73. C

Chapter 9
Practice Quiz 2
Page 498
1.

2.

f (x )

f (x )
O

f (x ) ϭ x Ϫ 1 xϪ4 x

x

O

f (x ) ϭ

3. 49

Ϫ2 x Ϫ 6x ϩ 9
2

4. 4.4

5. 112

Lesson 9-5 Classes of Functions
Pages 501–504

2. constant (y ϭ 1), direct variation (y ϭ 2x), identity (y ϭ x)

P

O

d

This graph is a rational function. It has an asymptote at x ϭ Ϫ1.
3. The equation is a greatest integer function. The graph looks like a series of steps.

4. greatest integer

5. inverse variation or rational

6. constant

259

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7. c

8. b

9. identity or direct variation

y

y

y ϭ Ϫx 2 ϩ 2

x

O

yϭx x O

12. A ϭ ␲r 2; quadratic; the graph is a parabola

11. absolute value y yϭ xϩ2

O

x

13. absolute value

14. square root

15. rational

16. direct variation

18. constant

19. b

20. e

21. g

22. a

23. constant

24. direct variation y y

y ϭ 2.5x x O

O

x

y ϭ Ϫ1.5

260

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25. square root

26. inverse variation or rational

y

y yϭ4 x

y ϭ ͙9x

x

O

x

O

AA

27. rational

C A BLACK

28. greatest integer

y

y

2 yϭx Ϫ1

xϪ1

y ϭ 3[x ]

x

O

x

O

29. absolute value

30. quadratic y y

y ϭ 2x
O

y ϭ 2x 2 x O

x

31. C ϭ 4.5 m

32. direct variation

33. a line slanting to the right and passing through the origin

34. similar to a parabola

261

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35.

36. The graph is similar to the graph of the greatest integer function because both graphs look like a series of steps. In the graph of the postage rates, the solid dots are on the right and the circles are on the left. However, in the greatest integer function, the circles are on the right and the solid dots are on the left.

y

Cost (cents)

160
120
80
40
0

x
2

4
6
Ounces

8

10

37a. absolute value
37c. greatest integer
37d. square root

38. A graph of the function that relates a person’s weight on
Earth with his or her weight on a different planet can be used to determine a person’s weight on the other planet by finding the point on the graph that corresponds with the weight on Earth and determining the value on the other planet’s axis. Answers should include the following.
• The graph comparing weight on Earth and Mars represents a direct variation function because it is a straight line passing through the origin and is neither horizontal nor vertical.
• The equation V ϭ 0.9E compares a person’s weight on Earth with his or her weight on Venus.
V

Venus

80
60
40
20
0

262

E
20

40 60
Earth

80

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39. C

40. D

41. 22

42.

f (x )
3
f (x ) ϭ x ϩ 2

x

O

43.

44.

f (x )

f (x )
2
f (x ) ϭ x Ϫ 5x ϩ 4

xϪ4

x

O

f (x ) ϭ (

8 x Ϫ 1)(x ϩ 3)

O

46. aϪ3 , 1b ; aϪ2 , 1b;

45. (8, Ϫ1); a8, Ϫ b ; x ϭ 8; yϭ 14
12
10
8
6
4
2
Ϫ2
Ϫ2

7
8
1
1
Ϫ1 ; up;
8
2

x

1
4

y ϭ 1; x

unit

1
4
1 ϭ Ϫ4 ;
4

right; 4 units

y

y

1

1

x ϭ 4 y2 Ϫ 2 y Ϫ 3
1(
y ϩ 1) ϭ (x Ϫ 8)2
2

O
2 4 6

10 12

O

x

x

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47.

48. c

(5, Ϫ4); a5 , Ϫ4b ; y ϭ Ϫ4;
3
4

1
4

Ϫ25
23 Ϫ54 d 66 Ϫ26
57

x ϭ 4 ; right; 3 units y O

x

3x Ϫ y 2 ϭ 8y ϩ 31

49. impossible

50. (7, Ϫ5)

51. a , 2b

52. (2, Ϫ2)

53. 1

54. 12

1
3

17
6

55. Ϫ

56. 60a3b 2c 2

57. 45x 3y 3

58. 15(d Ϫ 2)

59. 3(x Ϫ y)(x ϩ y)

60. (a Ϫ 3)(a ϩ 1)(a ϩ 2)

61. (t Ϫ 5)(t ϩ 6)(2t ϩ 1)

Lesson 9-6

Solving Rational Equations and Inequalities
Pages 509–511

1
5

ϩ

2 aϩ2 2. 2(x ϩ 4); Ϫ4

ϭ1

3. Jeff; when Dustin multiplied by 3a, he forgot to multiply the 2 by 3a.

4. 3

5. 2, 6

6.

7. Ϫ6, Ϫ2

8. Ϫ2 Ͻ c Ͻ 2

264

2
3

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2
9

1
6

10. 2 h

9. v Ͻ 0 or v Ͼ 1

4
3

11. 2

12. Ϫ

13. Ϫ6, 1

14. Ϫ3, 2

15. Ϫ1 Ͻ a Ͻ 0

16. Ϫ1 Ͻ m Ͻ 1

17. 11

18. 3

19. t Ͻ 0 or t Ͼ 3

20. 0 Ͻ b Ͻ 1

21. 0 Ͻ y Ͻ 2

22. p Ͻ 0 or p Ͼ 2

23. 14

24.

Ϫ3 Ϯ 3 22
2

1
2

3
2
1 Ϯ 2145
4

25. л

26. л

27. 7

28.

29.

30.

7
3

31. 32

32. 2 or 4

33. band, 80 members; chorale,
50 members

34. 4.8 cm/g

35. 24 cm

36. 15 km/h

37. 5 mL

38. 5

39. 6.15

40.

41. If something has a general fee and cost per unit, rational equations can be used to determine how many units a person must buy in order for the actual unit price to be a given number. Answers should include the following.

42. B

• To solve

500 ϩ 5x x b bc ϩ 1

ϭ 6,

multiply each side of the equation by x to eliminate the rational expression.

265

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Then subtract 5x from each side. Therefore, 500 ϭ x.
A person would need to make 500 minutes of long distance calls to make the actual unit price 6¢.
• Since the cost is 5¢ per minute plus \$5.00 per month, the actual cost per minute could never be 5¢ or less.

43. C

y

y ϭ 2x 2 ϩ 1 x O

46. direct variation

45. square root

y

y

O

y ϭ 0.8x

y ϭ 2͙x

51. 2137
O

x

x

47. 36

48. 33.75

49. 22130

50. 225

53. 5x 0 0 Յ x Յ 46

54. e b ` Ϫ1 Ͻ b Ͻ 2 f

52. 5x 0 x Ͻ Ϫ11 or x Ͼ 36
1
2

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Chapter 10 Esponential and Logarithmic Relations
Lesson 10-1 Exponential Functions
Pages 527–530
2a.
2b.
2c.
2d.

1. Sample answer: 0.8

quadratic exponential linear exponential 4. a

3. c

6. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}

5. b

y

y ϭ 3(4)x

O

7. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}

x

8. growth

y

( 1 )x
3

yϭ2

O

x

10. growth

9. decay

13. 2227 or 427
1 x
2

11. y ϭ 3a b

15. 3322 or 2722

16. Ϫ9

17. x Յ 0

18. 2

19. y ϭ 65,000(6.20)x

20. 22,890,495,000

12. y ϭ Ϫ18132 x
14. a4␲

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21. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}

22. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}

y

y

y ϭ 5(2)x y ϭ 2(3)x x O

x

O

23. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}

24. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}

y

y

(1)
3

yϭ4

)x

y ϭ 0.5(4

x

O

x

O

25. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͻ 0}

x

26. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͻ 0}

y

y

x
O

O

x

y ϭ Ϫ2.5(5)x yϭϪ ( 1 )x
5

27. growth

28. growth

29. decay

30. growth

31. decay

32. decay
1 x
4

33. y ϭ Ϫ2 a b

34. y ϭ 3(5)x

35. y ϭ 7(3)x

36. y ϭ Ϫ5 a b

37. y ϭ 0.2(4)x

38. y ϭ Ϫ0.3(2)x

1 x
3

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40. x115

41. 7412

42. y 2 13

43. n 2ϩ␲

44. 25␲

45. n ϭ 5

46.

47. 1

48. n Յ Ϫ2

39. 54 or 625

8
3

2
3

49. Ϫ

50. 0

51. n Ͻ 3

52.

53. Ϫ3

54. p Ն Ϫ2

55. 10

56. Ϫ3, 5

57. y ϭ 100(6.32)x

59. y ϭ 3.93(1.35)x

60. 9.67 million; 17.62 million;
32.12 million; These answers are in close agreement with the actual populations in those years.

61. 2144.97 million; 281.42 million; No, the growth rate has slowed considerably.
The population in 2000 was much smaller than the equation predicts it would be.

62. Exponential; the base, 1 ϩ , n is fixed, but the exponent, nt, is variable since the time t can vary.

63. A(t ) ϭ 1000(1.01)4t
65. s . 4x

64. \$2216.72

67. Sometimes; true when b Ͼ 1, but false when b Ͻ 1.

68. The number of teams y that could compete in a tournament with x rounds can be expressed as y ϭ 2x.
The 2 teams that make it to the final round got there as a result of winning games played with 2 other teams, for a total of 2 и 2 ϭ 22 or 4 games played in the previous rounds. Answers should include the following.

5
3

r

66. 1.5 three-year periods or 4.5 yr

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• Rewrite 128 as a power of
2, 27. Substitute 27 for y in the equation y ϭ 2x. Then, using the Property of
Equality for Exponents, x must be 7. Therefore, 128 teams would need to play
7 rounds of tournament play. • Sample answer: 52 would be an inappropriate number of teams to play in this type of tournament because 52 is not a power of 2.
69. A

70. 780.25

71.

72.

[Ϫ5, 5] scl: 1 by [Ϫ1, 9] scl: 1

[Ϫ5, 5] scl: 1 by [Ϫ1, 9] scl: 1

The graphs have the same shape. The graph of y ϭ 3xϩ1 is the graph of y ϭ 3x translated one unit to the left. The asymptote for the graph of y ϭ 3x and for y ϭ 3xϩ1 is the line y ϭ 0.
The graphs have the same domain, all real numbers, and range, y Ͼ 0. The y-intercept of the graph of y ϭ 3x is 1 and for the graph of y ϭ 3xϩ1 is 3.

The graphs have the same shape. The graph of y ϭ 2x ϩ 3 is the graph of y ϭ 2x translated three units up. The asymptote for the graph of y ϭ 2x is the line y ϭ 0 and for y ϭ 2x ϩ 3 is the line y ϭ 3. The graphs have the same domain, all real numbers, but the range of y ϭ 2x is y Ͼ 0 and the range of y ϭ 2x ϩ 3 is y Ͼ 3.
The y-intercept of the graph of y ϭ 2x is 1 and for the graph of y ϭ 2x ϩ 3 is 4.

270

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73.

74.

[Ϫ5, 5] scl: 1 by [Ϫ1, 9] scl: 1

[Ϫ5, 5] scl: 1 by [Ϫ3, 7] scl: 1

The graphs have the same shape. The graph of

The graphs have the same shape. The graph of yϭ 1 xϪ2 a b
5

1
4

x

1
4

x

y ϭ a b Ϫ1 is the graph of

is the graph of

x

y ϭ a b translated two units

y ϭ a b translated one

to the right. The asymptote

unit down. The asymptote

1
5

1
5

1
5

xϪ2

for y ϭ a b

for the graph of y ϭ a b is the line y ϭ 0 and

is the line

for the graph of y ϭ

y ϭ 0. The graphs have the same domain, all real numbers, and range, y Ͼ 0.
The y-intercept of the graph of y ϭ

1 x a b
5

graph of y ϭ

1 x a b
4

Ϫ1

is the line y ϭ Ϫ1. The graphs have the same domain, all real numbers, but the range of y ϭ

is 1 and for the
1 xϪ2 a b
5

x

1
4

x

for the graph of y ϭ a b and

1 x a b
4
1
4

x

is y Ͼ 0 and of y ϭ a b Ϫ 1

is 25.

is y Ͼ Ϫ1. The y-intercept
1
4

x

of the graph of y ϭ a b is 1 and for the graph of yϭ 75. For h Ͼ 0, the graph of y ϭ 2x is translated 0 h 0 units to the right. For h Ͻ 0, the graph of y ϭ 2x is translated
|h| units to the left. For k Ͼ
0, the graph of y ϭ 2x is translated 0k 0 units up. For k Ͻ 0, the graph of y ϭ 2x is translated 0k 0 units down.

1 x a b
4

Ϫ 1 is 0.

76. 1, 15

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13
3

77. 1, 6

78. Ϫ , 3

79. 0 Ͻ x Ͻ 3 or x Ͼ 6

80. square root

81. greatest integer

82. constant y yϭ8

O

83. B

1 0
R
0 1

85.

3
1
B
51 11

x

84. does not exist
Ϫ6
R
Ϫ5

86. about 23.94 cm
88. g [h(x)] ϭ x 2 ϩ 6x ϩ 9; h [g(x)] ϭ x 2 ϩ 3

87. g [h(x)] ϭ 2x Ϫ 6; h [g(x)] ϭ 2x Ϫ 11
89. g [h(x)] ϭ Ϫ2x Ϫ 2; h [g(x)] ϭ Ϫ2x ϩ 11

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Logarithms and Logarithmic Functions
Pages 535–538

1. Sample answer: x ϭ 5y and y ϭ log5 x

2. They are inverses.

3. Scott; the value of a logarithmic equation, 9, is the exponent of the equivalent exponential equation, and the base of the logarithmic expression, 3, is the base of the exponential equation. Thus x ϭ 39 or
19,683.

4. log5 625 ϭ 4

5. log7

1
49

6. 34 ϭ 81

ϭ Ϫ2

1

8. 4

7. 362 ϭ 6
9. Ϫ3

10. 21

11. Ϫ1

12. 27

13. 1000

14.

15.

1
,
2

1
2

ՅxՅ5

16. x Ͼ 6

1

17. 3

18. 1013

19. 107.5

20. 105.5 or about 316,228 times

21. log8 512 ϭ 3

22. log3 27 ϭ 3

23. log5

1
125

1
24. log 3 9 ϭ Ϫ2

ϭ Ϫ3

25. log100 10 ϭ

1
2

26. log2401 7 ϭ
28. 132 ϭ 169

27. 53 ϭ 125
29. 4Ϫ1 ϭ

1
4

1
4

1

30. 100Ϫ2 ϭ
1 Ϫ2
5

31. 83 ϭ 4

32. a b

33. 4

1
10

34. 2

2

273

ϭ 25

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1
2

36.

5
2

37. Ϫ5

38. Ϫ4

39. 7

40. 45

41. n Ϫ 5

42. 3x ϩ 2

43. Ϫ3

44. 2x

45. 1018.8

46. 1010.67

47. 81

48. c Ͼ 256

49. 0 Ͻ y Յ 8

50. 125

51. 7

52. 0 Ͻ p Ͻ 1

53. x Ն 24

54. Ϯ3

55. 4

56. 11

57. 2

58. 25

59. 5

60. y Ն 3

61. a Ͼ 3

62. Ϯ8
?

63. log5 25 ϭ 2 log5 5
2 ?

1

log5 5 ϭ 2 log5 5
?

2 ϭ 2(1)

2 ϭ 2

64.

Original equation ?

log16 2 ؒ log2 16 ϭ 1

2

25 ϭ 5 and 5 ϭ 51

1

?

log16 164 ؒ log2 24 ϭ 1

Inverse
Prop. of
Exp. and
Logarithms
Simplify.

1
(4)
4

Original equation 1

2 ϭ 164 and 16 ϭ 24

?

ϭ 1 Inverse Prop. of Exp. and
Logarithms

1 ϭ 1

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66a.

65.
?

log7 [log3 (log2 8)] ϭ 0

Original equation ?

log7 [log3 (log2 23)] ϭ 0

8 ϭ 23

?

log7 (log3 3) ϭ 0

log7 (log3 31) ϭ 0
?

x

y ϭ log 1 x
2

66b. The graphs are reflections of each other over the line y ϭ x.

3 ϭ 31

log7 1 ϭ 0

( 1)
2

x

O

Inverse
Prop. of
Exp. and
Logarithms

?

?

y

Inverse
Prop. of
Exp. and
Logarithms

log7 70 ϭ 0
0 ϭ 0

1 ϭ 70
Inverse
Prop. of
Exp. and
Logarithms

68. 103 or 1000 times as great

67a. y y ϭ log2(x ϩ 2)

y ϭ log2x ϩ 3 y ϭ log2(x Ϫ 1)
O

x

y ϭ log2x Ϫ 4

67b. The graph of y ϭ log2 x ϩ 3 is the graph of y ϭ log2 x translated 3 units up. The graph of y ϭ log2 x Ϫ 4 is the graph of y ϭ log2 x translated 4 units down. The graph of log2 (x Ϫ 1) is the graph of y ϭ log2 x translated 1 unit to the right.
The graph of log2 (x ϩ 2) is the graph of y ϭ log2 x translated 2 units to the left.
69. 101.4 or about 25 times as great

70. 101.7 or about 50 times
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71. 2 and 3; Sample answer: 5 is between 22 and 23.

72. All powers of 1 are 1, so the inverse of y ϭ 1x is not a function. 73. A logarithmic scale illustrates that values next to each other vary by a factor of 10. Answers should include the following.
• Pin drop: 1 ϫ 100;
Whisper: 1 ϫ 102; Normal conversation: 1 ϫ 106;
Kitchen noise: 1 ϫ 1010;
Jet engine: 1 ϫ 1012

74. B

Pin drop Whisper
(4 feet)

Normal conversation 2 ϫ 10 11

0

4 ϫ 10 11

Kitchen noise 6 ϫ 10 11

Jet engine 8 ϫ 10 11 1 ϫ 10 12

• On the scale shown above, the sound of a pin drop and the sound of normal conversation appear not to differ by much at all, when in fact they do differ in terms of the loudness we perceive.
The first scale shows this difference more clearly.
76. x 216

75. D
5 Ϯ 273
4

77. b12

79. Ϫ3,
81.
83.

78. л
7
3

14
5

80. Ϯ
82.

6x Ϫ 58
(x Ϫ 3)(x ϩ 3)(x ϩ 7)

43
30y

84. \$2400, CD; \$1600, savings

85. x10

86. y 24

87. 8a6b 3

88. an6

89.

x3 y 2z 3

90. 1

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Chapter 10
Practice Quiz 1
Page–538
1. growth

2. y ϭ 2(4)x

3. log4 4096 ϭ 6

4. 92 ϭ 27

3

5.

4
3

6. 15

7.

3
5

8. n Յ Ϫ1

9. x Ͼ 26

10. 3

Lesson 10-3

Properties of Logarithms
Pages 544–546

1. properties of exponents

2log3 x ϩ log3 5; log3 5x 2

3. Umeko; Clemente incorrectly applied the product and quotient properties of logarithms. log7 6 ϩ log7 3 ϭ log7 (6 и 3) or log7 18 Product Property of

4. 1.1402

Logarithms

log7 18 Ϫ log7 2 ϭ log7 (18 Ϭ 2) or log7 9
Quotient Prop. of
Logarithms

5. 2.6310

6. Ϫ0.3690

7. 6

8. 2

9. 3

10. 4

11. pH ϭ 6.1 ϩ log10

B
C

12. 20:1

13. 1.3652

14. 1.2921

15. Ϫ0.2519

16. 0.2519

17. 2.4307

18. 2.1133

19. Ϫ0.4307

20. 0.0655

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21. 2

22. 3

23. 4

24. Ϫ2

25. 14

26. 12

27. 2

28. Ϯ4

29. л

30. 6

31. 10

32. 12

33.

x3
4

34.

35. False; log2 (22 ϩ 23) ϭ log2 12, log2 22 ϩ log2 23 ϭ 2 ϩ 3 or 5, and log2 12 Z 5, since 25 Z 12.

1
1x
2

Ϫ 12

36.
?
n logb x ϩ m logb x ϭ (n ϩ m)logb x
?
logb x n ϩ logb x m ϭ (n ϩ m)logb x

Power Prop. of Logarithms
?

logb (x n ؒ x m) ϭ (n ϩ m)logb x
Product Prop. of Logarithms
?

logb (x nϩm) ϭ (n ϩ m)logb x
Product of Powers Prop.

(n ϩ m)logb x ϭ (n ϩ m)logb 
Power Prop. of Logarithms

C2
C1

37. 2

38. E ϭ 1.4 log

39. about 0.4214 kilocalories per gram 40. about 0.8429 kilocalories per gram 41. 3

42. 3

43. About 95 decibels;
L ϭ 10 log10 R, where L is the loudness of the sound in decibels and R is the relative intensity of the sound. Since the crowd increased by a factor of 3, we assume that the intensity also increases by a factor of 3. Thus, we need to find the loudness of 3R.

44. 5

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L ϭ 10 log10 3R
L ϭ 10 (log10 3 ϩ log10 R )
L ϭ 10 log103 ϩ 10 log10 R
L ഠ 10(0.4771) ϩ 90
L ഠ 4.771 ϩ 90 or about 95
45. 7.5

47. Let b x ϭ m and b y ϭ n. Then logb m ϭ x and logb n ϭ y.

48. Since logarithms are exponents, the properties of logarithms are similar to the properties of exponents. The
Product Property states that to multiply two powers that have the same base, add the exponents. Similarly, the logarithm of a product is the sum of the logarithms of its factors. The Quotient Property states that to divide two powers that have the same base, subtract their exponents. Similarly, the logarithm of a quotient is the difference of the logarithms of the numerator and the denominator. The Power
Property states that to find the power of a power, multiply the exponents. Similarly, the logarithm of a power is the product of the logarithm and the exponent. Answers should include the following.
• Quotient Property:

bx by ϭ

b xϪy ϭ

m n m n Quotient Prop.

m
Prop. of n Equality for
Logarithmic
Equations m logb Inverse n Prop. of

logb b x Ϫy ϭ logb

xϪyϭ

Exp. and
Logarithms

logb m Ϫ logb n ϭ logb

m
Replace x n with log m b and y with logbn. log2

32 a b
8

ϭ log2

25 a 3b
2

ϭ log2 2(5Ϫ3) ϭ 5 Ϫ 3 or 2

279

Replace 32 with 25 and 8 with 23.
Quotient of
Powers
Inverse Prop. of Exp. and
Logarithms

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log2 32 Ϫ log2 8 ϭ log2 25 Ϫ log2 23

Replace 32 with
25 and 8 with 23.

ϭ 5 Ϫ 3 or 2

Inverse Prop. of
Exp. and Logarithms

So, log2 a

32 b 8

ϭ log2 32 Ϫ log2 8

Power Property: log3 94 ϭ log3 (32)4

Replace 9 with 32.

ϭ log3 3(2ؒ4) ϭ 2 ؒ 4 or 8

Power of a Power
Inverse Prop. of Exp. and
Logarithms

4 log3 9 ϭ (log3 9) ؒ 4

Comm (ϫ)

ϭ (log3 32) ؒ 4 ϭ 2 ؒ 4 or 8

Replace 9 with 32.
Inverse Prop. of Exp. and
Logarithms

So, log3 94 ϭ 4 log3 9.
• The Product of Powers
Property and Product
Property of Logarithms both involve the addition of exponents, since logarithms are exponents.
50. Let b x ϭ m, then logb m ϭ x.
(b x)p ϭ m p b xp ϭ m p Product of Powers logb b xp ϭ logb m p Prop. of Equality

49. A

for Logarithmic
Equations

xp ϭ logb m p

Inverse Prop. of Exp. and
Logarithms

plogb m ϭ logb m p

Replace x with logb m.

51. 4

52. Ϫ3

53. 2x

54. 6

55. Ϫ8

56. d Ͻ 4

57. odd; 3

58. even; 4

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59.

3b a 60. 2

61.

5
3x

62. 3.06 s
64. 5

63. 1
65. x Ͼ

3
4

5
3

66. Ϫ Ͻ x Ͻ 2

Lesson 10-4 Common Logarithms
Pages 549–551
1. 10; common logarithms

5x ϭ 2; x ഠ 0.4307

3. A calculator is not programmed to find base
2 logarithms.

4. 0.6021

5. 1.3617

6. Ϫ0.3010

8. {n 0 n Ͼ 0.4907}

7. 1.7325
9. 4.9824

10. Ϯ1.1615

12. {p 0 p Յ 4.8188}

11. 11.5665
13.

log 5
;
log 7

0.8271

14.

15.

log 9
;
log 2

3.1699

16. at most 0.00003 mole per liter log 42
;
log 3

3.4022

17. 0.6990

18. 1.0792

19. 0.8573

20. 0.3617

21. Ϫ0.0969

22. Ϫ1.5229

23. 11

24. 2.2

25. 2.1

26. 3.5

27. {x 0 x Ն 2.0860}

28. 2.4550

29. {a 0 a Ͻ 1.1590}

30. 0.5537

31. 0.4341

32. 4.8362

33. 4.7820

34. 8.0086

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36. Ϯ2.6281

35. Ϯ1.1909

37. {n 0 n Ͼ Ϫ1.0178}

38. 1.0890

40. {p 0 p Յ 1.9803}

39. 3.7162

42. 4.7095

41. 0.5873

44. 2.7674

43. Ϫ7.6377
45.

log 13 log 2

47.

log 3 log 7

49.

2log 1.6 log 4

46.

Ϸ 0.6781

log 8 log 3

50.

Ϸ 0.5646

log 20 log 5

48.

Ϸ 3.7004

0.5 log 5 log 6

Ϸ 1.8614

Ϸ 1.8928
Ϸ 0.4491

51. between 0.000000001 and
0.000001 mole per liter

52. 8

53. Sirius

54. Sirius: 1.45, Vega: 0.58

55. Vega

56a. 3;

1
3
3 2
56b. ;
2 3

56c. conjecture: loga b ϭ proof: ?

1 log b a

?

1 logb a
1
logb a

loga b ϭ logb b logb a
1
logb a

ϭ

Original statement
Change of Base
Formula



Inverse Prop. of
Exponents and
Logarithms

58. about 11.64 yr or 11 yr, 8 mo

57. about 3.75 yr or 3 yr 9 mo

ϭ

1
;
logba

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59. Comparisons between substances of different acidities are more easily distinguished on a logarithmic scale. Answers should include the following.
Tomatoes: 6.3 ϫ 10Ϫ5 mole per liter
Milk: 3.98 ϫ 10Ϫ7 mole per liter Eggs: 1.58 ϫ 10Ϫ8 mole per liter
• Those measurements correspond to pH measurements of 5 and 4, indicating a weak acid and a stronger acid. On the logarithmic scale we can see the difference in these acids, whereas on a normal scale, these hydrogen ion concentrations would appear nearly the same. For someone who has to watch the acidity of the foods they eat, this could be the difference between an enjoyable meal and heartburn.

60. A

61. C

62. 1.4248

63. 1.6938

64. 1.8416

65. 64

66. z Յ

67. 62

68. Ϫ22

69. (d ϩ 2)(3d Ϫ 4)

70. (7p ϩ 3)(6q Ϫ 5)

71. prime

72. 2x ϭ 3

73. 32 ϭ x

74. 53 ϭ 125

75. log5 45 ϭ x

76. log7 x ϭ 3

1
64

77. logb x ϭ y

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Base e and Natural Logarithms
Pages 557–559

1. the number e

2. e x ϭ 8

3. Elsu; Colby tried to write each side as a power of 10.
Since the base of the natural logarithmic function is e, he should have written each side as a power of e;
10ln 4x Z 4x.

4. 403.4288

5. 0.0334

6. 0.1823

7. Ϫ2.3026

8. x ϭ ln 4

9. e0 ϭ 1

10. 3

11. 5x

12. x Ͼ 3.4012

13. 1.0986

14. Ϫ0.8047

15. 0 Ͻ x Ͻ 403.4288

16. 2.4630

17. Ϯ90.0171

18. h ϭ Ϫ26200 ln

19. about 15,066 ft

20. 54.5982

21. 148.4132

22. 0.3012

23. 1.6487

24. 1.0986

25. 2.3026

26. 1.6901

27. Ϫ3.5066

28. \$183.21

29. about 49.5 cm

30. Ϫx ϭ ln 5

31. 2 ϭ ln 6x

32. e1 ϭ e

33. e x ϭ 5.2

34. 0.2

35. y

36. Ϫ4x

37. 45

38. 0.2877

39. Ϫ0.6931

40. x Ͻ 1.5041

41. x Ͼ 0.4700

42. 0.2747

43. 0.5973

44. x Ն 0.6438

45. x Ն Ϫ0.9730

46. 27.2991

284

P
101.3

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47. 49.4711

48. 1.7183

49. 14.3891

50. 232.9197

51. 45.0086

52. 2, 6

53. 1

54. about 19.8 yr

55. t ϭ

100 ln 2 r 56. 100 ln 2 Ϸ 70

57. t ϭ

110 r 58. about 7.33 billion

59. about 55 yr

60. about 32 students

61. about 21 min

62. always; log x log y

?

ϭ

In x
In y

Original statement log x logx ? log e ϭ log y logy log e

log x log y

log x log y

63. The number e is used in the formula for continuously compounded interest,
A ϭ Pe rt. Although no banks actually pay interest compounded continually, the equation is so accurate in computing the amount of money for quarterly compounding or daily compounding, that it is often used for this purpose.

ϭ

?

log x log e

ϭ

?

log x log y

Change of
Base Formula

ؒ

log e log y

Multiply log x by the log e reciprocal of log y
.
log e
Simplify.

64. B

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• If you know the annual interest rate r and the principal P, the value of the account after t years is calculated by multiplying P times e raised to the r times t power. Use a calculator to find the value of e rt.
• If you know the value A you wish the account to achieve, the principal P, and the annual interest rate r, the time t needed to achieve this value is found by first taking the natural logarithm of A minus the natural logarithm of P.
Then, divide this quantity by r.
66.

67.

log 0.047 log 6

ϭ Ϫ1.7065

log 68 log 4

ϭ 3.0437

68.

65. 1946, 1981, 2015; It takes between 34 and 35 years for the population to double.

log 23 log 50

ϭ 0.8015

69. 5

70. 4

71. inverse; 4

72. joint; 1

73. direct; Ϫ7

74. x ϭ

75. 3.32

76. 1.54

77. 1.43

78. 323.49

79. 13.43

80. 9.32

286

1 2 y 20

Ϫ5

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Chapter 10
Practice Quiz 2
Page 559
1.

log 5
;
log 4

2. e 2 ϭ 3x

1.1610

3. 3

4. x Ͼ 5.3219

5. 1.3863

Lesson 10-6

Exponential Growth and Decay
Pages 563–565

1. y ϭ a(1 ϩ r)t, where r Ͼ 0 represents exponential growth, and r Ͻ 0 represents exponential decay.

2. Take the common logarithm of each side, use the Power
Property to write log (1 ϩ r)t as t log(1 ϩ r), and then divide each side by the quantity log(1 ϩ r).

3. Sample answer: money in a bank 4. Decay; the exponent is negative. 5. about 33.5 watts

6. about 402 days

7. y ϭ 212,000e0.025t

8. about 349,529 people

9. C

10. \$1600

11. at most \$108,484.93

12. about 8.1 days

13. No; the bone is only about
21,000 years old, and dinosaurs died out
63,000,000 years ago.

14. more than 44,000 years ago

16. y ϭ ae0.0347t

17. \$12,565 billion

19. after the year 2182

20. 4.7%

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21. Never; theoretically, the amount left will always be half of the previous amount.

22. Answers should include the following. • Find the absolute value of the difference between the price of the car for two consecutive years. Then divide this difference by the price of the car for the earlier year.
• Find 1 minus the rate of decrease in the value of the car as a decimal.
Raise this value to the number of years it has been since the car was purchased, and then multiply by the original value of the car.

23. about 19.5 yr

24. D

25. ln y ϭ 3

26. ln 29 ϭ 4n Ϫ 2

27. 4x 2 ϭ e8

28. 1.5323

29. p Ͼ 3.3219

30. 9

31.

0.5 (0.08 p)
6

33.

p
150

ϩ

0.5 (0.08 p)
4

32.

p
60

34. hyperbola

35. ellipse

36. parabola

37. circle

38. 2.06 ϫ 108

39. 8 ϫ 107

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Chapter 11 Sequences and Series
Lesson 11-1 Arithmetic Sequences
Pages 580–582
1. The differences between the terms are not constant.

2. 95

3. Sample answer: 1, Ϫ4, Ϫ9,
Ϫ14, . . .

4. 24, 28, 32, 36

5. Ϫ3, Ϫ5, Ϫ7, Ϫ9

6. 5, 8, 11, 14, 17

7. 14, 12, 10, 8, 6

8. 43
10. 79

9. Ϫ112
11. 15

12. an ϭ 11n Ϫ 37

13. 56, 68, 80

14. \$12,000

15. 30, 37, 44, 51

16. 10, 3, Ϫ4, Ϫ11

17. 6, 10, 14, 18

18. 1, 4, 7, 10

19.

7
,
3

3,

11 13
,
3 3

20.

12
,
5

8 6
5 5

2, ,

21. 5.5, 5.1, 4.7, 4.3

22. 8.8, 11.3, 13.8, 16.3

23. 2, 15, 28, 41, 54

24. 41, 46, 51, 56, 61

25. 6, 2, Ϫ2, Ϫ6, Ϫ10

26. 12, 9, 6, 3, 0

27.

4
,
3

2 1
3 3

28.

1, , , 0

5
,
8

1,

11 7 17
, ,
8 4 8

29. 28

30. Ϫ49

31. 94

32. Ϫ175

33. 335

34. 340

35.

26
3

25
2

36. Ϫ

37. 27

38. Ϫ47

39. 61

40. 173

41. 37.5 in.

42. 304 ft

43. 30th

44. 19th

45. 82nd

46. an ϭ 9n Ϫ 2

47. an ϭ Ϫ7n ϩ 25

48. an ϭ Ϫ2n Ϫ 1

Glencoe/McGraw-Hill

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49. 13, 17, 21

50. pn ϭ 4n Ϫ 3

51. Yes; it corresponds to n ϭ 100.

52. 70, 85, 100

53. 4, Ϫ2

54. Ϫ5, Ϫ2, 1, 4

55. 7, 11, 15, 19, 23

56. z ϭ 2y Ϫ x

57. Arithmetic sequences can be used to model the numbers of shingles in the rows on a section of roof. Answers should include the following.
• One additional shingle is needed in each successive row.
• One method is to successively add 1 to the terms of the sequence: a8 ϭ
9 ϩ 1 or 10, a9 ϭ 10 ϩ 1 or 11, a10 ϭ 11 ϩ 1 or 12, a11 ϭ 12 ϩ 1 or 13, a12 ϭ
13 ϩ 1 or 14, a13 ϭ
14 ϩ 1 or 15, a14 ϭ 15 ϩ 1 or 16, a15 ϭ 16 ϩ 1 or 17.
Another method is to use the formula for the nth term: a15 ϭ 3 ϩ (15 Ϫ 1)1 or 17.

58. B

59. B

61. Ϫ0.4055

62. 0.4621

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63. 146.4132

64. 15

65. 2, 5, 8, 11

66. 5, 4, 3, 2

67. 11, 15, 19, 23, 27

Lesson 11-2 Arithmetic Series
Pages 586–587
1. In a series, the terms are added. In a sequence, they are not.

2. Sample answer: 0 ϩ 1 ϩ 2 ϩ
3ϩ4

3. Sample answer: a (3n ϩ 4)

4. 1300

5. 230

6. 1932

7. 552

8. 800

4

nϭ1

9. 260

10. 63

11. 95

12. 11, 20, 29

13. Ϫ6, 0, 6

14. 28

15. 344

16. 663

17. 1501

18. 2646

19. Ϫ9

20. Ϫ88

21. 104

22. 182

23. Ϫ714

24. 225

25. 14

26. Ϫ

27. 10 rows

28. 8 days

29. 721

30. 735

31. 162

32. Ϫ204

33. 108

34. Ϫ35

35. Ϫ195

36. 510

37. 315,150

38. 24,300

39. 1,001,000

40. 166,833

41. 17, 26, 35

42. Ϫ13, Ϫ8, Ϫ3

245
6

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43. Ϫ12, Ϫ9, Ϫ6

44. 13, 18, 23

45. 265 ft

46. True; for any series, 2a1 ϩ
2a2 ϩ 2a3 ϩ p ϩ 2an ϭ
2(a1 ϩ a2 ϩ a3 ϩ p ϩ an).

47. False; for example, 7 ϩ 10 ϩ
13 ϩ 16 ϭ 46, but 7 ϩ 10 ϩ
13 ϩ 16 ϩ 19 ϩ 22 ϩ 25 ϩ
28 ϭ 140.

48. Arithmetic series can be used to find the seating capacity of an amphitheater.
Answers should include the following. • The sequence represents the numbers of seats in the rows. The sum of the first n terms of the series is the seating capacity of the first n rows.
• One method is to write out the terms and add them:
18 ϩ 22 ϩ 26 ϩ 30 ϩ 34 ϩ
38 ϩ 42 ϩ 46 ϩ 50 ϩ 54 ϭ
360. Another method is to use the formula n Sn ϭ [2a1 ϩ (n Ϫ 1)d ]:
S10

2
10
ϭ
2

[2(18) ϩ

(10 Ϫ 1)4] or 360.
49. C

50. C

51. 5555

52. 3649

53. 6683

54. 111

60. 23

3 Ϯ 289
2

56. about 3.82 days

55. Ϫ135

62. 2 22

61. 26 221

16
3

9
2

58. Ϫ

57. Ϫ
59.

64. Ϫ54

63. 16
65.

2
27

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Lesson 11-3 Geometric Sequences
Pages 590–592
1a. Geometric; the terms have a common ratio of Ϫ2.
1b. Arithmetic; the terms have a common difference of Ϫ3.

3. Marika; Lori divided in the wrong order when finding r.

4. 67.5, 101.25

5. 2, Ϫ4

6. Ϫ2, Ϫ6, Ϫ18, Ϫ54, Ϫ162

7.

2 4 8
,
3 9 27

1, , ,

15
64

p

8. 56
10. an ϭ 4 и 2nϪ1

9. Ϫ4
11. 3, 9

12. A

13. 15, 5

14. 192, 256

15. 54, 81

16. 48, 32

17.

20 40
,
27 81

18.

125 625
,
24 48

19. Ϫ2.16, 2.592

20. Ϫ21.875, 54.6875

21. 2, Ϫ6, 18, Ϫ54, 162

22. 1, 4, 16, 64, 256

23. 243, 81, 27, 9, 3

24. 576, Ϫ288, 144, Ϫ72, 36

25.

3
16

26. 2592

27. 729

28. 1024

29. 243

30.

31. 1

32. 192

33. 78,125

34. 2

35. Ϫ8748

36.

37. 655.36 lb

38. \$46,794.34

39. an ϭ

nϪ1

5
72

1
4

1
36 a b
3

nϪ1

40. an ϭ 64 a b

42. an ϭ 4(Ϫ3)n Ϫ1

41. an ϭ Ϫ2(Ϫ5)n Ϫ1

1
4

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43. Ϯ18, 36, Ϯ72

44. Ϯ12, 36, Ϯ108

45. 16, 8, 4, 2

46. 6, 12, 24, 48

47. 8 days

48. 5 mg

49. False; the sequence 1, 4, 9,
16, p, for example, is neither arithmetic nor geometric. 50. False, the sequence 1, 1, 1,
1, p, for example, is arithmetic (d ϭ 0) and geometric (r ϭ 1).

51. The heights of the bounces of a ball and the heights from which a bouncing ball falls each form geometric sequences. Answers should include the following.
• 3, 1.8, 1.08, 0.648, 0.3888
• The common ratios are the same, but the first terms are different. The sequence of heights from which the ball falls is the sequence of heights of the bounces with the term 3 inserted at the beginning.

52. A

53. C

54. 632.5

55. 203

56. 19, 23

57. Ϫ12, Ϫ16, Ϫ20

58. 5 22 ϩ 3 210 units

59. 127

60.

61.

63
32

61
81

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Chapter 11
Practice Quiz 1
Page 592
11
2

1. 46

2.

3. 187

4. 816

5. 1

Lesson 11-4 Geometric Series
Pages 596–598
2. The polynomial is a geometric series with first term 1, common ratio x, and
4 terms. The sum is

4ϩ2ϩ1ϩ

1
2

1(1 Ϫ x 4) x4 Ϫ 1 ϭ .
1Ϫx
xϪ1

4. 732

3. Sample answer: The first term is a1 ϭ 2. Divide the second term by the first to find that the common ratio is r ϭ 6. Therefore, the nth term of the series is given by
2 ؒ 6nϪ1. There are five terms, so the series can be written as a 2 ؒ 6nϪ1.
5

nϭ1

5. 39,063

6. 81,915

7. 165

8.

9. 129

10.

11.

1093
9

1330
9
31
4

12. 3

13. 3

14. 93 in. or 7 ft 9 in.

15. 728

16. 765

17. 1111

18. 300

295

Algebra 2

Chapter 11

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19. 244

20. 1,328,600

21. 2101

22. 1441

23.

728
3

24.

215
4

25. 1040.984

26. 7.96875

27. 6564

28. Ϫ118,096

29. 1,747,625

30. \$10,737,418.23

31. 3641

32. 206,668

33.

182
9

5461
16

34. Ϫ

35. 2555
37.

36. Ϫ364

387
4

38.

58,975
256

39. 3,145,725

40. 86,093,440

41. 243

42. 1024

43. 2

44. 6

45. 80

46. 8

47. about 7.13 in.

48. If the first term and common ratio of a geometric series are integers, then all the terms of the series are integers. Therefore, the sum of the series is an integer.

49. If the number of people that each person sends the joke to is constant, then the total number of people who have seen the joke is the sum of a geometric series. Answers should include the following.
• The common ratio would change from 3 to 4.
• Increase the number of days the joke circulates so that it is inconvenient to find and add all the terms of the series.

50. A

296

Algebra 2

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51. C

52. Ϫ1,048,575

53. 3.99987793

54. 6.24999936
9

1 3
4 2

27

81

55. Ϯ , , Ϯ9

56. Ϫ3, Ϫ , Ϫ , Ϫ
2
4
8

57. 232

58. 192

59. Drive-In Movie Screens

60. Sample answer using
(1, 826) and (3, 750): y ϭ Ϫ38x ϩ 864

Screens

1000
900
800
700
600
0

0

1 2 3 4 5
Years Since 1995

6

61. Sample answer: 294

62. 2

63. 2

64.

65.

2
3

1
4

66. Ϫ2

67. 0.6

Lesson 11-5

Infinite Geometric Series
Pages 602–604

1 a b a 2 nϭ1 ϱ

n

2. 0.999999 . . . can be written as the infinite geometric
9
9
9
ϩ ϩ ϩp. series 10

100

1000

The first term of this series is
9
and the common ratio is
10
9
1
10
, so the sum is
1 or 1.
10
1 Ϫ 10

297

Algebra 2

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3. Beth; the common ratio for the infinite geometric series
4
3

4. 108

4
3

is Ϫ . Since ` Ϫ ` Ն 1, the series does not have a sum a and the formula S ϭ 1
1Ϫr
does not apply.
5. does not exist
7.

6. does not exist

3
4

8.
10.

73
99

5
9

12.

9. 100
11.

30
7

175
999

13. 96 cm

14. 14

15. does not exist

16. 7.5

17. 45

18. 64

19. Ϫ16

20. does not exist

21.

54
5

22. 3

23. does not exist

24. 1

25. 1

26. 7.5

27.

2
3

28. 144

29.

3
2

30. 6

33. 40 ϩ 20 22 ϩ 20 ϩ p

32. 30 ft

31. 2

34. 80 ϩ 4022 or about
136.6 cm

35. 900 ft

36. 27, 18, 12

37. 75, 30, 12

38. 24, 16 , 11 , 7

1
5

1
2

7
25

64
125

40.

39. Ϫ8, Ϫ3 , Ϫ1 , Ϫ

298

11
32

409
512

7
9

Algebra 2

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41.

1
9

42.

4
11

43.

82
99

44.

82
333

45.

427
999

46.

5
11

47.

229
990

48.

S ϭ a1 ϩ a1r ϩ a1r 2 ϩ a1r 3 ϩ p

(Ϫ)rS ϭ a1r ϩ a1r 2 ϩ a1r 3 ϩ a1r 4 ϩ p
S Ϫ rS ϭ a1 ϩ 0 ϩ 0 ϩ0 ϩ0 ϩ p
S(1 Ϫ r )ϭ a1 a1 Sϭ
1Ϫr

49. The total distance that a ball bounces, both up and down, can be found by adding the sums of two infinite geometric series. Answers should include the following.
• an ϭ a1 ؒ r nϪ1, Sn ϭ a1(1 Ϫ r n)
,
1Ϫr

or S ϭ

50. D

a1
1Ϫr

• The total distance the ball falls is given by the infinite geometric series 3 ϩ
3(0.6) ϩ 3(0.6)2 ϩ p . The sum of this series is
3
or 7.5. The total
1 Ϫ 0.6

distance the ball bounces up is given by the infinite geometric series 3(0.6) ϩ
3(0.6)2 ϩ 3(0.6)3 ϩ p .
The sum of this series is
3(0.6)
1 Ϫ 0.6

or 4.5. Thus, the total distance the ball travels is 7.5 ϩ 4.5 or 12 feet.
52. Ϫ182

51. C
53.

8744
81

54. 32.768%
56. Ϫ

55. 3

299

3
2
Algebra 2

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58.

59.

Ϫx ϩ 7
(x Ϫ 3)(x ϩ 1)

61. (x Ϫ 2)2 ϩ (y Ϫ 4)2 ϭ 36
1
2

63. Ϫ ,

Ϫ2a ϩ 5b
2
a b

60.

57. x Ն 5

3x ϩ 7
(x ϩ 4)(x ϩ 2)

62. (x Ϫ 3)2 ϩ (y ϩ 1)2 ϭ 32

3 7
,
2 2

1
2

1
3

64. Ϫ , Ϫ , 0,

1
2

65. x 2 Ϫ 36 ϭ 0

66. x 2 ϩ 9x ϩ 14 ϭ 0

67. x 2 Ϫ 10x ϩ 24 ϭ 0

68. about Ϫ180,724 visitors per year 69. The number of visitors was decreasing. 70. 2

71. 3

72. 2

1
2

74. 4

73.

75. Ϫ4

Lesson 11-6

Recursion and Special Sequences
Pages 608–610

1. an ϭ anϪ1 ϩ d; an ϭ r ؒ anϪ1

2. Sample answer: an ϭ 2anϪ1 ϩ anϪ2

3. Sometimes; if f(x) ϭ x 2 and x1 ϭ 2, then x2 ϭ 22 or 4, so x2 x1. But, if x1 ϭ 1, then x2 ϭ 1, so x2 ϭ x1.

4. 12, 9, 6, 3, 0

5. Ϫ3, Ϫ2, 0, 3, 7

6. 0, Ϫ4, 4, Ϫ12, 20

7. 1, 2, 5, 14, 41

8. 5, 11, 29

9. 1, 3, Ϫ1

10. 3, 11, 123

11. bn ϭ 1.05bnϪ1 Ϫ 10

12. \$1172.41

13. Ϫ6, Ϫ3, 0, 3, 6

14. 13, 18, 23, 28, 33

15. 2, 1, Ϫ1, Ϫ4, Ϫ8

16. 6, 10, 15, 21, 28

17. 9, 14, 24, 44, 84

18. 4, 6, 12, 30, 84

19. Ϫ1, 5, 4, 9, 13

20. 4, Ϫ3, 5, Ϫ1, 9

300

Algebra 2

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21.

7 7 7 7 7
, , , ,
2 4 6 8 10

22.

3 3 15 25 425
, ,
,
,
4 2 4 2
8

23. 67

24. Ϫ2.1

25. 1, 1, 2, 3, 5, . . .

26. the Fibonacci sequence

27. \$99,921.21,
\$99,762.21,
\$99,601.29,
\$99,438.44,

28. 1, 3, 6, 10, 15

\$99,841.95,
\$99,681.99,
\$99,520.11,
\$99,356.28

29. tn ϭ tnϪ1 ϩ n

30. 20,100

31. 16, 142, 1276

32. 5, 17, 65

33. Ϫ7, Ϫ16, Ϫ43

34. Ϫ4, Ϫ19, Ϫ94

35. Ϫ3, 13, 333

36. Ϫ1, Ϫ1, Ϫ1

37.

5 37 1445
,
,
2 2
2

38.

4 10 76
,
,
3 3 3

39. \$75.77

40. No; according to the first two iterates, f(4) ϭ 4. According to the second and third iterates, f(4) ϭ 7. Since f(x) is a function, it cannot have two values when x ϭ 4.

41. Under certain conditions, the
Fibonacci sequence can be used to model the number of shoots on a plant. Answers should include the following.
• The 13th term of the sequence is 233, so there are 233 shoots on the plant during the 13th month.
• The Fibonacci sequence is not arithmetic because the differences (0, 1, 1, 2, . . .) of the terms are not constant. The Fibonacci sequence is not geometric because the ratios
3
Q1, 2, , . . .R of the terms

42. D

2

are not constant.

301

Algebra 2

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44. 27

43. C
1
6

46.

12
5

47. Ϫ5208

48.

1093
243

49. 3x ϩ 7 units

50. 120

51. 5040

52. 6

53. 20

54. 126

45.

55. 210

Lesson 11-7 The Binomial Theorem
Pages 615–617
1. 1, 8, 28, 56, 70, 56, 28, 8, 1

2. n

3. Sample answer: (5x ϩ y)4

4. 40,320

5. 17,160

6. 66

7. p5 ϩ 5p4q ϩ 10p3q 2 ϩ
10p 2q 3 ϩ 5pq 4 ϩ q 5

8. t 6 ϩ 12t 5 ϩ 60t 4 ϩ

160t 3 ϩ 240t 2 ϩ 192t ϩ 64

9. x 4 Ϫ 12x 3y ϩ 54x 2y 2 Ϫ
108xy 3 ϩ 81y 4

10. 56a5b3

11. 1,088,640a6b4

12. 10

13. 362,880

14. 6,227,020,800

15. 72

16. 210

17. 495

18. 2002

19. a 3 Ϫ 3a 2b ϩ 3ab 2 Ϫ b 3

20. m4 ϩ 4m 3n ϩ 6m 2n 2 ϩ
4mn3 ϩ n4

21. r 8 ϩ 8r 7s ϩ 28r 6s 2 ϩ
56r 5s 3 ϩ 70r 4s 4 ϩ
56r 3s 5 ϩ 28r 2s 6 ϩ 8rs 7 ϩ s 8

22. m 5 Ϫ 5m 4a ϩ 10m 3a 2 Ϫ
10m 2a 3 ϩ 5ma 4 Ϫ a 5

23. x 5 ϩ 15x 4 ϩ 90x 3 ϩ
270x 2 ϩ 405x ϩ 243

24. a4 Ϫ 8a3 ϩ 24a2 Ϫ 32a ϩ 16

302

Algebra 2

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25. 16b4 Ϫ 32b3x ϩ
24b 2x 2 Ϫ 8bx 3 ϩ x 4

26. 64a6 ϩ 192a5b ϩ 240a4b 2 ϩ
160a3b3 ϩ 60a2b4 ϩ
12ab5 ϩ b6
28. 81x 4 ϩ 216x 3y ϩ
216x 2y 2 ϩ 96xy 3 ϩ 16y 4

27. 243x 5 Ϫ 810x 4y ϩ 1080x 3y 2
Ϫ 720x 2y 3 ϩ 240xy 4 Ϫ 32y 5
29.

a5
32

ϩ

5a4
8

30. 243 ϩ 135m ϩ 30m 2 ϩ

ϩ 5a3 ϩ

10m 3
3

20a2 ϩ 40a ϩ 32

ϩ

5m 4
27

ϩ

31. 27x 3 ϩ 54x 2 ϩ 36x ϩ 8 cm3

32. 1, 4, 6, 4, 1

33. 45

34. Ϫ126x 4y 5

35. 924x 6y 6

36. 280x 4

37. 5670a4

38. 1,088,640a6b4

39. 145,152x 6y 3

40.

35 4 x 27

42.

12!
7!5!

m5
243

63
8

41. Ϫ x 5

and

12!
6!6!

represent the 6th

and 7th entries in the row for n ϭ 12 in Pascal’s triangle.
13!
represents the seventh
7!6!

entry in the row for n ϭ 13.
13!
12!
Since
is below and 7!6!
7!5!
12! in Pascal’s triangle,
6!6!
12!
12!
13! ϩ ϭ
.
7!5!
6!6!
7!6!

43. The coefficients in a binomial expansion give the numbers of sequences of births resulting in given numbers of boys and girls. Answers should include the following.
• (b ϩ g)5 ϭ b5 ϩ 5b4g ϩ
10b 3g 2 ϩ 10b 2g 3 ϩ
5bg 4 ϩ g 5;

44. D

303

Algebra 2

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There is one sequence of births with all 5 boys, five sequences with 4 boys and
1 girl, ten sequences with
3 boys and 2 girls, ten sequences with 2 boys and
3 girls, five sequences with
1 boy and 4 girls, and one sequence with all 5 girls.
• The number of sequences of births that have exactly k girls in a family of n children is the coefficient of bnϪkgk in the expansion of (b ϩ g)n. According to the Binomial Theorem, this n! . coefficient is
(n Ϫ k)!k!

45. C

46. 7, 5, 3, 1, Ϫ1

47. 3, 5, 9, 17, 33

48. 125 cm

49.

log 5
;
log 2

2.3219

50.

51.

log 8
;
log 5

1.2920

52. asymptotes: x ϭ Ϫ2, x ϭ Ϫ3

1
;
log 3

2.0959

53. asymptotes: x ϭ Ϫ4, x ϭ 1

54. hole: x ϭ Ϫ3

55. hyperbola

56. parabola

57. yes

58. no

59. True;

61. True;

1(1 ϩ 1)
2

ϭ

12(1 ϩ 1)2
4

1(2)
2

ϭ

60. False;

or 1.

1(4)
4

2(3)
2

(1 ϩ 1)(2 ؒ 1 ϩ 1)
2

ϭ

or 3.

62. True; 31 Ϫ 1 ϭ 2, which is even. or 1.

304

Algebra 2

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Chapter 11
Practice Quiz 2
Page 617
1. 1,328,600

2. Ϫ364

3. 24

4.

5. 1, 5, 13, 29, 61

6. 2, 4, 8, 14, 22

7. 5, Ϫ13, 41

8. 243x 5 ϩ 405x 4y ϩ 270x 3y 2 ϩ
90x 2y 3 ϩ 15xy 4 ϩ y 5

9. a6 ϩ 12a5 ϩ 60a4 ϩ
160a3 ϩ 240a 2 ϩ 192a ϩ 64

Lesson 11–8

25
4

10. 4032a5b4

Proof and Mathematical Induction
Pages 619–621

1. Sample answers: formulas for the sums of powers of the first n positive integers and statements that expressions involving exponents of n are divisible by certain numbers

2. Mathematical induction is used to show that a statement is true. A counter example is used to show that a statement is false.

3. Sample answer: 3n Ϫ 1

4. Step 1: When n ϭ 1, the left side of the given equation is
1(1 ϩ 1)

1. The right side is or 2
1, so the equation is true for n ϭ 1.
Step 2: Assume 1 ϩ 2 ϩ
3ϩ p ϩkϭ

k(k ϩ 1)
2

for

some positive integer k.
Step 3: 1 ϩ 2 ϩ 3 ϩ p ϩ k ϩ 1k ϩ 12 ϭ ϭ ϭ ©Glencoe/McGraw-Hill

305

k(k ϩ 1) ϩ (k ϩ 1)
2
k(k ϩ 1) ϩ 2(k ϩ 1)
2
(k ϩ 1) ϩ (k ϩ 2)
2

Algebra 2

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The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore, 1 ϩ 2 ϩ 3 ϩ p ϩ nϭ n(n ϩ 1)
2

for all

positive integers n.
5. Step 1: When n ϭ 1, the left side of the given equation is
1
.
2
1
,
2

The right side is 1 Ϫ

1
2

6. Step 1: 41 Ϫ 1 ϭ 3, which is divisible by 3. The statement is true for n ϭ 1.
Step 2: Assume that 4k Ϫ 1 is divisible by 3 for some positive integer k. This means that 4k Ϫ 1 ϭ 3r for some whole number r.
Step 3: 4k Ϫ 1 ϭ 3r
4k ϭ 3r ϩ 1
4kϩ1 ϭ 12r ϩ 4
4kϩ1 Ϫ 1 ϭ 12r ϩ 3
4kϩ1 Ϫ 1 ϭ 314r ϩ 12
Since r is a whole number,
4r ϩ 1 is a whole number.
Thus, 4kϩ1 Ϫ 1 is divisible by
3, so the statement is true for n ϭ k ϩ 1. Therefore,
4n Ϫ 1 is divisible by 3 for all positive integers n.

or

so the equation is true for

n ϭ 1.
1
1 ϩ 2ϩ
2
2
1
1
1
ϩ p ϩ k ϭ 1Ϫ k for some
3
2
2
2

Step 2: Assume

positive integer k.
Step 3:

1
1
ϩ 2ϩ
2
2
1
1
1
ϩ p ϩ k ϩ kϩ1
3
2
2
2
1
1 ϭ 1 Ϫ k ϩ kϩ1
2
2
2
1 ϭ 1 Ϫ kϩ1 ϩ kϩ1
2
2
1
ϭ 1 Ϫ kϩ1
2

The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore,
ϭ

1
2n

1
1
1
1
ϩ 2ϩ 3ϩ pϩ n
2
2
2
2

for all positive

integers n.

306

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7. Step 1: 51 ϩ 3 ϭ 8, which is divisible by 4. The statement is true for n ϭ 1.
Step 2: Assume that 5k ϩ 3 is divisible by 4 for some positive integer k. This means that 5k ϩ 3 ϭ 4r for some positive integer r.
Step 3: 5k ϩ 3 ϭ 4r
5k ϭ 4r Ϫ 3
5kϩ1 ϭ 20r Ϫ 15
5kϩ1 ϩ 3 ϭ 20r Ϫ 12
5kϩ1 ϩ 3 ϭ 415r Ϫ 32
Since r is a positive integer,
5r Ϫ 3 is a positive integer.
Thus, 5kϩ1 ϩ 3 is divisible by
4, so the statement is true for n ϭ k ϩ 1.
Therefore, 5n ϩ 3 is divisible by 4 for all positive integers n.

8. Sample answer: n ϭ 2

9. Sample answer: n ϭ 3

10. Step 1: After the first guest has arrived, no handshakes have taken place.

1(1 Ϫ 1)
2

ϭ 0,

so the formula is correct for n ϭ 1.
Step 2: Assume that after k guests have arrived, a total of k(k Ϫ 1)
2

handshakes have

take place, for some positive integer k.
Step 3: When the (k ϩ 1)st guest arrives, he or she shakes hands with the k guests already there, so the total number of handshakes that have then taken place is k(k Ϫ 1)
2

307

ϩ k.

Algebra 2

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k(k Ϫ 1) k(k Ϫ 1) ϩ 2k ϩkϭ 2
2
k [(k Ϫ 1) ϩ 2] ϭ 2 k (k ϩ 1)
(k ϩ 1)k or ϭ
2
2

The last expression is the formula to be proved, where n ϭ k ϩ 1. Thus, the formula is true for n ϭ k ϩ 1.
Therefore, the total number of handshakes is

for

all positive integers n.
12. Step 1: When n ϭ 1, the left side of the given equation is

11. Step 1: When n ϭ 1, the left side of the given equation is
1. The right side is 1[2(1) Ϫ 1] or 1, so the equation is true for n ϭ 1.
Step 2: Assume 1 ϩ 5 ϩ 9 ϩ p ϩ (4k Ϫ 3) ϭ k (2k Ϫ 1) for some positive integer k.
Step 3: 1 ϩ 5 ϩ 9 ϩ p ϩ
(4k Ϫ 3) ϩ [4(k ϩ 1) Ϫ 3] ϭ k (2k Ϫ 1) ϩ
[4(k ϩ 1) Ϫ 3] ϭ 2k 2 Ϫ k ϩ 4k ϩ 4 Ϫ 3 ϭ 2k 2 ϩ 3k ϩ 1 ϭ (k ϩ 1)(2k ϩ 1) ϭ (k ϩ 1)[2(k ϩ 1) Ϫ 1]
The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore, 1 ϩ 5 ϩ 9 ϩ p ϩ
(4n Ϫ 3) ϭ n(2n Ϫ 1) for all positive integers n.

n(n Ϫ 1)
2

1[3(1) ϩ 1]

2. The right side is
2
or 2, so the equation is true for n ϭ 1.
Step 2: Assume 2 ϩ 5 ϩ
8 ϩ p ϩ (3k Ϫ 1) ϭ

k(3k ϩ 1)
2

for some positive integer k.
Step 3: 2 ϩ 5 ϩ 8 ϩ p ϩ
(3k Ϫ 1) ϩ [3(k ϩ 1) Ϫ 1] k(3k ϩ 1) ϩ [3(k ϩ 1) Ϫ 1]
2
k(3k ϩ 1) ϩ 2[3(k ϩ 1) Ϫ 1] ϭ 2
2
3k ϩ k ϩ 6k ϩ 6 Ϫ 2 ϭ 2

ϭ

3k 2 ϩ 7k ϩ 4
2
(k ϩ 1)(3k ϩ 4) ϭ 2
(k ϩ 1)[3(k ϩ 1) ϩ 1] ϭ 2

ϭ

308

Algebra 2

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The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore, 2 ϩ 5 ϩ 8 ϩ p ϩ n(3n ϩ 1)

(3n Ϫ 1) ϭ for all
2
positive integers n.
14. Step 1: When n ϭ 1, the left side of the given equation is
12 or 1. The right side is

13. Step 1: When n ϭ 1, the left side of the given equation is
13 or 1. The right side is
12(1 ϩ 1)2
4

1[2(1) Ϫ 1][2(1) ϩ 1]
3

or 1, so the equation is true for n ϭ 1.
Step 2: Assume 13 ϩ 23 ϩ
33 ϩ p ϩ k 3 ϭ

k 2 1k ϩ 12 2
4

the equation is true for n ϭ 1.
Step 2: Assume
12 ϩ 32 ϩ 52 ϩ p ϩ(2k Ϫ 1)2 ϭ

for

some positive integer k.
Step 3: 13 ϩ 2 3 ϩ 33 ϩ p ϩ k 3 ϩ (k ϩ 1)3 ϭ ϭ ϭ ϭ ϭ ϭ

k (2k Ϫ 1)(2k ϩ 1)
3

for some

positive integer k.
Step 3:
12 ϩ 32 ϩ 52 ϩ p ϩ
(2k Ϫ 1)2 ϩ [2(k ϩ 1) Ϫ 1]2

k 2(k ϩ 1)2 ϩ (k ϩ 1)3
4
k 2(k ϩ 1)2 ϩ 4(k ϩ 1)3
4
2 2
(k ϩ 1) [k ϩ 4(k ϩ 1)]
4
2 2
(k ϩ 1) (k ϩ 4k ϩ 4)
4
2
(k ϩ 1) (k ϩ 2)2
4
2
(k ϩ 1) [(k ϩ 1) ϩ 1]2
4

ϭ

k(2k Ϫ 1)(2k ϩ 1) ϩ 3

[2(k ϩ 1) Ϫ 1]2 ϭ ϭ ϭ The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.

or 1, so

ϭ ϭ ϭ

309

k(2k Ϫ 1)(2k ϩ 1) ϩ 3(2k ϩ 1)2
3
(2k ϩ 1)[k (2k Ϫ 1) ϩ 3(2k ϩ 1)]
3
2
(2k ϩ 1)(2k Ϫ k ϩ 6k ϩ 3)
3
2
(2k ϩ 1)(2k ϩ 5k ϩ 3)
3
(2k ϩ 1)(k ϩ 1)(2k ϩ 3)
3
(k ϩ1)[2(k ϩ1) Ϫ 1][2(k ϩ 1) ϩ1]
3

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The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1. Therefore,
12 ϩ 32 ϩ 52 ϩ p ϩ

Therefore, 13 ϩ 23 ϩ 33 ϩ p ϩ
2

2

n (n ϩ 1)
4

n3 ϭ

for all

positive integers n.

(2n Ϫ 1)2 ϭ

n(2n Ϫ 1)(2n ϩ 1)
3

for all positive integers n.
16. Step 1: When n ϭ 1, the left side of the given equation is

15. Step 1: When n ϭ 1, the left side of the given equation is
1
1
. The right side is a1
2
3
1
or , so the equation is
3

p ϩ

ϭ

1 a1 2

3
3
1
Ϫ kb for
3

true

or , so the equation is true

1
3k

ϩ

1
3

ϩ

1
32

ϩ

1
33

The right side is a1 Ϫ b
1
4

for n ϭ 1.
1
1
Step 2: Assume ϩ 2 ϩ

3

1
43

some

ϩp ϩ

Step 3:

ϩ p ϩ

1
4k

1
3k ϩ 1

ϩ

1
4

ϭ ϭ ϭ ϭ ϩ

1
42

ϩ

1
43

ϩ p ϩ

1 k ϩ1

4

1
1
1 kb ϩ k ϩ 1
3
4
4
1
1
1
Ϫ
k ϩ k ϩ1
3
3ؒ4
4
4k ϩ 1 Ϫ 4 ϩ 3
3 ؒ 4k ϩ 1
4k ϩ 1 Ϫ 1
3 ؒ 4k ϩ 1
1 4k ϩ 1 Ϫ 1 a b
3
4k ϩ 1

1
1
1 a2 Ϫ kb ϩ k ϩ 1
2
3
3
1
1
1
Ϫ
k ϩ k ϩ1
2
2ؒ3
3
3k ϩ 1 Ϫ 3 ϩ 2
2 ؒ 3k ϩ 1
3k ϩ 1 Ϫ 1
2 ؒ 3k ϩ 1
1 3k ϩ 1 Ϫ 1 a b
2
3k ϩ 1

ϭ a1 Ϫ

1
2

ϭ

4
4
1
1
1 ϭ a1 Ϫ kb k 4
3
4

for some positive integer k.

positive integer k.
Step 3:

1
4

1
.
4

1
3

for n ϭ 1.
1
1
1
Step 2: Assume ϩ 2 ϩ 2 ϩ
1
3k

1
3

Ϫ b

ϭ a1 Ϫ

ϭ a1 Ϫ

ϭ ϭ ϭ ϭ 1
3

1

b
3k ϩ 1

310

1

k ϩ 1b

4

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The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
1
1
1
1
Therefore, ϩ 2 ϩ 3 ϩ p ϩ n

The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore,
ϭ

1 a1 2

Ϫ

1
1
1
1
ϩ 2ϩ 3ϩpϩ n
3
3
3
3
1
b for all positive
3n

4

1
3

ϭ a1 Ϫ

1 b 4n

4

4

4

for all positive

integers n.

integers n.
17. Step 1: 81 Ϫ 1 ϭ 7, which is divisible by 7. The statement is true for n ϭ 1.
Step 2: Assume that 8k Ϫ 1 is divisible by 7 for some positive integer k. This means that 8k Ϫ 1 ϭ 7r for some whole number r.
Step 3: 8k Ϫ 1 ϭ 7r
8k ϭ 7r ϩ 1
8k ϩ1 ϭ 56r ϩ 8
8kϩ1 Ϫ 1 ϭ 56r ϩ 7
8k ϩ1 Ϫ 1 ϭ 7(8r ϩ 1)
Since r is a whole number,
8r ϩ 1 is a whole number.
Thus, 8kϩ1 Ϫ 1 is divisible by
7, so the statement is true for n ϭ k ϩ 1.
Therefore, 8n Ϫ 1 is divisible by 7 for all positive integers n.

18. Step 1: 91 Ϫ 1 ϭ 8, which is divisible by 8. The statement is true for n ϭ 1.
Step 2: Assume that 9k Ϫ 1 is divisible by 8 for some positive integer k. This means that 9k Ϫ 1 ϭ 8r for some whole number r.
Step 3: 9k Ϫ 1 ϭ 8r
9k ϭ 8r ϩ 1
9kϩ1 ϭ 72r ϩ 9
9kϩ1 Ϫ 1 ϭ 72r ϩ 8
9kϩ1 Ϫ 1 ϭ 8(9r ϩ 1)
Since r is a whole number,
9r ϩ 1 is a whole number.
Thus, 9k ϩ1 Ϫ 1 is divisible by 8, so the statement is true for n ϭ k ϩ 1.
Therefore, 9n Ϫ 1 is divisible by 8 for all positive integers n.

19. Step 1: 121 ϩ 10 ϭ 22, which is divisible by 11. The statement is true for n ϭ 1.
Step 2: Assume that 12k ϩ 10 is divisible by 11 for some positive integer k. This means that 12k ϩ 10 ϭ 11r for some positive integer r.

20. Step 1: 131 ϩ 11 ϭ 24, which is divisible by 12. The statement is true for n ϭ 1.
Step 2: Assume that 13k ϩ11 is divisible by 12 for some positive integer k. This means that 13k ϩ 11 ϭ 12r for some positive integer r.

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Step 3:
12k ϩ 10 ϭ 11r
12k ϭ 11r Ϫ 10
12kϩ1 ϭ 132r Ϫ 120
12kϩ1 ϩ 10 ϭ 132r Ϫ 110
12kϩ1 ϩ 10 ϭ 11(12r Ϫ 10)
Since r is a positive integer,
12r Ϫ 10 is a positive integer. Thus, 12k ϩ1 ϩ 10 is divisible by 11, so the statement is true for n ϭ k ϩ 1.
Therefore, 12n ϩ 10 is divisible by 11 for all positive integers n.

Step 3:
13k ϩ 11 ϭ 12r
13k ϭ 12r Ϫ 11
13k ϩ1 ϭ 156r Ϫ 143
13k ϩ1 ϩ 11 ϭ 156r Ϫ 132
13k ϩ1 ϩ 11 ϭ 12(13r Ϫ 11)
Since r is a positive integer,
13r Ϫ 11 is a positive integer. Thus, 13kϩ1 ϩ 11 is divisible by 12, so the statement is true for n ϭ k ϩ 1.
Therefore, 13n ϩ 11 is divisible by 12 for all positive integers n.

21. Step 1: There are 6 bricks in the top row, and 12 ϩ 5(1) ϭ
6, so the formula is true for n ϭ 1.
Step 2: Assume that there are k 2 ϩ 5k bricks in the top k rows for some positive integer k.
Step 3: Since each row has
2 more bricks than the one above, the numbers of bricks in the rows form an arithmetic sequence. The number of bricks in the
(k ϩ 1)st row is
6 ϩ [(k ϩ 1) Ϫ 1](2) or
2k ϩ 6. Then the number of bricks in the top k ϩ 1 rows is k 2 ϩ 5k ϩ (2k ϩ 6) or k 2 ϩ 7k ϩ 6. k 2 ϩ 7k ϩ 6 ϭ (k ϩ 1)2 ϩ
5(k ϩ 1), which is the formula to be proved, where n ϭ k ϩ 1. Thus, the formula is true for n ϭ k ϩ 1.

22. Step 1: When n ϭ 1, the left side of the given equation is
1
a1. The right side is a1(1 Ϫ r )

1Ϫr

or a1, so the equation is true for n ϭ 1.
Step 2: Assume a1 ϩ a1r ϩ a1r 2 ϩ p ϩ a1r kϪ1 ϭ a1(1 Ϫ r k )
1Ϫr

for some positive

integer k.
Step 3: a1 ϩ a1r ϩ a1r 2 ϩ p ϩ a1r kϪ1 ϩ a1r k ϭ ϭ ϭ ϭ

a1(1 Ϫ r k) ϩ a1r k
1Ϫr
a1(1 Ϫ r k ) ϩ (1 Ϫ r )a1r k
1Ϫr
k ϩ a r k Ϫ a r kϩ1 a1 Ϫ a1r
1
1
1Ϫr
a1(1 Ϫ r k ϩ 1)
1Ϫr

The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.

312

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Thus, the equation is true for n ϭ k ϩ 1.
Therefore, a1 ϩ a1r ϩ a1r 2 ϩ p ϩ a1r nϪ1 ϭ

Therefore, the number of bricks in the top n rows is n 2 ϩ 5n for all positive integers n.

a1(1 Ϫ r n )
1Ϫr

for all positive integers n.
24. Step 1: The figure shows how to cover a 21 by 21 board, so the statement is true for n ϭ 1.

23. Step 1: When n ϭ 1, the left side of the given equation is
1
2

a1. The right side is [2a1 ϩ
(1 Ϫ 1)d ] or a1, so the equation is true for n ϭ 1.
Step 2: Assume a1ϩ
(a1 ϩ d ) ϩ (a1 ϩ 2d 2 ϩ p ϩ
[a1 ϩ (k Ϫ 1)d ] ϭ k
[2a1
2

ϩ (k Ϫ 1)d ] for

k
[2a1
2

ϩ (k Ϫ 1)d ] ϩ

k
[2a1
2

ϩ (k Ϫ 1)d ]

Step 2: Assume that a 2k by
2k board can be covered for some positive integer k.

some positive integer k.
Step 3: a1 ϩ (a1 ϩ d ) ϩ
(a1 ϩ 2d ) ϩ p ϩ [a1 ϩ
(k Ϫ 1)d ] ϩ
[a1 ϩ (k ϩ 1 Ϫ 1)d ] ϭ [a1 ϩ (k ϩ 1 Ϫ 1)d ] ϭ ϩ a1 ϩ kd

k [2a1 ϩ (k Ϫ 1)d ] ϩ 2(a1 ϩ kd )
2
2 k ؒ 2a1 ϩ (k Ϫ k)d ϩ 2a1 ϩ 2kd ϭ 2
(k ϩ 1)2a1 ϩ (k 2 Ϫ k ϩ 2k)d ϭ 2
(k ϩ 1)2a1 ϩ k(k ϩ 1)d ϭ 2 kϩ1 ϭ
(2a1 ϩ kd )
2
kϩ1
[2a1 ϩ (k ϩ 1 Ϫ 1)d ] ϭ 2

ϭ

Step 3: Divide a 2kϩ1 by
2kϩ1 board into four quadrants. By the inductive hypothesis, the first quadrant can be covered. Rotate the design that covers Quadrant
I 90Њ clockwise and use it to cover Quadrant II. Use the design that covers Quadrant
I to cover Quadrant III.

313

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The last expression is the right side of the formula to be proved, where n ϭ k ϩ 1.
Thus, the formula is true for n ϭ k ϩ 1.
Therefore, a1 ϩ (a1 ϩ d ) ϩ
(a1 ϩ 2d ) ϩ p ϩ [a1 ϩ
(n Ϫ 1)d ] ϭ

Rotate the design that covers
Quadrant I 90Њ counterclockwise and use it to cover Quadrant IV. This leaves three empty squares near the center of the board, as shown. Use one more
L-shaped tile to cover these
3 squares. Thus, a 2kϩ1 by
2kϩ1 board can be covered.
The statement is true for n ϭ k ϩ 1.
Therefore, a 2n by 2n checkerboard with the top right square missing can be covered for all positive integers n.

n
[2a1 ϩ(n Ϫ 1)d ]
2

for all positive integers n.

25. Sample answer: n ϭ 3

26. Sample answer: n ϭ 4

27. Sample answer: n ϭ 2

28. Sample answer: n ϭ 3

29. Sample answer: n ϭ 11

30. Sample answer: n ϭ 41

31. Write 7n as (6 ϩ 1)n. Then use the Binomial Theorem.
7n Ϫ 1 ϭ (6 ϩ 1)n Ϫ 1

32. An analogy can be made between mathematical induction and a ladder with the positive integers on the steps. Answers should include the following.
• Showing that the statement is true for n ϭ 1 (Step 1).
• Assuming that the statement is true for some positive integer k and showing that it is true for k ϩ 1 (Steps 2 and 3).

ϭ 6n ϩ n ؒ 6nϪ1 ϩ

n(n Ϫ 1)
ؒ
2

6nϪ2 ϩ p ϩ n ؒ 6 ϩ 1 Ϫ 1 ϭ 6n ϩ n ؒ 6nϪ1 ϩ

n(n Ϫ 1)
ؒ
2

6nϪ2 ϩ p ϩ n ؒ 6
Since each term in the last expression is divisible by 6, the whole expression is divisible by 6. Thus, 7n Ϫ 1 is divisible by 6.
33. C

34. A

35. x 6 ϩ 6x 5y ϩ 15x 4y 2 ϩ
20x 3y 3 ϩ 15x 2y 4 ϩ 6xy 5 ϩ y 6

36. a7 Ϫ 7a6b ϩ 21a5b 2 Ϫ
35a4b3 ϩ 35a3b4 Ϫ
21a 2b 5 ϩ 7ab6 Ϫ b7

314

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37. 256x 8 ϩ 1024x 7y ϩ
1792x 6y 2 ϩ 1792x 5y 3 ϩ
1120x 4y 4 ϩ 448x 3y 5 ϩ
112x 2y 6 ϩ 16xy 7 ϩ y 8

38. 4, 10, 28

39. 2, 14, 782

40. 12 h

41. 0, 1

42. Ϫ14

315

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Chapter 12 Probability and Statistics
Lesson 12-1 The Counting Principle
Pages 634–637
1. HHH, HHT, HTH, HTT, THH,
THT, TTH, TTT

2. Sample answer: buying a shirt that comes in 3 sizes and 6 colors

3. The available colors for the car could be different from those for the truck.

4. independent

5. dependent

6. 30

7. 256

8. 20
10. dependent

9. D
11. independent

12. independent

13. dependent

14. 6

15. 16

16. 6

17. 30

18. 48

19. 1024

20. 240

21. 10,080

22. 151,200

23. 362,880

24. 17

25. 27,216

26. 160

27. 800

28. See students’ work.

316

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29. The maximum number of license plates is a product with factors of 26s and 10s, depending on how many letters are used and how many digits are used.
Answers should include the following. 30. A

• There are 26 choices for the first letter, 26 for the second, and 26 for the third. There are 10 choices for the first number, 10 for the second, and 10 for the third. By the Fundamental
Counting Principle, there are 263 ؒ 103 or 17,576,000 possible license plates.
• Replace positions containing numbers with letters.
31. C

32. 45

33. 20 mi

34. Step 1: When n ϭ 1, the left side of the given equation is
4. The right side is

1[3(1) ϩ 5]
2

or 4, so the equation is true for n ϭ 1.
Step 2: Assume 4 ϩ 7 ϩ 10 ϩ p ϩ (3k ϩ 1) ϭ

k(3k ϩ 5)
2

for some positive integer k.
Step 3: 4 ϩ 7 ϩ 10 ϩ p ϩ
(3k ϩ 1) ϩ [3(k ϩ 1) ϩ 1] ϭ ϭ ϭ ϭ

317

k(3k ϩ 5) ϩ [3(k ϩ 1) ϩ 1]
2
k(3k ϩ 5) ϩ 2[3(k ϩ 1) ϩ 1]
2
2
3k ϩ 5k ϩ 6k ϩ 6 ϩ 2
2
2
3k ϩ 11k ϩ 8
2

Algebra 2

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ϭ ϭ (k ϩ 1)(3k ϩ 8)
2
(k ϩ 1)[3(k ϩ 1) ϩ 5]
2

The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore, 4 ϩ 7 ϩ 10 ϩ p n(3n ϩ 5)

ϩ (3n ϩ 1) ϭ
2
for all positive integers n.
35. 28x 6 y 2

36. 280a3b4

37. 7

38. 5

39.

1
2

40. Ϫ1

41. Ϫ

x x ϩ 5y

42. 36 mi

43. Ϯ1, Ϯ2

44. 0, Ϫ2

45. y ϭ (x Ϫ 3)2 ϩ 2

46. y ϭ Ϫ2(x ϩ 1)2 ϩ 4

1
2

47. y ϭ Ϫ x 2 ϩ 8

48. 4

49. 3

50. 4
5
R
4
54. y ϭ Ϫ2x Ϫ 2

Ϫ1
R
3
53. no inverse exists
51.

1 1
B
7 4

2
3

55. y ϭ x ϩ

52.

1
3

1 Ϫ1
B
6 Ϫ2

56. 60

57. 30

58. 840

59. 720

60. 6

61. 15

62. 56

63. 1

318

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Permutations and Combinations
Pages 641–643

1. Sample answer: There are six people in a contest. How many ways can the first, second, and third prizes be awarded? 2. C (n, n Ϫ r) ϭ ϭ ϭ n!
[n Ϫ (n Ϫ r )]!(n Ϫ r )! n! r !(n Ϫ r )! n! (n Ϫ r )!r !

ϭ C(n, r )
3. Sometimes; the statement is only true when r ϭ 1.

4. 60

5. 120

6. 6

7. 6

8. combination; 15
10. permutation; 90,720

9. permutation; 5040
11. 84

12. 56

13. 9
15. 665,280

14. 2520
16. 10

17. 70

18. 792

19. 210

20. 27,720

21. 1260

22. permutation; 5040

23. combination; 28

24. permutation; 2520

25. permutation; 120

26. combination; 220

27. permutation; 3360

28. combination; 45

29. combination; 455

30. 11,880

31. 60

32. 75,287,520

33. 111,540

34. 267,696

35. 80,089,128

36. 528

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38. Permutations and combinations can be used to find the number of different lineups. Answers should include the following.
• There are 9! different
9-person lineups available:
9 choices for the first player, 8 choices for the second player, 7 for the third player, and so on.
So, there are 362,880 different lineups.
• There are C(16, 9) ways to choose 9 players from

37. C (n Ϫ 1,r ) ϩC (n Ϫ 1, r Ϫ 1) ϭ ϭ

(n Ϫ 1)! ϩ (n Ϫ 1 Ϫ r )!r !
(n Ϫ 1)!
[n Ϫ 1 Ϫ (r Ϫ 1)] !(r Ϫ 1)!
(n Ϫ 1)! ϩ (n Ϫ r Ϫ 1 )!r !
(n

ϭ

ϭ ϭ ϭ ϭ (n
(n
(n
(n

(n Ϫ 1)!
Ϫ r )! (r Ϫ 1)!
(n Ϫ 1)! nϪr ؒ ϩ Ϫ r Ϫ 1)!r ! n Ϫ r
(n Ϫ 1)! r ؒ
Ϫ r )!(r Ϫ 1)! r
Ϫ 1)!(n Ϫ r )
(n Ϫ 1)!r ϩ (n Ϫ r )!r !
(n Ϫ r )!r !
Ϫ 1)!(n Ϫ r ϩ r )
(n Ϫ r )!r !

16: C (16, 9) ϭ
11,440.

16!
7!9!

or

(n Ϫ 1)!n
(n Ϫ r )!r ! n! (n Ϫ r )!r !

ϭ C(n, r )
39.

D

40. A

41. 24

42. 6

43. 120

44. 8

45. 80

46. Sample answer: n ϭ 3

47. Sample answer: n ϭ 2

48. Ϫ1.0986

49. x Ͼ 0.8047

50. 21.0855

51. 20 days
53.

(y Ϫ 4)2
9

ϩ

52.
(x Ϫ 4) 2
4

ϭ1

x2
16

ϩ

y2
9

ϭ1

7 53
2 2

54. Ϫ ;

60. 0 x 3 0 y 3 23

55. Ϫ4; 128

56. {Ϫ4, 4}

57. {Ϫ2, 5}

58. e Ϫ3, f

1
3

59. 822
61. 425

62. (Ϫ1, 3)
320

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4
3

63. (0, 2)

64. Ϫ

65. Ϫ

6
7

66. 0

67. {Ϫ7, 15}

68. л

69.

3
5

70.

1
2

71.

1
5

72.

1
3

Lesson 12-3 Probability
Pages 647–650
1. Sample answer: The event
July comes before June has a probability of 0. The event
June comes before July has a probability of 1.

2.

3
5

3. There are 6 ؒ 6 or 36 possible outcomes for the two dice. Only 1 outcome, 1 and 1, results in a sum of 2,

4.

1
7

6.

4
7

1

so P(2) ϭ . There are 2
36
outcomes, 1 and 2 as well as
2 and 1, that result in a sum
2
1 of 3, so P(3) ϭ or .
36

5.

18

2
7

7. 8:1

8. 1:5
10.

9. 2:7

6
11

11.

10
11

12.

2
7

13.

1
8

14.

3
8

321

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15.

1
10

16.

21
50

17.

2
25

18.

1
50

19.

6
55

20.

21
55

21.

28
55

22.

14
575

23.

11
115

24.

7
115

25.

6
115

26.

132
575

27.

24
115

28.

6
115

29. 0

30.

1
22,957,480

31. 0.007

32. 0.623

33. 0.109

34. 1:1

35. 3:5

36. 11:1

37. 5:3

38. 4:3

39. 1:4

40. 4:7

41. 3:1

42.

6
7

43.

3
10

44.

5
11

45.

4
9

46.

9
17

47.

1
9

48.

7
16

49.

3
5

50.

1
10

51. 2:23

52. 1:999

53. 1:4

54.

322

540
1771

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55.

1
20

56.

9
20

57.

9
20

58.

1
20

59.

9
20

60.

9
20

61.

1
120

62.

␲Ϫ1

64. C

63. Probability and odds are good tools for assessing risk.
Answers should include the following. • P(struck by lightning) ϭ
1
s
, so ϭ sϩf

750,000

odds ϭ 1:(750,000 Ϫ 1) or
1:749,999. P(surviving a lightning strike) ϭ s sϩf

3
4

ϭ , so odds ϭ

3:(4 Ϫ 3) or 3 :1.
• In this case, success is being struck by lightning or surviving the lightning strike. Failure is not being struck by lightning or not surviving the lightning strike. 1
36

65. D

66. theoretical;

67. experimental; about 0.307

68. experimental;

69. theoretical;

1
17

1
5

70. permutation; 120

71. permutation; 1260

72. combination; 35

73. 16

74. 24

75. direct variation

76. square root

77. (4, 4)

78. (1, 3)

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79.

6
35

80.

3
14

81.

1
4

82.

2
21

83.

9
20

Chapter 12
Practice Quiz 1
Page 650
1. 24

2. 756

3. 18,720

4. 1320

5. 56

6. permutation; 40,320

7. combination; 20,358,520

8.

1
221

10.

8
663

9.

13
102

Lesson 12-4 Multiplying Probabilities
Pages 654–657
1. Sample answer: putting on your socks, and then your shoes 2. P(A, B, C, and D) ϭ
P(A) ؒ P(B) ؒ P(C) ؒ P(D)

3. Mario; the probabilities of rolling a 4 and rolling a 2 are

4.

1
36

1
6

both .
5.

1
4

6.

1
17

7.

4
663

8.

7
30

9.

1
4

10.

1
16

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21
220

12. independent;

13.

1
12

14.

1
36

15.

25
36

16.

1
36

17.

1
6

18.

1
6

19.

5
6

20.

1
42

21.

1
49

22.

25
49

23.

10
21

1
36

24. 0
26.

25. 0

1
15
1
10

27.

2
15

28.

29.

2
15

30. dependent;

31. independent;

25
81

3
28

32. independent;

168
4913
1
32

33. dependent;

1
21

34. independent;

35. dependent;

81
2401

36.

1
9

First Spin

Second Spin
R

R
P(R,B) ϭ

B
1
3

ϫ

1
3

or

1
9

B

Y
R
B
Y
R
B

Y

Y

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38.

First Spin
Blue Yellow Red
1
1
1
3
3
3
Blue
1
3
Second
Spin

BB
1
9

BY
1
9

BR
1
9

Yellow YB
1
1
3
9

YY
1
9

YR
1
9

RY
1
9

1
3

RR
1
9

Red
1
3

RB
1
9

39.

1
9

40.

1
635,013,559,600

41.

19
1,160,054

42.

1
158,753,389,900

43.

6327
20,825

44. a

99 4 b 100
1
1320

46.

47. no

48. no

49. Sample answer: As the number of trials increases, the results become more reliable. However, you cannot be absolutely certain that there are no black marbles in the bag without looking at all of the marbles.

50. 21

51. Probability can be used to analyze the chances of a player making 0, 1, or 2 free throws when he or she goes to the foul line to shoot 2 free throws. Answers should include the following.

52. D

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• One of the decimals in the table could be used as the value of p, the probability that a player makes a given free throw. The probability that a player misses both free throws is
(1 Ϫ p)(1 Ϫ p) or (1 Ϫ p) 2.
The probability that a player makes both free throws is p ؒ p or p 2.
Since the sum of the probabilities of all the possible outcomes is 1, the probability that a player makes exactly 1 of the 2 free throws is
1 Ϫ (1 Ϫ p) 2 Ϫ p 2 or
2p (1 Ϫ p).
• The result of the first free throw could affect the player’s confidence on the second free throw. For example, if the player makes the first free throw, the probability of making the second free throw might increase. Or, if the player misses the first free throw, the probability of making the second free throw might decrease.
54.

55.

3
340

57. 1440 ways

1
119

58. 6

59. 36

1
204

56.

53. C

60. x 2 Ϫ 4x ϩ 2

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62.

61. x, x Ϫ 4

y

x

O

y ϭ x2 ϩ x Ϫ 2

63.

64.

y

O

y

x

O

x y ϭ x 2 Ϫ 3x

y ϭ x2 Ϫ 4

65. 153

66. Ϫ9

69. (1, 2)

70. (13, 9)

67. 0 b 0

68. 5a 4 0 b 3 0
72.

71. (Ϫ2, 4)

2
3
5
4

73.

5
6

74.

75.

11
12

76. 1

1
6

5
12

77. 1

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Lesson 12-5 Adding Probabilities
Pages 660–663
1. Sample answer: mutually exclusive events: tossing a coin and rolling a die; inclusive events: drawing a 7 and a diamond from a standard deck of cards

2.

3. The events are not mutually exclusive, so the chance of rain is less than 100%.

4.

1
3

French and Algebra

French
150

5.

1
3

6.

1
2

8.

5
6

9.

2
3

Algebra
300

1
3

7.

400

11. inclusive;

10. mutually exclusive;
12.

13. 1

13
16

14.

4
13

2
13

1
6

15.

25
42

16.

37
42

17.

35
143

18.

105
143

19.

3
143

20.

84
143

21.

38
143

22.

32
39

23. mutually exclusive;
25. inclusive;

7
9

24. inclusive;

21
34

1
2

26. mutually exclusive;

329

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13

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27.

4
13

28.

2
3

29.

55
221

30.

11
221

31.

188
663

32.

63
221

33.

1
8

34.

1
8

35.

1
4

36.

3
4

37.

1
780

38.

145
156

39.

9
130

40.

1
26

41.

11
780

42.

1
78

43.

3
5

44.

53
108

45.

17
27

46.

17
162

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47. Subtracting P(A and B) from each side and adding P (A or
B) to each side results in the equation P (A or B) ϭ P(A) ϩ
P(B) Ϫ P(A and B). This is the equation for the probability of inclusive events. If A and B are mutually exclusive, then P (A and B) ϭ 0, so the equation simplifies to P(A or B) ϭ
P(A) ϩ P(B), which is the equation for the probability of mutually exclusive events.
Therefore, the equation is correct in either case.

48. Probability can be used to estimate the percents of people who do the same things before going to bed.
Answers should include the following. • The events are inclusive because some people brush their teeth and set their alarm. Also, you know that the events are inclusive because the sum of the percents is not
100%.
• According to the information in the text and the table, P (read book) ϭ
38
and P (brush teeth) ϭ
100
81
. Since the events
100

are inclusive, P (read book and brush teeth) ϭ P (read book) ϩ P (brush teeth) Ϫ
P (read book and brush
38
81 ϩ Ϫ teeth) ϭ
100
1200
59
ϭ
.
2000
100

49. C

100

50. A

51.

1
216

52.

125
216

53.

1
216

54.

1
8

55. 4:1

56. 1:8

57. 2:5

58. 5:3

59. 254

60. 24

61. (Ϯ8, Ϫ10)

62. (Ϯ12, Ϯ5)

63. (x ϩ 1)2(x Ϫ 1)(x 2 ϩ 1)

64. min: (0, Ϫ5); max: (Ϫ1.33, Ϫ3.81)

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65. min: (Ϫ0.42, 0.62); max: (Ϫ1.58, 1.38)

66.

y

x

O

(0, 2), (2, 0), (0, Ϫ2); max: f(2, 0) ϭ 6; min: f (0, Ϫ2) ϭ Ϫ2
67.

68. d ϭ 12.79t

y

x

O

(1, 3), (1, Ϫ1), (3, 3), (3, 5); max: f(3, 5) ϭ 23; min: f (1, Ϫ1) ϭ Ϫ3
69. direct variation

70. 323.4, 298, no mode, 143

71. 35.4, 34, no mode, 72

72. 3.6, 3.45, 2.1, 3.6

73. 63.75, 65, 50 and 65, 30

74. 79.83, 89, 89, 57

75. 12.98, 12.9, no mode, 4.7

Lesson 12-6 Statistical Measures
Pages 666–670
1
a (x
B n iϭ1 i

2. Sample answer: The variance of the set {0, 1} is 0.25 and the standard deviation is 0.5.

{10, 10, 10, 10, 10, 10}
3. ␴ ϭ

n

Ϫ x )2

4. 40, 6.3

5. 8.2, 2.9

6. 424.3, 20.6

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7. \$7300.50, \$5335.25

8. The mean is more representative for the southwest central states because the data for the
Pacific states contains the most extreme value,
\$10,650.

9. 2500, 50

10. 1.6, 1.3

11. 3.1, 1.7

12. 4.8, 2.2

13. 37,691.2, 194.1

14. 569.4, 23.9

15. 82.9, 9.1

16. 43.6, 6.6

17. 114.5, 105, 23

18. The mean and median both seem to represent the center of the data.

19. Mean; it is highest.

20. Mode; it is lower and is what most employees make. It reflects the most representative worker.

21. \$1047.88, \$1049.50, \$695

22. Mode; it is the least expensive price.

23. Mean or median; they are nearly equal and are more representative of the prices than the mode.

24. 2,290,403; 2,150,000;
2,000,000

25. Mode; it is lowest.

26. Mean; it is highest.

27. 19.3

28. 28.9

29. 19.5

30. Washington; see students’ work. 31. 59.8, 7.7

32. 64%

33. 100%

34. Different scales are used on the vertical axes.

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35. Sample answer: The first graph might be used by a sales manager to show a salesperson that he or she does not deserve a big raise.
It appears that sales are steady but not increasing fast enough to warrant a big raise. 36. Sample answer: The second graph might be shown by the company owner to a prospective buyer of the company. It looks like there is a dramatic rise in sales.

37. A: 2.5, 2.5, 0.7, 0.8; B: 2.5,
2.5, 1.1, 1.0

38. The first histogram is lower in the middle and higher on the ends, so it represents data that are more spread out. Since set B has the greater standard deviation, set B corresponds to the first histogram and set A corresponds to the second.

39. The statistic(s) that best represent a set of test scores depends on the distribution of the particular set of scores.
Answers should include the following. • mean, 73.9; median, 76.5; mode, 94
• The mode is not representative at all because it is the highest score. The median is more representative than the mean because it is influenced less than the mean by the two very low scores of 34 and 19.

40. A

41. D

42. 3

43. 1.9

44. The mean deviations would be greater for the greater standard deviation and lower for the groups of data that have the smaller standard deviation. ©Glencoe/McGraw-Hill

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4
13

46. mutually exclusive;

47.

1
169

48.

4
663

49.

13
204

50.

3
7

1
16

51. 10, Ϯ92; 10, Ϯ21062; Ϯ

52. Ϫ3

53. 17

54. Ϫ2

9
5

56. 1Ϫ4, 62

55. 12 cm3

58. (3, 5)

57. (1, 5)
59. 136

60. 340

61. 380

62. 475

63. 396

64. 495

Chapter 12
Practice Quiz 2
Page 670
1.

3
20

2.

1
6

3.

2
9

4.

1
4

5.

1
6

6.

2
3

7.

3
4

8. 6.6, 2.6

9. 23.6, 4.9

10. 134.0, 11.6

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Lesson 12-7 The Normal Distribution
Pages 673–675

2. The mean of the three graphs is the same, but the standard deviations are different. The first graph has the least standard deviation, the standard deviation of the middle graph is slightly greater, and the standard deviation of the last graph is greatest. the use of cassettes since
CDs were introduced
3. Since 99% of the data is within 3 standard deviations of the mean, 1% of the data is more than 3 standard deviations from the mean. By symmetry, half of this, or
0.5%, is more than 3 standard deviations above the mean.

4. normally distributed

5. 68%

6. 13.5%

7. 95%

8. 6800

9. 250

10. 1600

11. 81.5%

12. positively skewed

13. normally distributed

14. Negatively skewed; the histogram is high at the right and has a tail to the left.

15. 68%

16. 34%

17. 0.5%

18. 16%

19. 50%

20. 50%

21. 95%

22. 50%

23. 815

24. 25

25. 16%

26. 652

27. The mean would increase by
25; the standard deviation would not change; and the

28. If a large enough group of athletes is studied, some of the characteristics may be

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graph would be translated 25 units to the right.

normally distributed; others may have skewed distributions. Answers should include the following.

10

Frequency

8
6
4
2
0

67 68 69 70 71 72 73 74 75 76 77 78 79 80
Height (in.)

• Since the histogram has two peaks, the data may not be normally distributed. This may be due to players who play certain positions tending to be of similar large sizes while players who play the other positions tend to be of similar smaller sizes.
29. A

30. D

31. 17.5, 4.2

32. 42.5, 6.5

33.

2
13

34.

35.

4
13

36. Ϫ5, 0, 1
38. 1, Ϫ1

37. Ϫ3, 2, 4
39.

1
,
4

4
13

40.

1

y
50
Ϫ2

1

O

Ϫ1

2t

Ϫ50
Ϫ100

y ϭ 216t 2 Ϫ 53

41. 0.76 h

42. 21a 5b 2

43. 56c 5d 3

44. 126x 5y 4

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Lesson 12-8 Binomial Experiments
Pages 678–680
1. Sample answer: In a 5-card hand, what is the probability that at least 2 cards are hearts? 2. RRRWW, RRWRW,
RRWWR, RWRRW,
RWRWR, RWWRR,
WRRRW, WRRWR,
WRWRR, WWRRR

3a. Each trial has more than two possible outcomes.
3b. The number of trials is not fixed. 3c. The trials are not independent. 4.

3
8

5.

1
8

6.

7
8

7.

1
28,561

8.

48
28,561

9.

27,648
28,561

12.

1
16

13.

1
16

14.

3
8

15.

1
4

16.

5
16

17.

11
16

18.

3125
7776

19.

125
3888

20.

625
648

21.

23
648

22.

243
1024

23.

1
1024

24.

15
1024

25.

135
512

26.

459
512

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27.

53
512

28.

105
512

29.

105
512

30.

319
512

31.

319
512

34.

560
2187

36.

1 3 15 5 15 3 1
, , , , , ,
64 32 64 16 64 32 64

37.

1
4

38. C (n, m)p m(1 Ϫ p)nϪm
40. 2

39. Getting a right answer and a wrong answer are the outcomes of a binomial experiment. The probability is far greater that guessing will result in a low grade than in a high grade. Answers should include the following.
• Use (r ϩ w)5 ϭ r 5 ϩ 5r 4w ϩ 10r 3w 2 ϩ
10r 2w 3 ϩ 5rw 4 ϩ w 5 and the chart on page 676 to determine the probabilities of each combination of right and wrong.
1 5
4

• P(5 right): r 5 ϭ P a b ϭ

1
1024

or about 0.098%; P (4 right,
1 wrong):

15
1024

1.5%; P (3 right, 2 wrong):
1 3 3 2
4
4

10r 2w 3 ϭ 10 a b a b ϭ

45
512

or about 8.8%; P (3 wrong,
2 right): 10r 2w 3 ϭ
1 2 3 3
4
4

10 a b a b ϭ

135
512

26.4%; P (4 wrong, 1 right):

339

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4

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3 4
4

405
1024

243
1024

5r w 4 ϭ 5 a b a b ϭ

or

about 39.6%; P (5 wrong):
3
4

5

w5 ϭ a b ϭ
23.7%.
41. B

42. See students’ work.

43. normal distribution

44. 68%

45. 10

46. 16%

47. Mean; it is highest.

48.

y x ϭ Ϫ3 x O

49.

y

50.

xϩyϭ4

y y ϭ |5x |

x
O

O

51. 0.1

52. 0.05

53. 0.039

54. 0.027

55. 0.041

x

56. 0.031

340

Algebra 2

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Lesson 12-9 Sampling and Error
Pages 683–685
1. Sample answer: If a sample is not random, the results of a survey may not be valid.

2. Sample answer for good sample: doing a random telephone poll to rate the mayor’s performance; sample answer for bad sample: conducting a survey on how much the average person reads at a bookstore

3. The margin of sampling error decreases when the size of the sample n increases. As n p (1 Ϫ p) increases, decreases.

4. Yes; last digits of social security numbers are random. 5. No; these students probably study more than average.

n

9. The probability is 0.95 that the percent of Americans ages 12 and older who listen to the radio every day is between 72% and 82%.

11. No; you would tend to point toward the middle of the page. 12. Yes; all seniors would have the same chance of being selected. 13. Yes; a wide variety of people would be called since almost everyone has a phone.

14. No; freshmen are more likely than older students to be still growing, so a sample of freshmen would not give representative heights for the whole school.

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28. 36 or 64

29. A political candidate can use the statistics from an opinion poll to analyze his or her standing and to help plan the rest of the campaign.
Answers should include the following. • The candidate could decide to skip areas where he or she is way ahead or way behind, and concentrate on areas where the polls indicate the race is close.
• The margin of error indicates that with a probability of 0.95 the percent of the Florida population that favored
Bush was between 43.5% and 50.5%. The margin of error for Gore was also about 3.5%, so with probability 0.95 the percent that favored Gore was between 40.5% and 47.5%.
Therefore, it was possible that the percent of the
Florida population that favored Bush was less than the percent that favored
Gore.

30. A

31. C

32.

1
32

5
32

34.

1
2

33.

35. 95%

36. 210

37. 97.5%

38. x Ϫ 2, x Ϫ 3

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Chapter 13 Trigonometry
Lesson 13-1 Right Triangle Trigonometry
Pages 706–708
2.

1. Trigonometry is the study of the relationships between the angles and sides of a right triangle. ␪

hypotenuse

opposite

8
15
; cos ␪ ϭ ;
17
17
8
17 tan ␪ ϭ ; csc ␪ ϭ ;
15
8
17
15 sec ␪ ϭ ; cot ␪ ϭ
15
8

3. Given only the measures of the angles of a right triangle, you cannot find the measures of its sides.

5. sin ␪ ϭ tan ␪ ϭ sec ␪ ϭ

285
;
11

285
;
6
11
;
6

7. cos 23Њ ϭ

csc ␪ ϭ

32
;
x

11 285
;
85
6
;
11

5 211
;
11

5
6

6 285
85

cos ␪ ϭ

cot ␪ ϭ

4. sin ␪ ϭ

6 211
;
11

6. sin ␪ ϭ ; cos ␪ ϭ tan ␪ ϭ sec ␪ ϭ

x Ϸ 34.8

8. tan x Њ ϭ

9. B ϭ 45Њ, a ϭ 6, c ϭ 8.5

15
;
21

211
;
6

211
5

6
5

csc ␪ ϭ ; cot ␪ ϭ

x Ϸ 36

10. A ϭ 34Њ, a Ϸ 8.9, b Ϸ 13.3

2105
;
11

11. a Ϸ 16.6, A Ϸ 67Њ, B Ϸ 23Њ

12. c Ϸ 19.1, A Ϸ 47Њ, B Ϸ 43Њ

13. 1660 ft

14. B

15. sin ␪ ϭ tan ␪ ϭ sec ␪ ϭ

Glencoe/McGraw-Hill

4 2105
; csc
105

4
;
11

cos ␪ ϭ

112105
;
105

␪ϭ

3
5
3 tan ␪ ϭ ; csc
4
5 sec ␪ ϭ ; cot
4

2105
4

11
;
4

cot ␪ ϭ

4
5
5 ϭ ;
3
4 ϭ 3

16. sin ␪ ϭ ; cos ␪ ϭ ;

343

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17. sin ␪ ϭ

27
;
4

27
;
3

tan ␪ ϭ

4
3

tan ␪ ϭ

3 27
7

csc ␪ ϭ

sec ␪ ϭ ; cot ␪ ϭ

19. sin ␪ ϭ

4 27
;
7

3
4

18. sin ␪ ϭ

cos ␪ ϭ ;

25
;
2

csc ␪ ϭ

x
,
10

23. sin 54Њ ϭ

17.8
,
x

25. cos x Њ ϭ

15
,
36

27a. sin 30Њ ϭ sin 30Њ ϭ

sec ␪ ϭ
20. sin ␪ ϭ tan ␪ ϭ

cos 30Њ

x Ϸ 22.0

24. tan 17.5Њ ϭ
26. sin x Њ ϭ

16
,
22

28a. sin 45Њ ϭ

sine ratio
Replace opp with x and hyp with
2x.

sin 45Њ ϭ

1
Simplify.
2 adj cosine ratio ϭ hyp
23x Replace adj with 13x and ϭ 2x hyp with 2x.

cos 30Њ ϭ

215
;
7

8 215
;
15
7
8

7 215
15

cos ␪ ϭ ; csc ␪ ϭ

sin 45Њ ϭ
28b. cos 45Њ

cos 45Њ ϭ

Simplify.

cos 45Њ ϭ

344

x
;
23.7

x Ϸ 7.5

x Ϸ 47

opp hyp x
12x
1
12

12
2

sin 45Њ ϭ

cos 45Њ ϭ

5
9

3 x x Ϸ 65

23
2

215
;
8

cot ␪ ϭ

22. cos 60Њ ϭ , x ϭ 6

sin 30Њ ϭ
27b. cos 30Њ

2106
;
5

sec ␪ ϭ ; cot ␪ ϭ

x Ϸ 5.8

opp hyp x
2x

2106
;
9

9
5

tan ␪ ϭ ;

8
7

cot ␪ ϭ 2

21. tan 30Њ ϭ

5 2106
;
106

cos ␪ ϭ

25
2 25
;
; cos ␪ ϭ
5
5
1
; csc ␪ ϭ 25;
2

sec ␪ ϭ

9 2106
;
106

x
12x
1
12
12
2

Replace opp with x and hyp with 12x.

sine ratio

Simplify.
Rationalize the denominator. Replace adj with x and hyp with 12x. cosine ratio

Simplify.
Rationalize the denominator. Algebra 2

Chapter 13

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27c. sin 60Њ ϭ sin 60Њ ϭ sin 60Њ ϭ

23x
2x

opp hyp 23
2

Replace opp with 13x and hyp with 2x.

28c. tan 45Њ ϭ

sine ratio

tan 45Њ ϭ

opp adj tangent ratio

x x Replace opp with x and adj with x.

tan 45Њ ϭ 1

Simplify.

Simplify.

29. B ϭ 74Њ, a Ϸ 3.9, b Ϸ 13.5

30. A ϭ 63Њ, a Ϸ 13.7, c Ϸ 15.4

31. B ϭ 56Њ, b Ϸ 14.8, c Ϸ 17.9

32. A ϭ 75Њ, a Ϸ 24.1, b Ϸ 6.5

33. A ϭ 60Њ, a Ϸ 19.1, c ϭ 22

34. B ϭ 45Њ, a ϭ 7, b ϭ 7

35. A ϭ 72Њ, b Ϸ 1.3, c Ϸ 4.1

36. B ϭ 80Њ, a Ϸ 2.6, c Ϸ 15.2

37. A Ϸ 63Њ, B Ϸ 27Њ, a Ϸ 11.5

38. A Ϸ 26Њ, B Ϸ 64Њ, b Ϸ 8.1

39. A Ϸ 49Њ, B Ϸ 41Њ, a ϭ 8, c Ϸ 10.6

40. A Ϸ 19Њ, B Ϸ 71Њ, b Ϸ 14.1, c ϭ 15

41. about 300 ft

42. about 142.8 ft

44. about 3.2 in.

45. 93.54 units2

46. about 1.72 km high

47. The sine and cosine ratios of acute angles of right triangles each have the longest measure of the triangle, the hypotenuse, as their denominator. A fraction whose denominator is greater than its numerator is less than 1. The tangent ratio of an acute angle of a right triangle does not involve the measure of the hypotenuse,

48. When construction involves right triangles, including building ramps, designing buildings, or surveying land before building, trigonometry is likely to be used. Answers should include the following.
• If you view the ramp from the side then the vertical rise is opposite the angle that the ramp makes with the horizontal. Similarly, the horizontal run is the adjacent side. So the tangent of the angle is the ratio of the rise to the run or the slope of the ramp.
• Given the ratio of the slope of ramp, you can find the angle of inclination by calculating tan–1 of this ratio. opp
.

If the measure of the opposite side is greater than the measure of the adjacent side, the tangent ratio is greater than 1. If the measure of the opposite side is less than the measure of the adjacent side, the tangent ratio is less than 1.

345

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49. C

50. 7.7

51. No; band members may be more likely to like the same kinds of music.
3
53.
8
15
55.
16
57. {Ϫ2, Ϫ1, 0, 1, 2}

52. Yes; this sample is random since different kinds of people go to the post office.
1
54.
16

59. 20 qt

60. 35,904 ft

61. 12 m2

62. 48 L

Lesson 13-2

56. {Ϯ222, 2i 22}
58. {121}

Angles and Angle Measure
Pages 712–715

1. reals

2. In a circle of radius r units, one radian is the measure of an angle whose rays intercept an arc length of r units. 3.

4.

y

y

290˚

70˚ x O

x

O

Ϫ70˚

5.

6.

y

y

300˚

570˚
O

x

O

346

x

Algebra 2

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8.

y

O

Ϫ45˚

13␲
18

10.

7.

97␲
36

x

18

9. Ϫ

11. 135Њ

12. Ϫ30Њ

13. 1140Њ

14. 420Њ, Ϫ300Њ

15. 785Њ, Ϫ295Њ

16.

17. 21 h

18. 2 h

19.

7␲
,
3

5␲
3

Ϫ

20.

y
235˚

y
270˚

x

O

21.

22.

y

O

24.

y

x

O

y

O

Ϫ150˚

x

380˚

790˚

23.

y

x

O

x

O

347

Ϫ50˚

x

Algebra 2

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25.

26.

y

y

O

x

x

O
2␲
Ϫ
3

27.

2␲
3

28.

3
5␲
4

12

29. Ϫ

30. Ϫ

31.

11␲
3

32.

19␲
6

33.

79␲
90

34.

13␲
9

35. 150Њ

36. 495Њ

37. Ϫ45Њ

38. Ϫ60Њ

39. 1305Њ

40. 510Њ

41.

1620

Ϸ 515.7Њ

42.

540

Ϸ 171.9Њ

43. Sample answer: 585Њ, Ϫ135Њ

44. Sample answer: 390Њ, Ϫ330Њ

45. Sample answer: 345Њ, Ϫ375Њ

46. Sample answer: 220Њ, Ϫ500Њ

47. Sample answer: 8Њ, Ϫ352Њ

48. Sample answer: 400Њ, Ϫ320Њ

11␲
,
4

3␲
,
4

13␲
,
2

5␲
4

19␲
,
6

13␲
4

4␲
,
3

3␲
2

25␲
,
4

Ϫ

Ϫ

Ϫ

5␲
6

Ϫ

8␲
3

Ϫ

7␲
4

Ϫ

55. 2689Њ per second; 47 radians per second

56. 209.4 in2

57. about 188.5 m2

58. number 17

59. about 640.88 in2

60a. a 2 ϩ (Ϫb)2 ϭ a 2 ϩ b 2 ϭ 1
60b. b 2 ϩ a 2 ϭ a 2 ϩ b 2 ϭ 1
60c. b 2 ϩ (Ϫa)2 ϭ a 2 ϩ b 2 ϭ 1

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61. Student answers should include the following.
• An angle with a measure of more than 180Њ gives an indication of motion in a circular path that ended at a point more than halfway around the circle from where it started.
• Negative angles convey the same meaning as positive angles, but in an opposite direction. The standard convention is that negative angles represent rotations in a clockwise direction.
• Rates over 360Њ per minute indicate that an object is rotating or revolving more than one revolution per minute.

62. C

63. D

64. a Ϸ 3.4, c Ϸ 6.0, B ϭ 56Њ

65. A ϭ 22Њ, a Ϸ 5.9, c Ϸ 15.9

66. A ϭ 35Њ, a Ϸ 9.2, b Ϸ 13.1

67. c Ϸ 0.8, A ϭ 30Њ, B ϭ 60Њ

70. permutation, 17,100,720

71. combination, 35

72. [g ‫ ؠ‬h](x) ϭ 6x Ϫ 8,
[h ‫ ؠ‬g](x) ϭ 6x Ϫ 4

73. [g ‫ ؠ‬h](x) ϭ 4x 2 Ϫ 6x ϩ 23,
[h ‫ ؠ‬g](x) ϭ 8x 2 ϩ 34x ϩ 44

74. 1041.8

75. 1418.2 or about 1418; the number of sports radio stations in 2008

76.

77.

78.

79.
81.

3 25
5
210
2

210
4

80.

349

2 23
3
2 26
3
214
2

Algebra 2

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Chapter 13
Practice Quiz 1
Page 715

10 2149
;
149
7 2149
;
149

1. A ϭ 42Њ, a Ϸ 13.3, c Ϸ 17.9

2. A Ϸ 59Њ, B Ϸ 31Њ, b Ϸ 10.8

3.

4. sin ␪ ϭ

y

cos ␪ ϭ
O

Ϫ60˚

x

tan ␪ ϭ sec ␪ ϭ

5.

19␲
18

6.

2149
;
7

10
;
7

csc ␪ ϭ

2149
;
10

cot ␪ ϭ

7
10

5␲
2

8. Ϫ396Њ

7. 210Њ

10.

9. 305Њ; Ϫ415Њ

5␲
;
3

3

Ϫ

Lesson 13-3 Trigonometric Functions of General Angles
Pages 722–724
1. False; sec 0Њ ϭ tan 0Њ ϭ

0 r r r or 1 and

2. Sample answer: 190Њ

or 0.
8
15
, cos ␪ ϭ Ϫ ,
17
17
8
17 tan ␪ ϭ Ϫ , csc ␪ ϭ ,
15
8
17
15 sec ␪ ϭ Ϫ , cot ␪ ϭ Ϫ
15
8

3. To find the value of a trigonometric function of ␪, where ␪ is greater than 90Њ, find the value of the trigonometric function for ␪Ј, then use the quadrant in which the terminal side of ␪ lies to determine the sign of the trigonometric function value of ␪.

4. sin ␪ ϭ

350

Algebra 2

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22
,
2

22
,
2

tan ␪ ϭ 1, csc ␪ ϭ 22, sec ␪ ϭ 22, cot ␪ ϭ 1

5. sin ␪ ϭ 0, cos ␪ ϭ Ϫ1, tan ␪ ϭ 0, csc ␪ ϭ undefined, sec ␪ ϭ Ϫ1, cot ␪ ϭ undefined

6. sin ␪ ϭ

7. 55Њ

8.

y

cos ␪ ϭ

4

y
7␲
4

235˚
O

x

␪'

9. 60Њ

10. Ϫ

y

O

23
2

␪'

x

␪'
O

2 23
3

x

Ϫ240˚

11. Ϫ1

12. Ϫ23

13. Ϫ

14. sin ␪ ϭ
26
,
3

sec ␪ ϭ 23

15. sin ␪ ϭ Ϫ

cos ␪ ϭ

23
,
3

tan ␪ ϭ Ϫ22, cos ␪ ϭ Ϫ

24
,
25
24
tan ␪ ϭ ,
7
25 sec ␪ ϭ ,
7

17. sin ␪ ϭ

csc ␪ ϭ

23
,
2

2 23
,
3

cot ␪ ϭ Ϫ

26
,
2

23
3

tan ␪ ϭ Ϫ23, sec ␪ ϭ Ϫ2,

16. about 12.4 ft

7
,
25
25
csc ␪ ϭ ,
24
7 cot ␪ ϭ
24

25
, cos
5
1 ϭ , csc ␪
2

18. sin ␪ ϭ

cos ␪ ϭ

tan ␪

sin ␪ ϭ

351

25
,
5

2 25
,
5

ϭ 25,

␪ϭ

cos ␪ ϭ 2,

Algebra 2

Chapter 13

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19. cos ␪ ϭ cos ␪ ϭ

8 289
,
89
5 289
,
89
289
,
5
8
5

tan ␪ ϭ Ϫ , csc ␪ ϭ Ϫ sec ␪ ϭ

3
4
5
5
3
5
tan ␪ ϭ Ϫ , csc ␪ ϭ Ϫ ,
4
3
5
4 sec ␪ ϭ , cot ␪ ϭ Ϫ
4
3

289
,
8

20. sin ␪ ϭ Ϫ , cos ␪ ϭ ,

5
8

cot ␪ ϭ Ϫ

21. sin ␪ ϭ Ϫ1, cos ␪ ϭ 0, tan ␪ ϭ undefined, csc ␪ ϭ Ϫ1, sec ␪ ϭ undefined, cot ␪ ϭ 0

22. sin ␪ ϭ 0, cos ␪ ϭ Ϫ1, tan ␪ ϭ 0, csc ␪ ϭ undefined, sec ␪ ϭ Ϫ1, cot ␪ ϭ undefined

23. sin ␪ ϭ Ϫ

24. sin ␪ ϭ Ϫ

22
,
2

22
,
2

26
,
3

tan ␪ ϭ 22, csc ␪ ϭ Ϫ

sec ␪ ϭ 22, cot ␪ ϭ Ϫ1 cos ␪ ϭ

tan ␪ ϭ Ϫ1, csc ␪ ϭ Ϫ22,

sec ␪ ϭ Ϫ23, cot ␪ ϭ
25. 45Њ

26. 60Њ

y

23
,
3

26
,
2
22
2

cos ␪ ϭ Ϫ

y
240˚

315˚
O

27. 30Њ

␪'

x

x

O
␪'

28. 55Њ

y

y

␪'
O

x

␪'

Ϫ210˚

352

x

O
Ϫ125˚

Algebra 2

Chapter 13

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29.

4

y

30.

6

5␲
4

7

33. Ϫ

␪'

32.

13␲
7

O

␪'

x

O

y

23
2

5␲
6

x

O

␪'

31.

y

3

y

x

O

␪'

x
2␲
Ϫ
3

34. Ϫ2

39. 23

22
2

35. Ϫ23

36. Ϫ23

37. undefined

38.

23
2

40.

1
2

41. undefined

42. 2

43.

44. Ϫ1
46. 6092.5 ft

45. 0.2, 0, Ϫ0.2, 0, 0.2, 0, and
Ϫ0.2; or about 11.5Њ, 0Њ,
Ϫ11.5Њ, 0Њ, 11.5Њ, 0Њ, and
Ϫ11.5Њ
4
5
5 csc ␪ ϭ Ϫ ,
4
3 cot ␪ ϭ Ϫ
4

sec ␪ ϭ

226
,
26

5 226
,
26

csc ␪ ϭ 226, sec ␪ ϭ Ϫ

4
3
5
,
3

47. sin ␪ ϭ Ϫ , tan ␪ ϭ Ϫ ,

48. sin ␪ ϭ

cos ␪ ϭ Ϫ

226
,
5

cot ␪ ϭ Ϫ5

353

Algebra 2

Chapter 13

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2 22
,
3

49. cos ␪ ϭ Ϫ

22
,
4

2 25
,
5

3 22
,
4

tan ␪ ϭ Ϫ

25
,
5

50. sin ␪ ϭ Ϫ

csc ␪ ϭ 3, sec ␪ ϭ Ϫ

cos ␪ ϭ Ϫ

cot ␪ ϭ Ϫ222

csc ␪ ϭ Ϫ

3 210
,
10

51. sin ␪ ϭ Ϫ

cos ␪ ϭ Ϫ csc ␪ ϭ Ϫ

210
,
10

210
,
3

25
,
2

26
,
12

tan ␪ ϭ 2,

1
5

52. sin ␪ ϭ Ϫ , cos ␪ ϭ tan ␪ ϭ 3, cot ␪ ϭ

tan ␪ ϭ Ϫ

1
3

2 26
,
5

sec ␪ ϭ Ϫ25

sec ␪ ϭ

5 26
,
12

cot ␪ ϭ Ϫ226

53. about 173.2 ft

54. 45Њ; 2 ϫ 45Њ or 90Њ yields the greatest value for sin 2␪.

55. 9 meters

56. I, II

57. II

58. III

59. Answers should include the following. • The cosine of any angle is x defined as , where x is r the x-coordinate of any point on the terminal ray of the angle and r is the distance from the origin to that point. This means that for angles with terminal sides to the left of the y-axis, the cosine is negative, and those with terminal sides to the right of the y-axis, the cosine is positive. Therefore, the cosine function can be used to model real-world data that oscillate between being positive and negative. 60. C

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• If we knew the length of the cable we could find the vertical distance from the top of the tower to the rider.
Then if we knew the height of the tower we could subtract from it the vertical distance calculated previously. This will leave the height of the rider from the ground.
5
2

5 23 b 2

61. a , Ϫ

62.

2

63. 300Њ

64.

900Њ

65. sin 28Њ ϭ
67. sin x Њ ϭ

x
,
12

5
,
13

Ϸ 286.5Њ

66. cos 43Њ ϭ

5.6

x
,
83

60.7

68. 635

23

69. (7, 2)

70. (Ϫ4, 3)

71. (5, Ϫ4)

72. 4.7

73. 15.1

74. 2.7

75. 32.9Њ

76. 20.6Њ

77. 39.6Њ

Lesson 13-4 Law of Sines
Pages 729–732
1. Sometimes; only when A is acute, a ϭ b sin A, or a Ͼ b and when A is obtuse, a Ͼ b.

2. Sample answer: A ϭ 42Њ, a ϭ 2.6 cm, b ϭ 3.2 cm
C
3.2 cm

A

2.6 cm

3.9 cm

B
C

3.2 cm

A

0.9 cm

355

2.6 cm

B
Algebra 2

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3. Gabe; the information given is of two sides and an angle, but the angle is not between the two sides, therefore the area formula involving sine cannot be used.

4. 57.5 in2

5. 6.4 cm2

6. C ϭ 30Њ, a Ϸ 2.9, c Ϸ 1.5

7. B ϭ 80Њ, a Ϸ 32.0, b Ϸ 32.6

8. B Ϸ 20Њ, A Ϸ 20Њ, a Ϸ 20.2
10. two; B Ϸ 42Њ, C Ϸ 108Њ, c Ϸ 5.7; B Ϸ 138Њ, C Ϸ 12Њ, c Ϸ 1.2

9. no solution

11. one; B ϭ 24Њ, C Ϸ 101Њ, c Ϸ 12.0

12. one; B Ϸ 19Њ, C Ϸ 16Њ, c Ϸ 8.7

13. 5.5 m

14. 43.1 m2

15. 19.5 yd2

16. 572.8 ft2

17. 62.4 cm2

18. 4.2 m2

19. 14.6 mi2

20. B ϭ 101Њ, c Ϸ 3.0, b Ϸ 3.4

21. C ϭ 73Њ, a Ϸ 55.6, b Ϸ 48.2

22. B Ϸ 21Њ, C Ϸ 37Њ, b Ϸ 13.1

23. B Ϸ 47Њ, C Ϸ 68Њ, c Ϸ 5.1

24. C ϭ 97Њ, a Ϸ 5.5, b Ϸ 14.4

25. A Ϸ 40Њ, B Ϸ 65Њ, b Ϸ 2.8

26. C Ϸ 67Њ, B Ϸ 63Њ, b Ϸ 2.9

27. A ϭ 20Њ, a Ϸ 22.1, c Ϸ 39.8

28. no

29. one; B Ϸ 36Њ, C Ϸ 45Њ, c Ϸ 1.8

30. two; B Ϸ 72Њ, C Ϸ 75Њ, c Ϸ 3.5; B Ϸ 108Њ, C Ϸ 39Њ, c Ϸ 2.3

31. no

32. one; B ϭ 90Њ, C ϭ 60Њ, c Ϸ 24.2

33. one; B Ϸ 18Њ, C Ϸ 101Њ, c Ϸ 25.8

34. two; B Ϸ 56Њ, C Ϸ 72Њ, c Ϸ 229.3; B Ϸ 124Њ,
C Ϸ 4Њ, c Ϸ 16.8

35. two; B Ϸ 85Њ, C Ϸ 15Њ, c Ϸ 2.4; B Ϸ 95Њ, C Ϸ 5Њ, c Ϸ 0.8

36. one; B Ϸ 23Њ, C Ϸ 129Њ, c Ϸ 14.1

37. two; B Ϸ 65Њ, C Ϸ 68Њ, c Ϸ 84.9; B Ϸ 115Њ, C Ϸ 18Њ, c Ϸ 28.3

38. 4.6 and 8.5 mi

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39. 7.5 mi from Ranger B,
10.9 mi from Ranger A

40. 690 ft

41. 107 mph

42a. 14.63 Ͻ b Ͻ 20
42b. b ϭ 14.63 or b Ն 20
42c. b Ͻ 14.63

43. Answers should include the following. • If the height of the triangle is not given, but the measure of two sides and their included angle are given, then the formula for the area of a triangle using the sine function should be used.
• You might use this formula to find the area of a triangular piece of land, since it might be easier to measure two sides and use surveying equipment to measure the included angle than to measure the perpendicular distance from one vertex to its opposite side.
1
• The area of ᭝ABC is ah.

44. D

2

C b a h A

sin B ϭ
Area ϭ
Area ϭ

B

c h or h ϭ c sin c 1 ah or
2
1 a (c sin B)
2

B

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47.

46.

49. 660Њ, Ϫ60Њ
51.

17␲
,
6

53.

23
2

48. 22

23
3

45. B ϭ 78Њ, a Ϸ 50.1, c Ϸ 56.1

50. 407Њ, Ϫ313Њ

7␲
6

55
221

52.

Ϫ

3
68

54. 780 ft

55. 5.6

56. 7.8

57. 39.4Њ

58. 136.0Њ

Lesson 13-5 Law of Cosines
Pages 735–738
1. Mateo; the angle given is not between the two sides, therefore the Law of Sines should be used.

2a. Use the Law of Cosines to find the measure of one angle. Then use the Law of
Sines or the Law of Cosines to find the measure of a second angle. Finally, subtract the sum of these two angles from 180Њ to find the measure of the third angle. 2b. Use the Law of Cosines to find the measure of the third side. Then use the Law of
Sines or the Law of Cosines to find the measure of a second angle. Finally, subtract the sum of these two angles from 180Њ to find the measure of the third angle. 358

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4. cosines; A Ϸ 76Њ, B Ϸ 69Њ, c Ϸ 6.5

15

9

13

5. sines; B ϭ 70Њ, a Ϸ 9.6, b ϭ 14

6. sines; C Ϸ 101Њ, B Ϸ 37Њ, c Ϸ 92.5

7. cosines; A ϭ 23Њ, B Ϸ 67Њ,
C Ϸ 90Њ

8. 19.5 m
10. sines; A ϭ 60Њ, b Ϸ 14.3, c Ϸ 11.2

9. 94.3Њ
11. cosines; A Ϸ 48Њ, B Ϸ 62Њ,
C Ϸ 70Њ

12. cosines; A Ϸ 46Њ, B Ϸ 74Њ,
C Ϸ 59.6

13. sines; B Ϸ 102Њ, C Ϸ 44Њ, b Ϸ 21.0

14. cosines; A Ϸ 56.8Њ, B Ϸ 82Њ, c Ϸ 11.5

15. sines; A ϭ 80Њ, a Ϸ 10.9, c Ϸ 5.4

16. cosines; A Ϸ 55Њ, C Ϸ 78Њ, b Ϸ 17.9

17. cosines; A Ϸ 30Њ, B Ϸ 110Њ,
C Ϸ 40Њ

18. no

19. sines; C Ϸ 77Њ, b Ϸ 31.7, c Ϸ 31.6

20. cosines; A Ϸ 103Њ, B Ϸ 49Њ,
C Ϸ 28Њ

21. no

22. cosines; A Ϸ 15Њ, B Ϸ 131Њ,
C Ϸ 34Њ

23. cosines; A Ϸ 52Њ, C Ϸ 109Њ, b Ϸ 21.0

24. sines; C ϭ 102Њ, b Ϸ 5.5, c Ϸ 14.4

25. cosines; A Ϸ 24Њ, B Ϸ 125Њ,
C Ϸ 31Њ

26. cosines; A Ϸ 107Њ, B Ϸ 35Њ, c Ϸ 13.8

27. cosines; B Ϸ 82Њ, C Ϸ 58Њ, a Ϸ 4.5

30. Since the step angle for the carnivore is closer to 180Њ, it appears as though the carnivore made more forward progress with each step than the herbivore did.

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31. 4.4 cm, 9.0 cm

32. about 1362 ft; about 81,919 ft 2

33. 91.6Њ

34. Since cos 90Њ ϭ 0, a 2 ϭ b 2 ϩ c 2 Ϫ 2bc cos A becomes a 2 ϭ b 2 ϩ c 2.

35. Answers should include the following. • The Law of Cosines can be used when you know all three sides of a triangle or when you know two sides and the included angle. It can even be used with two sides and the nonincluded angle. This set of conditions leaves a quadratic equation to be solved. It may have one, two, or no solution just like the SSA case with the
Law of Sines.
• Given the latitude of a point on the surface of
Earth, you can use the radius of the Earth and the orbiting height of a satellite in geosynchronous orbit to create a triangle. This triangle will have two known sides and the measure of the included angle. Find the third side using the Law of Cosines and then use the Law of
Sines to determine the angles of the triangle.
Subtract 90 degrees from the angle with its vertex on
Earth’s surface to find the angle at which to aim the receiver dish.

36. B

37. A

38. 100.0Њ

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39. Sample answer: 100.2Њ

40. By finding the measure of angle C in one step using the Law of Cosines, only the given information was used.
By finding this angle measure using the Law of
Cosines and then the Law of
Sines, a calculated value that was not exact was introduced; 100.0Њ.

41. one; B ϭ 46Њ, C ϭ 79Њ, c ϭ 9.6

42. no solution

12
5
, cos ␪ ϭ ,
13
13
12
13 tan ␪ ϭ , csc ␪ ϭ
5
12
13
5 sec ␪ ϭ , cot ␪ ϭ
5
12

44. sin ␪ ϭ

43. sin ␪ ϭ

45. sin ␪ ϭ

26
,
4

tan ␪ ϭ sec ␪ ϭ

215
,
5

2 210
,
5

cos ␪ ϭ

210
,
4

csc ␪ ϭ

csc ␪ ϭ cot ␪ ϭ

2 26
,
3

cot ␪ ϭ

47. {x 0 x Ͼ Ϫ0.6931}

cos ␪ ϭ

215
3

7 265
,
65

4 265
,
65

265
,
7
4
7

265
,
4
7
4

tan ␪ ϭ ,

sec ␪ ϭ

46. 1.3863

48. 4.3891

49. 405, Ϫ315Њ

50. 390Њ, Ϫ330Њ

51. 540Њ, Ϫ180Њ

52.

5␲
,
2

54.

10␲
,
3

53.

19␲
,
6

5␲
6

Ϫ

361

3␲
2

Ϫ

2␲
3

Ϫ

Algebra 2

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1. sin ␪ ϭ

3 213
,
13

2 213
,
13

cos ␪ ϭ Ϫ

213
,
2

3
2

tan ␪ ϭ Ϫ , csc ␪ ϭ sec ␪ ϭ Ϫ

2 23
3

Chapter 13
Practice Quiz 2
Page 738
213
,
3

2. Ϫ

2
3

cot ␪ ϭ Ϫ

3. 27.7 m2

4. two; B Ϸ 27Њ; C Ϸ 131Њ; c Ϸ 30.2; B Ϸ 153Њ;
C Ϸ 5Њ; c Ϸ 3.5

5. cosines; c Ϸ 15.9, A Ϸ 59Њ,
B Ϸ 43Њ

Lesson 13-6 Circular Functions
Pages 742–745
1. The terminal side of the angle ␪ in standard position must intersect the unit circle at P (x, y).

2. Sample answer: the motion of the minute hand on a clock; 60 s

3. Sample answer: The graphs have the same shape, but cross the x-axis at different points. 4. sin ␪ ϭ Ϫ , cos ␪ ϭ

5. sin ␪ ϭ

6.

22
;
2

cos ␪ ϭ

22
2

1
2

7. Ϫ

23
2

12
13

5
13

8. 720Њ

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10.

9. 2 s

h
3

O

2

1

3

Ϫ3

4
5

5
13

3
5

15. sin ␪ ϭ

23
;
2

15
;
17

cos ␪ ϭ

8
17
1
2

12
13

23
2

16. sin ␪ ϭ 0.8; cos ␪ ϭ 0.6

cos ␪ ϭ Ϫ

23
2

1
2

18.

19. Ϫ1

20.

21. 1

22. Ϫ

25.

1
2

14. sin ␪ ϭ Ϫ ; cos ␪ ϭ

17. Ϫ

23.

13
2

t

12. sin ␪ ϭ Ϫ ; cos ␪ ϭ Ϫ

11. sin ␪ ϭ ; cos ␪ ϭ Ϫ
13. sin ␪ ϭ

4

22
2

26. 23

1 Ϫ 23
2
1
4

9
4

24.

27. Ϫ323

28. 1

29. 6

30. 9

31. 2␲

32. 8

33.

1
440

34.

s

y
1

O
Ϫ1

363

1
440

1
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Algebra 2

x

Chapter 13

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1 13
1 13 b, aϪ , b, 2 2
2 2

35. a ,

13 b 2

1
2

(Ϫ1, 0), aϪ , Ϫ
1
a ,
2

Ϫ

13 b, 2

36. The population is around 425 near the 60th day of the year.
It rises to around 625 in
May/June. It falls to around 425 again by August/September.
It continues to drop to around
225 in November/December.

y x 41. 23

38. tan ␪

43. sine: D ϭ {all reals},
R ϭ {Ϫ1 Յ y Յ 1}; cosine: D ϭ {all reals},
R ϭ {Ϫ1 Յ y Յ 1}

44. Answers should include the following. • Over the course of one period both the sine and cosine function attain their maximum value once and their minimum value once.
From the maximum to the minimum the functions decrease slowly at first, then decrease more quickly and return to a slow rate of change as they come into the minimum. Similarly, the functions rise slowly from their minimum. They begin to increase more rapidly as they pass the halfway point, and then begin to rise more slowly as they increase into the maximum. Annual temperature fluctuations behave in exactly the same manner.

37.

23
3

40. Ϫcot ␪

39. Ϫ

42. Ϫ

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• The maximum value of the sine function is 1 so the maximum temperature would be 50 ϩ 25(1) or
75Њ F. Similarly, the minimum value would be
50 ϩ 25(Ϫ1) or 25Њ F. The average temperature over this time period occurs when the sine function takes on a value of 0. In this case that would be
50Њ F.
23
3

45. A

46.

47. cosines; c Ϸ 12.4, B Ϸ 59Њ,
A Ϸ 76Њ

48. cosines; A Ϸ 34Њ, B Ϸ 62Њ,
C Ϸ 84Њ

49. 27.0 in2

50. 12.5 m2

51. 6800

52. 9500

53. 5000

54. 5000

55. 250

56. 50

57. does not exist

58.

59. 8

60. 4x Ϫ 5

61. 2x ϩ 9

62. 5y 2 Ϫ 4y ϩ 4 Ϫ

63. 2y ϩ 7 ϩ

5 yϪ3 64
3

11 yϩ1 64. 20Њ

65. 110Њ

66. 73Њ

67. 80Њ

68. 56Њ

69. 89Њ

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Lesson 13-7

Inverse Trigonometric Functions
Pages 749–751

1. Restricted domains are denoted with a capital letter.

22
2

22
;
2

Cos 45Њ ϭ
CosϪ1

ϭ 45Њ

3. They are inverses of each other. 4. ␪ ϭ Arctan x

5. ␣ ϭ Arccos 0.5

6. 45Њ

7. 0Њ

8. Ϫ Ϸ Ϫ0.52

6

9. ␲ Ϸ 3.14

10. 0.22

11. 0.75

12. 0.66

13. 0.58

14. 30Њ

15. ␤ ϭ Arcsin ␣

16. a ϭ Arctan b

17. y ϭ Arccos x

18. 30Њ ϭ Arcsin

19. Arccos y ϭ 45Њ

Ϫ
20. Arctan a b ϭ x

21. 60Њ

22. 30Њ

23. 45Њ

24. 30Њ

25. 45Њ

26. 90Њ

27. 2.09

28. does not exist

29. 0.52

30. 0.52

31. 0.5

32. 0.66

33. 0.60

34. 0.5

35. 0.8

36. 0.81

37. 0.5

38. 3

39. Ϫ0.5

40. 1.57

41. 0.71

42. does not exist

43. 0.96

44. 0.87

1
2

4
3

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45. 60Њ south of west

46. 83Њ

47. No; with this point on the terminal side of the throwing angle ␪, the measure of ␪ is found by solving the equation

48. 60Њ

tan ␪ ϭ

17
. Thus
18

␪ ϭ tanϪ1

or about 43.4Њ, which is greater than the 40Њ requirement. 17
18

49. 31Њ

50. 102Њ

51. Suppose P (x 1, y 1) and
Q (x 2, y 2) lie on the line y ϭ mx ϩ b. Then m ϭ y2 Ϫ y1
. The tangent of

52. Trigonometry is used to determine proper banking angles. Answers should include the following.
• Knowing the velocity of the cars to be traveling on a road and the radius of the curve to be built, then the banking angle can be determined. First find the ratio of the square of the velocity to the product of the acceleration due to gravity and the radius of the curve. Then determine the angle that had this ratio as its tangent. This will be the banking angle for the turn.
• If the speed limit were increased and the banking angle remained the same, then in order to maintain a safe road the curvature would have to be decreased. That is, the radius of the curve would also have to increase, which would make the road less curved.

x2 Ϫ x1

the angle ␪ the line makes with the positive x-axis is opp equal to the ratio or adj

y2 Ϫ y1
.
x2 Ϫ x 1

Thus tan ␪ ϭ m.

y

Q (x 2, y 2)
P (x , y 1)
O

x2 Ϫ x1

y2 Ϫ y1 x y ϭ mx ϩ b m 53. 37Њ

54. D
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23
1 22 23
1 22
Ϫ
Ϫ1
1 Ϫ Ϫ
2 2
2
2
2
2
␲ ␲ ␲
␲ ␲ ␲

y
2 2 2
2 2 2
2
2
2

56. SinϪ1 x ϩ CosϪ1 x ϭ for
2
all values of x.

55.

x 0

23
2

57. From a right triangle perspective, if an acute angle
␪ has a given sine x, then the complementary angle

Ϫ ␪ has that same value
2
as its cosine. This can be verified by looking at a right triangle. Therefore, the sum of the angle whose sine is x and the angle whose cosine

is x should be .

58.

59. Ϫ1

60. 1

61. sines; B Ϸ 69Њ, C Ϸ 81Њ, c Ϸ 6.1 or B Ϸ 111Њ, C Ϸ 39Њ,
C Ϸ 3.9

62. cosines; A ϭ 13Њ, B ϭ 77Њ,
C ϭ 90Њ

63. 46, 39

64. Ϫ22, Ϫ57

65. 11, 109

66. 2.5 s

2

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Chapter 14 Trigonetmetric Graphs and Identities
Lesson 14-1 Graphing Trigonometric Functions
Pages 766–768
1. Sample answer: Amplitude is half the difference between the maximum and minimum values of a graph; y ϭ tan ␪ has no maximum or minimum value.

2. Sample answer: The graph repeats itself every 180Њ.

3. Jamile; The amplitude is 3 and the period is 3␲.

4. amplitude: ; period 360Њ or 2␲

1
2

y
2.5
2
1.5
1
0.5
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ1
Ϫ1.5
Ϫ2
Ϫ2.5

y
2
1.5
1
0.5

y ϭ 2 sin ␪

O

90˚ 180˚ 270˚

6. amplitude: ; period 360Њ or 2␲

y

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

O

2
3

5. amplitude: 2; period: 360Њ or 2␲
5
4
3
2
1

1

y ϭ 2 sin ␪

90˚ 180˚ 270˚

Ϫ0.5
Ϫ1
Ϫ1.5
Ϫ2

369

O

2 cos ␪
3

90˚ 180˚ 270˚ 360˚

Algebra 2

Chapter 14

PQ245-6457F-P14[369-410].qxd

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8. amplitude: does not exist; period: 180Њ or ␲

7. amplitude: does not exist; period: 180Њ or ␲

y

y

2
1.5
1
0.5

2
1.5
1
0.5
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ1
1
Ϫ1.5
y ϭ 4 tan ␪
Ϫ2

O

90˚ 180˚ 270˚

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ1
Ϫ1.5 y ϭ csc 2␪
Ϫ2

10. amplitude: 4; period: 480Њ or

9. amplitude: 4; period: 180Њ or ␲ y 5
4
3
2
1

5
4
3
2
1

90˚ 180˚ 270˚ 360˚

y
1.25
1
0.75
0.5
0.25

30˚

60˚

O
Ϫ90˚
Ϫ0.5
Ϫ0.75
Ϫ1
Ϫ1.25

90˚ 120˚ 150˚
1

y ϭ 2 sec 3␪

13. 12 months; Sample answer:
The pattern in the population will repeat itself every 12 months. ©Glencoe/McGraw-Hill

3
4

2␲
3

2
1.5
1
0.5
Ϫ60˚ Ϫ30˚
Ϫ1
Ϫ1.5
Ϫ2

90˚ 180˚ 270˚ 360˚ 450˚

12. amplitude: ; period: 720Њ or 4␲

y

O

3
4

y ϭ 4 cos ␪

O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5

11. amplitude: does not exist; period: 120Њ or

8␲
2

y

y ϭ 4 sin 2␪

O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5

90˚ 180˚ 270˚

3

1

y ϭ 4 cos 2 ␪

90˚ 180˚ 270˚ 360˚ 450˚

14. 4250; June 1

370

Algebra 2

Chapter 14

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15. amplitude: 3; period: 360Њ or 2␲

16. amplitude: 5; period: 360Њ or 2␲

y
5
4
3
2
1
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

y
5
4
3
2
1

y ϭ 3 sin ␪

O

90˚ 180˚ 270˚

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

y

y
10
8
6
4
2

5
4
3
2
1
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
y ϭ 2 csc ␪
Ϫ3
Ϫ4
Ϫ5

90˚ 180˚ 270˚

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
y ϭ 2 tan ␪
Ϫ10

1
5

y
1
0.8
0.6
0.4
0.2

90˚ 180˚ 270˚

y
10
8
6
4
2

1 y ϭ 5 sin ␪

O

O

20. amplitude: does not exist; period: 360Њ or 2␲

19. amplitude: ; period: 360Њ or 2␲

90˚ 180˚ 270˚

18. amplitude: does not exist; period: 180Њ or ␲

17. amplitude: does not exist; period: 360Њ or 2␲

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ0.4
Ϫ0.6
Ϫ0.8
Ϫ1

y ϭ 5 cos ␪

90˚ 180˚ 270˚

O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ4
Ϫ6
1 y ϭ 3 sec ␪
Ϫ8
Ϫ10

371

90˚ 180˚ 270˚

Algebra 2

Chapter 14

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21. amplitude: 1; period 90Њ or

2

22. amplitude: 1; period: 180Њ or ␲ y y
5
4
3
2
1
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

5
4
3
2
1

y ϭ sin 4␪

O

90˚ 180˚ 270˚

period: 36Њ or

O
Ϫ60˚ Ϫ30˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

5

y

30˚

60˚

O
Ϫ72˚ Ϫ54˚ Ϫ36˚ Ϫ18˚
Ϫ2
Ϫ3
Ϫ4
y ϭ cot 5␪
Ϫ5

25. amplitude: does not exist; period: 540Њ or 3␲

18˚

36˚

54˚

72˚

26. amplitude: does not exist; period: 360Њ or 2␲ y y

10
8
6
4
2

10
8
6
4
2

5
4
3
2
1

5
4
3
2
1

Ϫ810˚ Ϫ540˚Ϫ270˚
Ϫ4
Ϫ6
1
y ϭ 4 tan 3 ␪
Ϫ8
Ϫ10

90˚ 180˚ 270˚

24. amplitude: does not exist;

2␲
3

y

y ϭ sec 3␪

O

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

23. amplitude: does not exist; period: 120Њ or

y ϭ sin 2␪

O

270˚ 540˚ 810˚

O
Ϫ540˚ Ϫ360˚ Ϫ180˚
Ϫ4
Ϫ6
1
y ϭ 2 cot 2 ␪
Ϫ8
Ϫ10

372

180˚ 360˚ 540˚

Algebra 2

Chapter 14

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27. amplitude: 6; period: 540Њ or 3␲

28. amplitude: 3; period: 720Њ or 4␲

y
10
8
6
4
2
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10

y
2
y ϭ 6 sin 3 ␪

O

90˚ 180˚ 270˚

10
8
6
4
2
O
Ϫ540˚Ϫ360˚
Ϫ180˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10

29. amplitude: does not exist; period: 720Њ or 4␲

2

y

10
8
6
4
2
O
Ϫ540˚Ϫ360˚Ϫ180˚
Ϫ4
1 y ϭ 3 csc 2 ␪
Ϫ6
Ϫ8
Ϫ10

180˚ 360˚ 540˚

10
8
6
4
2

Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ4
Ϫ6
1
y ϭ 2 cot 2␪
Ϫ8
Ϫ10

31. amplitude: does not exist; period: 180Њ or ␲

O

45˚

90˚ 135˚

8
9

32. amplitude: ; period: 600Њ or
10␲
3

y
10
8
6
4
2

180˚ 360˚ 540˚

30. amplitude: does not exist;

period: 90Њ or

y

O
Ϫ270˚Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
2y ϭ tan ␪
Ϫ8
Ϫ10

1

y ϭ 3 cos 2 ␪

y

90˚ 180˚ 270˚

5
4
3
2
1

Ϫ540˚Ϫ360˚Ϫ180˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

373

3
2
3 y ϭ 3 sin 5 ␪
4

O

180˚ 360˚ 540˚

Algebra 2

Chapter 14

PQ245-6457F-P14[369-410].qxd

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33.

34. y y
5
4
3
2
1
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

35.

3
5

5
4
3
2
1

3

y ϭ 5 sin 4␪
O

45˚

90˚ 135˚

7

y ϭ 8 cos 5␪

O

45˚

Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

sin 4␪

1
107

7
8

90˚ 135˚

cos 5␪

36. y ϭ 0.25 sin 128␲t, y ϭ 0.25 sin 512␲t, y ϭ 0.25 sin 1024␲t

37. Sample answer: The amplitudes are the same. As the frequency increases, the period decreases.

38. f (x ) ϭ cos x and f (x ) ϭ sec x f (x )
5
4
3
2
1

f (x ) ϭ cos x f (x ) ϭ cos (Ϫx )

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

x

90˚ 180˚ 270˚

f (x ) f (x ) ϭ sec x f (x ) ϭ sec (Ϫx )

5
4
3
2
1

O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

374

90˚ 180˚ 270˚

Algebra 2

x

Chapter 14

PQ245-6457F-P14[369-410].qxd

7/24/02

39. y ϭ 2 sin

5

2:03 PM

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40.

t

y
2.5
2
1.5
1
0.5
Ϫ0.5
Ϫ1
Ϫ1.5
Ϫ2
Ϫ2.5

y ϭ 2 sin 5 t

O

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

41. about 1.9 ft

42. Sample answer: Tides display periodic behavior.
This means that their pattern repeats at regular intervals.
Answers should include the following information.
• Tides rise and fall in a periodic manner, similar to the sine function.
• In f (x ) ϭ a sin bx, the amplitude is the absolute value of a.

43. A

t

44. C

22
2

22
2

45. 90Њ

46. Ϫ90Њ

47. 45Њ

48.

49.

50.

51.

13
16

1
2

52. 3, 11, 27, 59, 123

375

Algebra 2

Chapter 14

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53.

2:03 PM

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54.

y
15
13
11
9
7
5
3
1
Ϫ8

y ϭx

15
13
11
9
7
5
3
1

2

y ϭ 3x 2

O

Ϫ4

y

4

8

x

Ϫ8

Ϫ4

56.

y

y ϭ 2(x ϩ 1)2
Ϫ8

Ϫ4

15
13
11
9
7
5
3
1
O

y ϭ 2x 2

4

4

8

x

Lesson 14-2

x

8

y ϭ 3x 2 Ϫ 4

y

y ϭ x2 ϩ 2

Ϫ8

Ϫ4

15
13
11
9
7
5
3
1
O

y ϭ (x Ϫ 3)2 ϩ 2
4

x

8

Ϫ3
Ϫ5

Ϫ3
Ϫ5

Translations of Trigonometric Graphs
Pages 774–776

1. vertical shift: 15; amplitude: 3; period: 180Њ; phase shift: 45Њ

O
Ϫ3
Ϫ5

Ϫ3
Ϫ5

55.

y ϭ 3x 2

2. The midline of a trigonometric function is the line about which the graph of the function oscillates after a vertical shift.

376

Algebra 2

Chapter 14

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3. Sample answer: y ϭ sin (␪ ϩ 45Њ)

4. 1; 2␲;

2
y

(

y ϭ sin ␪ Ϫ 2

Ϫ

5. no amplitude; 180Њ; Ϫ60Њ

3␲
2

)

Ϫ␲ Ϫ ␲
2

5
4
3
2
1
O

2

Ϫ2
Ϫ3
Ϫ4
Ϫ5

3␲
2

6. 1; 360Њ; 45Њ y y
5
4
3
2
1

0.75

y ϭ cos (␪ Ϫ 45˚)

0.5

O
Ϫ270˚Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3 y ϭ tan (␪ ϩ 60˚) Ϫ4
Ϫ5

90˚ 180˚ 270˚

0.25

O
Ϫ45˚
Ϫ0.25

45˚ 90˚ 135˚ 180˚ 225˚

Ϫ0.5
Ϫ0.75

3

8.

7. no amplitude; 2␲; Ϫ
(

y ϭ sec ␪ ϩ 3

Ϫ

3␲
2

)

1
4

y ϭ ; 1; 360Њ y y
4
3
2
1

O
Ϫ␲ Ϫ ␲ Ϫ1
2
Ϫ2
Ϫ3
Ϫ4

1
;
4

2

3␲
2

5
4
3
2
1

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

377

1

y ϭ cos ␪ ϩ 4

90˚ 180˚ 270˚

Algebra 2

Chapter 14

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9. Ϫ5; y ϭ Ϫ5; no amplitude;
360Њ

10. 4; y ϭ 4; no amplitude; 180Њ y 7

y

6

10
8
6
4
2

5
4

O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ4
Ϫ6
Ϫ8 y ϭ sec ␪ Ϫ 5
Ϫ10

1
Ϫ135˚ Ϫ90˚ Ϫ45˚

y ϭ sin ␪ ϩ 0.25

0.5
90˚

180˚

270˚

45˚

y ϭ 3 sin [2(␪ Ϫ 30˚)] ϩ 10
14
12
10
8
6
4
2

1

Ϫ0.5

O
Ϫ1

90˚

135˚

12. 10; 3; 180Њ; 30Њ

y

O

2

90˚ 180˚ 270˚

11. 0.25; y ϭ 0.25; 1; 360Њ
1.5

3

y ϭ tan ␪ ϩ 4

360˚

y

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4

Ϫ1

90˚ 180˚ 270˚

Ϫ1.5

␲ ␲
2 4

14. 1; no amplitude; ;

13. Ϫ6; no amplitude; 60Њ; Ϫ45Њ y y

1
Ϫ45˚

O
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8
Ϫ9
Ϫ10
Ϫ11

45˚

4
3
2
1

Ϫ

3␲
8

Ϫ

4

Ϫ

␲ Ϫ1O
8

Ϫ2
Ϫ3

1
Ϫ4 y ϭ 2 sec 4 ␪ Ϫ 4 ϩ 1

y ϭ 2 cot (3␪ ϩ 135˚) Ϫ 6

[(

378

8

4

3␲
8

)]

Algebra 2

Chapter 14

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2
3

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Page 379 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:

6

15. Ϫ2; ; 4␲; Ϫ

16. 4; 1; 4 s

y
1
O
Ϫ3␲ Ϫ2␲ Ϫ␲
Ϫ1

2␲

3␲

Ϫ2
Ϫ3

2

[ 1(

y ϭ 3 cos 2 ␪ ϩ 6

)] Ϫ 2

17. h ϭ 4 Ϫ cos t or
2
h ϭ 4 Ϫ cos 90Њt

18.

h

h ϭ 4 Ϫ cos 2 t

6
5
4
3
2
1
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4

19. 1; 360Њ; Ϫ90Њ

1

2

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

4

t

20. no amplitude; 180Њ; 30Њ y y
5
4
3
2
1

3

5
4
3
2
1

y ϭ cos (␪ ϩ 90˚)

90˚ 180˚ 270˚

O
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

379

45˚

90˚ 135˚

y ϭ cot (␪ Ϫ 30˚)

Algebra 2

Chapter 14

PQ245-6457F-P14[369-410].qxd

21. 1; 2␲;

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3

4

22. 1; 2␲; Ϫ

y

y
5
4
3
2
1
Ϫ

3␲
2

(

y ϭ sin ␪ Ϫ 4

5
4
3
2
1

)

O
Ϫ␲ Ϫ ␲
2 Ϫ2
Ϫ3
Ϫ4
Ϫ5

2

3␲
2

Ϫ

3␲
2

O
Ϫ␲ Ϫ ␲
2 Ϫ2
Ϫ3
Ϫ4
Ϫ5

5
4
3
2
1

5
4
3
2
1
45˚

90˚ 135˚

3␲
2

y ϭ 3 sin (␪ Ϫ 75˚)

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

1 y ϭ 4 tan (␪ ϩ 22.5˚)

90˚ 180˚ 270˚

26. 2; y ϭ 2; no amplitude; 360Њ

25. Ϫ1; y ϭ Ϫ1; 1; 360Њ y y

5
4
3
2
1

2

)

y

y

O
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

24. 3; 360Њ; 75Њ

23. no amplitude; 180Њ; Ϫ22.5Њ

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3 y ϭ sin ␪ Ϫ 1
Ϫ4
Ϫ5

(

y ϭ cos ␪ ϩ 3

5
4
3
2
1

O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
Ϫ3
y ϭ sec ␪ ϩ 2
Ϫ4
Ϫ5

90˚ 180˚ 270˚

380

90˚ 180˚ 270˚

Algebra 2

Chapter 14

PQ245-6457F-P14[369-410].qxd

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Page 381 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:

3
4

y

360Њ

2
1

y

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8

29.

1
;
2

90˚ 180˚ 270˚

5
4
3
2
1

O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
Ϫ3
3 y ϭ csc ␪ Ϫ 4
Ϫ4
Ϫ5

y ϭ cos ␪ Ϫ 5

1 1
2 2

y

y
5
4
3
2
1

1

10
8
6
4
2

1

y ϭ 2 sin ␪ ϩ 2

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

90˚ 180˚ 270˚

32.

3␲

Ϫ2
4

O

Ϫ4

(

4

2

3␲
4

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

)

2 y ϭ 3 cos (␪ Ϫ 50˚) ϩ 2

90˚ 180˚ 270˚

translation 50Њ right and
2 units up with an amplitude
2
of unit

translation units left and
4
5 units up

90˚ 180˚ 270˚

y
5
4
3
2
1

y

y ϭ 5 ϩ tan ␪ ϩ 4

y ϭ 6 cos ␪ ϩ 1.5

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10

31.

Ϫ

90˚ 180˚ 270˚

30. 1.5; y ϭ 1.5; 6; 360Њ

y ϭ ; ; 360Њ

18
16
14
12
10
8
6
4
2

3
4

28. Ϫ ; y ϭ Ϫ ; no amplitude;

27. Ϫ5; y ϭ Ϫ5; 1; 360Њ

3

381

Algebra 2

Chapter 14

PQ245-6457F-P14[369-410].qxd

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Page 382 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:

33. 1; 2; 120Њ; 45Њ

34. Ϫ5; 4; 180Њ; Ϫ30Њ y 5
4
3
2
1

y
10
8
6
4
2

y ϭ 2 sin [3(␪ Ϫ 45˚)] ϩ 1

O

90˚ 180˚ 270˚

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10

35. Ϫ3.5; does not exist; 720Њ;
Ϫ60Њ

90˚ 180˚ 270˚

y ϭ 4 cos [2(␪ ϩ 30˚)] Ϫ 5

36. 0.75; does not exist; 270Њ; 90Њ y 20
16
12
8
4

y
8
6
4
2
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12

O

90˚ 180˚ 270˚

[1 (

Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ8
Ϫ12
Ϫ16
Ϫ20

)]

y ϭ 3 csc 2 ␪ ϩ 60˚ Ϫ 3.5

O

90˚ 180˚ 270˚

[2(

)]

y ϭ 6 cot 3 ␪ Ϫ 90˚ ϩ 0.75

1
4

38. Ϫ4; does not exist; 30Њ;Ϫ22.5Њ

37. 1; ; 180Њ; 75Њ

y

y
5
4
3
2
1
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

2
1

1 y ϭ 4 cos (2␪ Ϫ 150˚) ϩ 1

Ϫ22.5˚
90˚ 180˚ 270˚

O

22.5˚

Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8

2 y ϭ 5 tan (6␪ ϩ 135˚) Ϫ 4

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2␲
3

4

40. 4; does not exist; 6␲; Ϫ

39. 3; 2; ␲; Ϫ

y

y
8
7
6
5
4
3
2
1
Ϫ

3␲
2

[(

O

Ϫ␲ Ϫ ␲

3␲
2

2

2 Ϫ2

16
14
12
10
8
6
4
2

y ϭ 3 ϩ 2 sin 2 ␪ ϩ 4

)]

O
Ϫ4␲ Ϫ2␲
Ϫ4

2␲

[1(

4␲
2␲

y ϭ 4 ϩ sec 3 ␪ ϩ 3

41.

42.

1

5
4
3
2
1
Ϫ

3␲
2

Ϫ␲ Ϫ ␲
2

y y ϭ 3 Ϫ 2 cos ␪
1
y ϭ 3 ϩ cos (␪ ϩ ␲)

2

Ϫ2
Ϫ3
Ϫ4
Ϫ5

3␲
2

)]

y
5
4
3
2
1

2

O

O
Ϫ4␲ Ϫ2␲
Ϫ2
Ϫ3
Ϫ4
Ϫ5

The graphs are identical.

[ 1 ( ␲ )]
1
3␲ y ϭ cos [ 4 (␪ ϩ 2 )] y ϭ Ϫsin 4 ␪ Ϫ 2

2␲

4␲

The graphs are identical.
43. c

44. 180; 5 yr

45. 300; 14.5 yr

46. Sample answer: When the prey (mouse) population is at its greatest the predator will consume more and the predator population will grow while the prey population falls.

47. h ϭ 9 ϩ 6 sin c (t Ϫ 1.5)d

48. a ϭ Ϫ1, b ϭ 1, h ϭ

49. Sample answer: You can use changes in amplitude and period along with vertical and horizontal shifts to show an animal population’s starting point and display changes to that population over a period of

50. B

9

383

2

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time. Answers should include the following information.
• The equation shows a rabbit population that begins at 1200, increases to a maximum of 1450, then decreases to a minimum of 950 over a period of 4 years.
• Relative to y ϭ a cos bx, y ϭ a cos bx ϩ k would have a vertical shift of k units, while y ϭ a cos
[b (x Ϫ h)] has a horizontal shift of h units.
52. amplitude: does not exist; period: 360Њ or 2␲

51. D

y
5
4
3
2
1
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
Ϫ3
y ϭ 3 csc ␪
Ϫ4
Ϫ5

period: 270Њ or

y

3␲
2

y

5
4
3
2
1

10
8
6
4
2

␪ y ϭ sin 2

O

90˚ 180˚ 270˚

Ϫ360˚ Ϫ180˚
Ϫ4
Ϫ6
2
y ϭ 3 tan ␪
3
Ϫ8
Ϫ10

O

180˚ 360˚

56. 0.57

55. 0.75

54. amplitude: does not exist;

53. amplitude: 1; period: 720Њ or
4␲

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

90˚ 180˚ 270˚

384

Algebra 2

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57. 0.83

58. 0.8

59. 35

60. 2.29

61. 0.66

62. 0.66

5a Ϫ 13
(a Ϫ 2)(a Ϫ 3)

64. Ϫ

65.

3y 2 ϩ 10y ϩ 5
2(y Ϫ 5)(y ϩ 3)

66. Ϫ

23
3

67. Ϫ1
69.
71.

23
2

1
4

63.

22
2

68. Ϫ1

1
2

70. 0
72. Ϫ

73. 1

Lesson 14-3 Trigonometric Identities
Pages 779–781
1. Sample answer: The sine function is negative in the third and fourth quadrants.
Therefore, the terminal side of the angle must lie in one of those two quadrants.

2. Sample answer: Pythagorean identities are derived by applying the Pythagorean
Theorem to trigonometric concepts. 3. Sample answer: Simplifying a trigonometric expression means writing the expression as a numerical value or in terms of a single trigonometric function, if possible.

4. Ϫ

7. 22
5
4

6.

5. Ϫ

3
5

8. 1

9. tan2 ␪

10. sec ␪
12. sin ␪ ϭ cos ␪

11. csc ␪

23
3

385

v2 gR Algebra 2

Chapter 14

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13.

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14.

23
2

15. Ϫ25
17.
19.
21.

5
4

3
4

25
3

16. 2 22

1
2

18.

3
5

3 25
5

20. Ϫ

4 27
7

22.

4
5

4 217
17

23. Ϫ

24. Ϫ

25. cot ␪

26. 1

27. cos ␪

28. sin ␪

29. 2

30. Ϫ3

31. cot2 ␪

32. tan ␪

33. 1

34. cot2 ␪

35. csc2 ␪

36. 1

38. about 4 m/s

40. E ϭ

I tan ␪ cos ␪
E
I sin ␪ simplifies to E ϭ
.
R2

42. P ϭ I 2R sin2 2␲ft

41. No; R 2 ϭ

43. P ϭ I 2R Ϫ

I 2R
.
1 ϩ tan2 2␲ft

44.

45. Sample answer: You can use equations to find the height and the horizontal distance of a baseball after it has been hit. The equations involve using the initial angle the ball makes with the ground with the sine function.
Answers should include the following information.

I cos ␪
R2

9
16

46. B

386

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• Both equations are quadratic in nature with a leading negative coefficient.
Thus, both are inverted parabolas which model the path of a baseball.
• model rockets, hitting a golf ball, kicking a rock
47. A

48. Ϫ1; y ϭ Ϫ1; 1; 360Њ y 5
4
3
2
1
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

49. 12; y ϭ 12; no amplitude; 180Њ
20

y ϭ sin ␪ Ϫ 1

y
5
4
3
2
1

15
10
5

50. amplitude: does not exist; period: 180Њ or ␲

y

O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ5
y ϭ tan ␪ ϩ 12

90˚ 180˚ 270˚

90˚ 180˚ 270˚

O
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3 y ϭ csc 2␪
Ϫ4
Ϫ5

387

45˚

90˚ 135˚

Algebra 2

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51. amplitude: 1; period: 120Њ or

2␲
3

52. amplitude: does not exist; period: 36Њ or

y
5
4
3
2
1

5

y
5
4
3
2
1

y ϭ cos 3␪

O
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

45˚

90˚ 135˚

1

y ϭ 3 cot 5␪

O

Ϫ22.5˚

22.5˚

Ϫ2
Ϫ3
Ϫ4
Ϫ5

1
6

53. 93

54. y ϭ Ϫ (x Ϫ 11)2 ϩ

55. Symmetric (ϭ)

56. Substitution (ϭ)

57. Multiplication (ϭ)

1
2

58. Substitution (ϭ)

Chapter 14
Practice Quiz 1
Page 781
1.

3
,
4

2. Ϫ5, 2, 8␲,

720Њ or 4␲ y 5
4
3
2
1

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

y
3

6
4
2

1

y ϭ 4 sin 2 ␪

O

90˚ 180˚ 270˚

O
Ϫ4␲ Ϫ2␲
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
Ϫ14

25
2
3
5

3. Ϫ
5.

4

4. Ϫ

388

213
3

2␲

4␲

[1 (

y ϭ 2 cos 4 ␪ Ϫ 4

Algebra 2

)] Ϫ 5

Chapter 14

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Lesson 14-4

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Verifying Trigonometric Identities
Pages 784–785

?

1. sin ␪ tan ␪ ϭ sec ␪ Ϫ cos ␪
?

sin ␪ tan ␪ ϭ

1 cos ␪

2. Sample answer: Use various identities, multiply or divide terms to form an equivalent expression, factor, and simplify rational expressions.

Ϫ cos ␪

sec ␪ ϭ

1 cos ␪

1 cos2 ␪
Ϫ
, cos ␪ cos ␪
Multiply by the LCD, cos ␪.
? 1 Ϫ cos2 ␪ tan ␪ ϭ cos ␪
Subtract.
1 Ϫ cos 2 ␪
? sin2 ␪ tan ␪ ϭ ϭ sin2 ␪ cos ␪ sin ␪
?
tan ␪ ϭ sin ␪ ؒ cos ␪
Factor.
?

sin ␪ tan ␪ ϭ sin ␪ sin ␪ sin ␪

sin ␪ tan ␪ ϭ sin ␪ tan ␪

sin ␪ ϭ tan ␪ cos ␪
?

3. Sample answer: sin2 ␪ ϭ 1 ϩ cos2 ␪; it is not an identity because sin 2 ␪ ϭ 1 Ϫ cos2 ␪.
5.

4. tan ␪(cot ␪ ϩ tan ␪) ϭ sec 2 ␪
?

1 ϩ tan2 ␪ ϭ sec2 ␪ sec2 ␪ ϭ sec2 ␪

?

tan2 ␪ cos2 ␪ ϭ 1 Ϫ cos2 ␪ sin2 ␪ cos2 ␪

6.

?

ؒ cos2 ␪ ϭ sin2 ␪ sin2 ␪ ϭ sin2 ␪

cos2 ␪
1 Ϫ sin ␪
1 Ϫ sin2 ␪
1 Ϫ sin ␪
(1 Ϫ sin ␪)(1 ϩ sin ␪)
1 Ϫ sin ␪

?

ϭ 1 ϩ sin ␪
?

ϭ 1 ϩ sin ␪
?

ϭ 1 ϩ sin ␪

1 ϩ sin ␪ ϭ 1 ϩ sin ␪

389

Algebra 2

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1 ϩ tan2 ␪ csc2 ␪

ϭ tan2 ␪

sec2 ␪ csc2 ␪

7.

Page 390 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:

ϭ tan2 ␪

?

8.

?

cos ␪

1 cos 2 ␪ ? ϭ 1
?

2

tan2 ␪

sin ␪ cos ␪

sin ␪ sec ␪ sin ␪ sec ␪ sin ␪ sec ␪

2

ؒ sin ␪ ϭ tan ␪ tan2 ␪ ϭ tan2 ␪

9.

?

ϭ
?

ϭ
?

ϭ ϭ ?

ϭ ϭ sin ␪ cos ␪ sin2 ␪ ϩ cos2 ␪ sin ␪ cos ␪
1
sin ␪ sec ␪

10. D

?

?

12. cot ␪ (cot ␪ ϩ tan ␪) ϭ csc2 ␪

?

cot2 ␪ ϩ cot ␪ tan ␪ ϭ csc2 ␪

cos2 ␪ ϩ tan2 ␪ cos2 ␪ ϭ 1 cos2 ␪ ϩ

sin2 ␪ cos2 ␪
2

?

ؒ cos2 ␪ ϭ 1
2

cot 2 ␪ ϩ

?

cos ␪ ϩ sin ␪ ϭ 1
1ϭ1

13.

?

ϭ

tan ␪ sec ␪ ϩ 1
ؒ
sec ␪ Ϫ 1 sec ␪ ϩ 1 tan ␪ ؒ (sec ␪ ϩ 1) sec2 ␪ Ϫ 1 tan ␪ ؒ (sec ␪ ϩ 1) tan2 ␪ sec ␪ ϩ 1 tan ␪

ϭ

sec ␪ ϩ 1 tan ␪ sec ␪ ϩ 1 tan ␪ sec ␪ ϩ 1 tan ␪ sec ␪ ϩ 1 tan ␪

11.

tan ␪ sec ␪ Ϫ 1

?

sec ␪ ϩ 1 tan ␪

cos ␪

sin ␪ ?
1
ϭ sin2 ␪ ϩ cos2 ␪ sec ␪

sin 2 ␪

1 cos2 ␪

1 sin ␪ ? ϭ sec ␪ tan ␪ ϩ cot ␪ sin ␪ ?
1
ϭ sin2 ␪ sin ␪ sec ␪ ϩ 14.

?

ؒ sin2 ␪ ϭ sec2 ␪

?

ϭ csc 2 ␪

?

sin ␪ sec ␪ cot ␪ ϭ 1 sin ␪ ؒ

1 cos ␪

ؒ

cos ␪ sin ␪

?

ϭ1

1ϭ1

?

1 ϩ tan2 ␪ ϭ sec2 ␪ sec2 ␪ ϭ sec2 ␪

cos ␪ sin ␪

?

?

1 cos2 ␪

ؒ

cot2 ␪ ϩ 1 ϭ csc2 ␪ csc2 ␪ ϭ csc2 ␪

1 ϩ sec2 ␪ sin2 ␪ ϭ sec2 ␪

sin ␪ cos ␪

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1 Ϫ 2 cos2 ␪ ? ϭ tan ␪ Ϫ cot ␪ sin ␪ cos ␪
(1 Ϫ cos2 ␪) Ϫ cos2 ␪ ? ϭ tan ␪ Ϫ cot ␪ sin ␪ cos ␪ sin2 ␪ Ϫ cos2 ␪ ? ϭ tan ␪ Ϫ cot ␪ sin ␪ cos ␪ sin2 ␪ cos2 ␪
?
Ϫ ϭ tan ␪ Ϫ cot ␪ sin ␪ cos ␪ sin ␪ cos ␪ sin ␪ cos ␪ ?
Ϫ
ϭ tan ␪ Ϫ cot ␪ cos ␪ sin ␪ tan ␪ Ϫ cot ␪ ϭ tan ␪ Ϫ cot ␪

16.

15.
1 Ϫ cos ␪ ? ϭ (csc ␪ Ϫ cot ␪)2
1 ϩ cos ␪
1 Ϫ cos ␪ ? ϭ csc2 ␪ Ϫ 2 cot ␪ csc ␪
1 ϩ cos ␪ ϩ cot2 ␪
1 Ϫ cos ␪ ? 1 cos ␪ ϭ ؒ
Ϫ2ؒ
2
1 ϩ cos ␪ sin ␪ sin ␪
1
cos2 ␪ ϩ sin ␪ sin2 ␪ cos2 ␪
1 Ϫ cos ␪ ? 1
2cos ␪ ϭ Ϫ ϩ 1 ϩ cos ␪ sin2 ␪ sin2 ␪ sin2 ␪
1 Ϫ cos ␪ ? 1 Ϫ 2 cos ␪ ϩ cos2 ␪ ϭ 1 ϩ cos ␪ sin2 ␪
1 Ϫ cos ␪ ? (1 Ϫ cos ␪)(1 Ϫ cos ␪) ϭ 1 ϩ cos ␪
1 Ϫ cos2 ␪
1 Ϫ cos ␪ ? (1 Ϫ cos ␪)(1 Ϫ cos ␪) ϭ 1 ϩ cos ␪ (1 Ϫ cos ␪)(1 ϩ cos ␪)
1 Ϫ cos ␪
1 Ϫ cos ␪ ϭ 1 ϩ cos ␪
1 ϩ cos ␪

17.
?

cot ␪ csc ␪ ϭ

cos ␪
? sin ␪

cot ␪ csc ␪ ϭ

cot ␪ csc ␪

cot ␪ csc ␪

18.

cot ␪ ϩ csc ␪ sin ␪ ϩ tan ␪

?

sin ␪ ϩ cos ␪ ϭ

1

ϩ sin ␪

?

sin ␪ ϩ cos ␪ ϭ

sin ␪ sin ␪ ϩ cos ␪ cos ␪ Ϫ 1 sin ␪
?
ϭ sin ␪ cos ␪ ϩ sin ␪ cos ␪ cos ␪ ϩ 1 sin ␪
?
ϭ sin ␪ (cos ␪ ϩ 1) cos ␪
?

cos ␪ sin ␪

1

sin ␪

ϩ cos ␪

1 cos ␪ sin ␪ ϩ cos ␪ cos ␪
?
ϭ
1
cos ␪
?

sin ␪ ϩ cos ␪ ϭ

sin ␪ ϩ cos ␪ cos ␪

и cos ␪

sin ␪ ϩ cos ␪ ϭ sin ␪ ϩ cos ␪

cos ␪ ϩ 1
ؒ
sin ␪ cos ␪ sin ␪(cos ␪ ϩ 1)

?

sin ␪ ϩ cos ␪

1 ϩ tan ␪ sec ␪

cot ␪ csc ␪ ϭ

cot ␪ csc ␪ ϭ

ؒ

1 sin ␪

cot ␪ csc ␪ ϭ cot ␪ csc ␪

391

Algebra 2

Chapter 14

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19.

sec ␪ sin ␪
1
cos ␪ sin ␪

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sin ␪ cos ␪

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?

20.

ϭ cot ␪

sin ␪
1 Ϫ cos ␪ ? ϩ ϭ 2csc ␪
1 Ϫ cos ␪ sin ␪ sin ␪ sin ␪
1 Ϫ cos ␪ 1 Ϫ cos ␪ ?
ؒ
ϩ
ؒ
ϭ 2csc ␪ sin ␪ 1 Ϫ cos ␪
1 Ϫ cos ␪ sin ␪

sin ␪ ?

Ϫ cos ␪ ϭ cot ␪

1 sin ␪ cos ␪

Ϫ

sin2 ␪ sin ␪ cos ␪
1 Ϫ sin2 ␪ sin ␪ cos ␪ cos2 ␪ sin ␪ cos ␪ cos ␪ sin ␪

1 Ϫ 2cos ␪ ϩ cos2 ␪ ? sin2 ␪ ϩ ϭ 2csc ␪ sin ␪(1 Ϫ cos ␪) sin ␪ (1 Ϫ cos ␪)

?

ϭ cot ␪

sin2 ␪ ϩ cos2 ␪ ϩ 1 Ϫ 2cos ␪ ? ϭ 2csc ␪ sin ␪(1 Ϫ cos ␪)
2 Ϫ 2cos ␪
?
ϭ 2csc ␪ sin ␪(1 Ϫ cos ␪)

?

ϭ cot ␪

2(1 Ϫ cos ␪) ? ϭ 2csc ␪ sin ␪(1 Ϫ cos ␪)

?

ϭ cot ␪

2 ? ϭ 2csc ␪ sin ␪

?

ϭ cot ␪

?

2csc ␪ ϭ 2csc ␪

cot ␪ ϭ cot ␪

21.

1 ϩ sin ␪ sin ␪
1 ϩ sin ␪ sin ␪
1 ϩ sin ␪ sin ␪
1 ϩ sin ␪ sin ␪
1 ϩ sin ␪ sin ␪
1 ϩ sin ␪ sin ␪
1 ϩ sin ␪ sin ␪
1

?

ϭ
?

ϭ
?

ϭ
?

ϭ

cot2 ␪ csc ␪ Ϫ 1 cot2 ␪ csc ␪ ϩ 1
ؒ
csc ␪ Ϫ 1 csc ␪ ϩ 1 cot2 ␪(csc ␪ ϩ 1) csc2 ␪ Ϫ 1 cot2 ␪ (csc ␪ ϩ 1) cot2 ␪

1 ϩ tan ␪
1 ϩ cot ␪

22.

1

?

ϭ ϭ sin ␪ ϩ cos ␪ cos ␪

1 sin ␪ ϩ sin ␪ sin ␪
1 ϩ sin ␪ sin ␪
1

sec2 ␪

ϩ

csc2 ␪

?

24. 1 ϩ

?

23.

ϭ1

cos2 ␪ ϩ sin2 ␪ ϭ 1
1ϭ1

sin ␪ cos ␪

?

sin ␪ cos ␪

sin ␪

ϩ cos ␪

ϭ

cos ␪ sin ␪ sin ␪ ϩ cos ␪ cos ␪
? sin ␪ ϭ cos ␪ sin ␪ ϩ cos ␪ sin ␪

1

ϭ csc ␪ ϩ 1
?

?

ϭ

392

1 cos ␪
1
cos ␪
1
cos ␪
1
cos ␪
1
cos ␪

1 cos ␪
?

ϭ
?

ϭ
?

ϭ

ؒ

ϩ

sin ␪ sin ␪ ϩ cos ␪ sin ␪ cos ␪

?

ϭ ϭ sin ␪ cos ␪ sin ␪ cos ␪

tan2 ␪ sec ␪ Ϫ 1 tan2 ␪ sec ␪ ϩ 1
ؒ
sec ␪ Ϫ 1 sec ␪ ϩ 1 tan2 ␪(sec ␪ ϩ 1) sec 2 ␪ Ϫ 1
2 ␪(sec ␪ ϩ 1) tan ?

ϭ

tan2 ␪

?

ϭ sec ␪ ϩ 1 ϭ1ϩ 1 cos ␪

Algebra 2

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25. 1 Ϫ tan4 ␪

26.

?

cos4 ␪ Ϫ sin4 ␪ ϭ cos2 ␪ Ϫ sin2 ␪
(cos2 ␪ Ϫ sin2 ␪)(cos2 ␪ ϩ sin2 ␪)

?

ϭ cos2 ␪ Ϫ sin2 ␪
(cos2 ␪ Ϫ sin2 ␪) ؒ 1

?

ϭ cos2 ␪ Ϫ sin2 ␪ cos2 ␪ Ϫ sin2 ␪ ϭ cos2 ␪ Ϫ sin2 ␪

?

ϭ 2 sec2 ␪ Ϫ sec4 ␪
(1 Ϫ tan2 ␪)(1 ϩ tan2 ␪)

?

ϭ sec2 ␪(2 Ϫ sec2 ␪)
[1 Ϫ (sec2 ␪ Ϫ 1)](sec2 ␪)

?

ϭ (2 Ϫ sec2 ␪)(sec2 ␪)
(2 Ϫ sec2 ␪)(sec2 ␪) ϭ (2 Ϫ sec2 ␪)(sec2 ␪)

1 Ϫ cos ␪ sin ␪

sin ␪
1 ϩ cos ␪

?

sin ␪
1 ϩ cos ␪

?

sin ␪
1 ϩ cos ␪

ϭ

ؒ

1 ϩ cos ␪
1 ϩ cos ␪

ϭ

1 Ϫ cos2 ␪ sin ␪ (1 ϩ cos ␪) sin2 ␪ sin ␪ (1 ϩ cos ␪) sin ␪
1 ϩ cos ␪

29.

?

1 Ϫ cos ␪ sin ␪

27.

ϭ
?

ϭ ϭ cos ␪ ? cos ␪ ϩ ϭ 2sec ␪
1 ϩ sin ␪
1 Ϫ sin ␪ cos ␪
1 Ϫ sin ␪ cos ␪
1 ϩ sin ␪ ?
ؒ
ϩ
ؒ
ϭ 2sec ␪
1 ϩ sin ␪ 1 Ϫ sin ␪ 1 Ϫ sin ␪ 1 ϩ sin ␪

28.

cos ␪11 Ϫ sin ␪2 ϩ cos ␪11 ϩ sin ␪2
11 ϩ sin ␪211 Ϫ sin ␪2

cos ␪ Ϫ sin ␪ cos ␪ ϩ cos ␪ ϩ sin ␪ cos ␪ ? ϭ 2 sec ␪
1 Ϫ sin 2 ␪
2cos ␪ ? ϭ 2sec ␪ cos2 ␪
2
? ϭ 2sec ␪ cos2 ␪
2sec ␪ ϭ 2sec ␪

sin ␪
1 ϩ cos ␪ sin ␪
1 ϩ cos ␪
?

tan ␪ sin ␪ cos ␪ csc2 ␪ ϭ 1 sin ␪ cos ␪

ؒ sin ␪ ؒ cos ␪ ؒ

1
2

sin2 ␪
1 Ϫ cos ␪

30.

sin2 ␪
1 ϩ cos ␪
ؒ
1 Ϫ cos ␪ 1 ϩ cos ␪ sin2 ␪ (1 ϩ cos ␪)
1 Ϫ cos2 ␪ sin2 ␪(1 ϩ cos ␪)

?

sin ␪

?

ϭ 2sec ␪

ϭ1

1ϭ1

sin2 ␪

?

ϭ 1 ϩ cos ␪
?

ϭ 1 ϩ cos ␪
?

ϭ 1 ϩ cos ␪
?

ϭ 1 ϩ cos ␪

1 ϩ cos ␪ ϭ 1 ϩ cos ␪
31.

2 v0 tan2 ␪

2g sec2 ␪

ϭ

sin2 ␪ cos2 ␪
1
2g cos2 ␪
2
v0

2 v0 ؒ

sin2 ␪ cos2 ␪

2g v 2 sin2 ␪
0

32. 598.7 m

ؒ

cos 2 ␪
1

2g

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Trigonometric identities are verified in a similar manner to proving theorems in geometry before using them.
Answers should include the following. • The expressions have not yet been shown to be equal, so you could not use the properties of equality on them.
• To show two expressions you must transform one, or both independently.
• Graphing two expressions could result in identical graphs for a set interval, that are different elsewhere. 33. Sample answer: Consider a right triangle ABC with right angle at C. If an angle A has a sine of x, then angle B must have a cosine of x.
Since A and B are both in a right triangle and neither is the right angle, their sum

2

must be .

35. D

36. B

37.

38.

[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1

[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1

may be

is not

40.

39.

[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1

[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1

may be

may be

394

Algebra 2

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41.

42.

25
2

25
3

[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1

[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1

may be

is not

2193
12

43.
45.

44. Ϫ
46. Ϫ

27
4

48. 1: 360Њ; 45Њ

47. 1: 360Њ; 30Њ y 5
4
3
2
1

y
5
4
3
2
1

y ϭ cos (␪ Ϫ 30˚)

O

90˚ 180˚ 270˚

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

O

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5

2

49. 3; 2␲; Ϫ

50.

y ϭ sin (␪ Ϫ 45˚)

90˚ 180˚ 270˚

5
6

y
5
4
3
2
1
Ϫ

51.
53.

3␲
2

26
4

26
4

O
Ϫ␲ Ϫ ␲
2 Ϫ2
Ϫ3
Ϫ4
Ϫ5

ϩ

(

y ϭ 3 cos ␪ ϩ 2

2

3␲
2

)

22
2

52.
54.

395

22
4

2 Ϫ 23
4

Algebra 2

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Sum and Difference of Angles Formulas
Pages 788–790
2. Use the formula sin(␣ ϩ ␤) ϭ sin ␣ cos ␤ ϩ cos ␣ sin ␤.
Since sin 105Њ ϭ sin(60Њ ϩ 45Њ), replace ␣ with 60Њ and ␤ with 45Њ to get sin 60Њ cos 45Њ ϩ cos 60Њ sin 45Њ. By finding the sum of the products of the values,

1. sin (␣ ϩ ␤) Ϫ sin ␣ ϩ sin ␤ sin ␣ cos ␤ ϩ cos ␣ sin ␤ sin ␣ ϩ ␤

26 ϩ 22
4

the result is

26 ϩ 22
4

26 Ϫ 22
4

3. Sometimes; sample answer:
The cosine function can equal 1.

4.

5.

6.

7.

23
2

8.

1
2

22 Ϫ 26
4

23
2

10. cos (270Њ Ϫ ␪)

9. Ϫ

?

ϭ cos 270Њ cos ␪ ϩ sin 270Њ sin ␪
?

ϭ 0 ϩ (Ϫ1 sin ␪) ϭ Ϫsin ␪ sin a␪ ϩ b ϭ cos ␪

2

11. sin ␪ cos

2

ϩ cos ␪ sin

2

?

12.
?

5 Ϫ 23
1 ϩ 5 23

23 sin ␪
2

ϭ sin ␪ cos 30Њ ϩ cos ␪ cos 30Њ ϩ

?

cos ␪ cos 60Њ Ϫ sin ␪ sin 60Њ

ϭ cos ␪

?

ϭ

?

sin ␪ ؒ 0 ϩ cos ␪ ؒ 1 ϭ cos ␪ cos ␪ ϭ cos ␪

13.

sin(␪ ϩ 30Њ) ϩ cos(␪ ϩ 60Њ)

1 cos 2
?

1
2

23 sin 2

22
2

␪Ϫ

1
2

ϩ cos ␪ ϩ

1
2

ϭ cos ␪ ϩ cos ␪ ϭ cos ␪

14.

396

Algebra 2

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15.
17.
19.

25.
27.

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22 Ϫ 26
4

Ϫ 26 Ϫ 22
4

16.

Ϫ 26 Ϫ 22
4

18.

22
2

Ϫ 26 Ϫ 22
4

Ϫ 26 Ϫ 22
4

22.

22 Ϫ 26
4

Ϫ 26 Ϫ 22
4

23
2

22
2

24. Ϫ
26.

22
2

22 Ϫ 26
4

20. Ϫ

22
2

21. Ϫ
23.

7/24/02

28. sin (270Њ Ϫ ␪)
?

ϭ sin 270Њ cos ␪ Ϫ cos 270Њ sin ␪
?

ϭ Ϫ1 cos ␪ Ϫ 0 ϭ Ϫcos ␪
30. cos (90Њ Ϫ ␪)

29. cos (90Њ ϩ ␪)

?

?

ϭ cos 90Њ cos ␪ Ϫ sin 90Њ sin ␪

?

ϭ 0 ؒ cos ␪ ϩ 1 ؒ sin ␪ ϭ sin ␪

ϭ cos 90Њ cos ␪ Ϫ sin 90Њ sin ␪

?

ϭ 0 Ϫ 1 sin ␪ ϭ Ϫsin ␪
31.

?

32.

sin(90Њ Ϫ ␪) ϭ cos ␪ sin 90Њ cos ␪ Ϫ cos 90Њ sin ␪

sin 1␪ ϩ sin ␪ cos

?

ϭ cos ␪

3␲
2

3␲
2
2

?

ϭ Ϫcos ␪

ϩ cos ␪ sin

3␲
2

?

?

ϭ Ϫcos ␪

?

sin ␪ ؒ 0 ϩ cos ␪(Ϫ1) ϭ Ϫcos ␪

1 ؒ cos ␪ Ϫ 0 ؒ sin ␪ ϭ cos ␪

?

cos ␪ Ϫ 0 ϭ cos ␪

?

0 ϩ (Ϫcos ␪) ϭ Ϫcos ␪

cos ␪ ϭ cos ␪

Ϫcos ␪ ϭ Ϫcos ␪

397

Algebra 2

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33.

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?

34.

cos (␲ Ϫ ␪) ϭ Ϫcos ␪ cos ␲ cos ␪ ϩ sin ␲ sin ␪

?

cos(2␲ ϩ ␪) ϭ cos ␪ cos 2␲ cos ␪ Ϫ [sin 2␲ sin ␪]

?

?

ϭ cos ␪

ϭ Ϫcos ␪
?

Ϫ1 ؒ cos ␪ ϩ 0 ؒ sin ␪ ϭ Ϫcos ␪
Ϫcos ␪ ϭ Ϫcos ␪

?

1 ؒ cos ␪ Ϫ [0 ؒ sin ␪] ϭ cos ␪
?

1 ؒ cos ␪ Ϫ 0 ϭ cos ␪ cos ␪ ϭ cos ␪
?

35.

?

ϭ sin 60Њ cos ␪ ϩ cos 60Њ sin ␪ ϩ sin 60Њ cos ␪ Ϫ cos 60Њ sin ␪

?

sin ␲ cos ␪ Ϫ [cos ␲ sin ␪] ϭ sin ␪

23
2

ϭ 23 cos ␪

?

0 ؒ cos ␪ Ϫ [Ϫ1 ؒ sin ␪] ϭ sin ␪

?

ϭ

?

0 Ϫ [Ϫsin ␪] ϭ sin ␪ sin ␪ ϭ sin ␪
37. sin a␪ ϩ b Ϫ cos a␪ ϩ b

3

23 sin ␪ ϩ cos 2
1
cos ␪ ϩ sin ␪
2
1
? 1 ϭ sin ␪ ϩ sin ␪
2
2
?

3
␲ cos 6

6

1 ϭ 2
?

ϩ sin ␪

23
2

3
␲ sin 6

␪Ϫ

cos ␪ ϩ

cos ␪ Ϫ

1
2

1
2

sin ␪ ϩ

sin ␪

38. sin ( ␣ ϩ ␤) sin ( ␣ Ϫ ␤) ϭ sin2 ␣ Ϫ sin2 ␤

ϭ sin ␪ cos ϩ cos ␪ sin Ϫ cos ␪

23
2

36. sin (60Њ ϩ ␪) ϩ sin (60Њ Ϫ ␪)

sin(␲ Ϫ ␪) ϭ sin ␪

?

ϭ (sin ␣ cos ␤ ϩ cos ␣ sin ␤)
(sin ␣ cos ␤ Ϫ cos ␣ sin ␤)
?

ϭ sin2 ␣ cos2 ␤ Ϫ cos2 ␣ sin2 ␤
?

ϭ sin2 ␣(1 Ϫ sin2 ␤) Ϫ
(1 Ϫ sin2 ␣) sin2 ␤
?

ϭ sin2 ␣ Ϫ sin2 ␣ sin2 ␤ Ϫ

ϭ sin ␪

sin2 ␤ ϩ sin2 ␣ sin2 ␤ ϭ sin2 ␣ Ϫ sin2 ␤

398

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39.

40.
?

cos (␣ ϩ ␤) ϭ
?

cos (␣ ϩ ␤) ϭ

1 Ϫ tan ␣ tan ␤ sec ␣ sec ␤
1

sin ␣

y ϭ 10 sin (2t Ϫ 30°) ϩ 10 cos (2t ϩ 60°)

sin ␤

Ϫ cos ␣ ؒ cos ␤
1
cos ␣

ؒ

Ϫ180° Ϫ90°

1 cos ␤

O

90°

180° t

Ϫ2
Ϫ4

?

cos (␣ ϩ ␤) ϭ sin ␣

1 Ϫ cos ␣ ؒ
1
cos ␣

y

4

ؒ

sin ␤ cos ␤

1 cos ␤

и

cos ␣ cos ␤ cos ␣ cos ␤

?

cos (␣ ϩ ␤) ϭ cos ␣ cos ␤ Ϫ sin ␣ sin ␤
1

cos (␣ ϩ ␤) ϭ cos (␣ ϩ ␤)
41. Destructive; the resulting graph has a smaller amplitude than the two initial graphs. 42. 0.3681 E

43. 0.4179 E

44. 0.6157 E

45. 0.5563 E

46. tan (␣ ϩ ␤) ϭ ϭ

sin(␣ ϩ ␤) cos(␣ ϩ ␤) sin ␣ cos ␤ ϩ cos ␣ sin ␤ cos ␣ cos ␤ Ϫ sin ␣ sin␤

399

sin ␣ cos ␤ cos ␣ cos ␤ cos ␣ cos ␤ cos ␣ cos ␤

cos ␣ sin ␤

tan ␣ ϩ tan ␤
1 Ϫ tan ␣ tan ␤

ϩ cos ␣ cos ␤ sin ␣ sin ␤

Ϫ cos ␣ cos ␤

Algebra 2

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tan (␣ Ϫ ␤) ϭ ϭ

sin(␣ Ϫ ␤) cos(␣ Ϫ ␤) sin ␣ cos ␤ Ϫ cos ␣ sin ␤ cos ␣ cos ␤ ϩ sin ␣ sin ␤

ϭ

sin ␣ cos ␤ cos ␣ cos ␤ cos ␣ cos ␤ cos ␣ cos ␤

cos ␣ sin ␤

ϭ

tan ␣ Ϫ tan ␤
1 ϩ tan ␣ tan ␤

Ϫ cos ␣ cos ␤ sin ␣ sin ␤

ϩ cos ␣ cos ␤

47. Sample answer: To determine communication interference, you need to determine the sine or cosine of the sum or difference of two angles. Answers should include the following information. • Interference occurs when waves pass through the same space at the same time. When the combined waves have a greater amplitude, constructive interference results and when the combined waves have a smaller amplitude, destructive interference results. 48. ϭA

49. C

50. cot ␪ ϩ sec ␪
?

ϭ
?

ϭ
?

ϭ

cos2 ␪ ϩ sin ␪ sin ␪ cos ␪ cos2 ␪ sin ␪ ϩ sin ␪ cos ␪ sin ␪ cos ␪ cos ␪ sin ␪

ϩ

1 cos ␪

ϭ cot ␪ ϩ sec ␪

400

Algebra 2

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51. sin2 ␪ ϩ tan2 ␪

52.

?

ϭ (1 Ϫ cos2 ␪) ϩ
?

ϭ sin2 ␪ ϩ
?

ϭ sin2 ␪ ϩ
?

ϭ sin2 ␪ ϩ

sec2 ␪ csc2 ␪
1
cos2 ␪ sin2 ␪ cos2 ␪

sec2

␪ csc ␪

?

sin ␪ (sin ␪ ϩ csc ␪) ϭ 2 Ϫ cos2 ␪

2

?

sin2 ␪ ϩ 1 ϭ 2 Ϫ cos2 ␪
Ϭ

?

1 Ϫ cos2 ␪ ϩ 1 ϭ 2 Ϫ cos2 ␪
2 Ϫ cos2 ␪ ϭ 2 Ϫ cos2 ␪

1 sin2 ␪

ϭ sin2 ␪ ϩ tan2 ␪
53.
1
Ϭ
cos ␪
1
ؒ cos ␪

sec ␪ tan ␪ sin ␪ cos ␪ cos ␪ sin ␪
1
sin ␪

?

ϭ csc ␪

54. 1

?

ϭ csc ␪
?

ϭ csc ␪
?

ϭ csc ␪

csc ␪ ϭ csc ␪

3 234
,
34

5 234
,
34

55. 4

56. sec ␪

57. 2 sec ␪

58. sin ␪ ϭ Ϫ cos ␪ ϭ

sec ␪ ϭ
59.

4
5

3
5

sin ␪ ϭ Ϫ , cos ␪ ϭ Ϫ ,
4
3

5
3

5
3

cot ␪ ϭ Ϫ

60. sin ␪ ϭ 1, cos ␪ ϭ 0, tan ␪ ϭ undefined, csc ␪ ϭ 1, sec ␪ ϭ undefined, cot ␪ ϭ 0

5
4
3
4

tan ␪ ϭ , csc ␪ ϭ Ϫ , sec ␪ ϭ Ϫ , cot ␪ ϭ

234
,
3

234
,
5

csc ␪ ϭ Ϫ

3
5

tan ␪ ϭ Ϫ ,

61. 360

62. 3,991,680

2 25
2

63. 56

64. 210

65. about 228 mi

66.

67. Ϯ

68. Ϯ

y2
34

Ϫ

x2
6

ϭ1

3
5

401

Algebra 2

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25
5

3 25
5

26
2

69. Ϯ
71. Ϯ
73. Ϯ

3
4

70. Ϯ

2 16Ϫ 12
2

72. Ϯ
74. Ϯ

Lesson 14-6

22Ϫ2 12
2

Double-Angle and Half-Angle Formulas
Pages 794–797
12
2

or 22

1. Sample answer: If x is in the x third quadrant, then is
2
between 90Њ and 135Њ. Use the half-angle formula for cosine knowing that the value is negative.

2. Sample answer: 45Њ; cos 2(45Њ) ϭ cos 90Њ or 0,

3. Sample answer: The identity used for cos 2␪ depends on whether you know the value of sin ␪, cos ␪, or both values.

4.

24
,
25

5.

6.

22 ϩ 13
2

4 25
,
9

1 230
,
9
6

Ϫ ,

3 27
,
8

28 Ϫ 2 17
,
4

7. Ϫ

1
8

Ϫ ,

22 Ϫ 13
2

Ϫ

Ϫ

2 cos 45Њ ϭ 2 ؒ

26
6

28 ϩ 2 17
4

7 25 2 25
,
25 5
5

Ϫ ,

23 1 22 Ϫ 13
, ,
,
2 2
2

8. Ϫ

22 Ϫ 13
2

?

sin 2x
1 Ϫ cos 2x

?

2 sin x cos x
1 Ϫ (1 Ϫ 2 sin2 x)

?

2 sin x cos x
2 sin2 x

?

9.

cos x sin x

10. cot x ϭ ϭ ϭ ϭ ϭ cot x

402

Algebra 2

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?

11. cos2 2x ϩ 4 sin2 x cos2 x ϭ 1
120 119 5 226 226
,
,
,
169 169
26
26

4 22
,
9

13. Ϫ
15.

Ϫ

16.

17 215 221
,
18
6
6
1 26 230
,
9 6
6

22.

25. Ϫ

20.

Ϫ ,

Ϫ ,

215
,
8

210
10

Ϫ

215
5

7 210 26
,
8
4
4

Ϫ

Ϫ ,

120 119 5 226
,
,
,
169 169
26
215 7
, ,
8
8

226
26

28 ϩ 2 115
,
4

28 Ϫ 2 115
4

Ϫ

4 221 17
, ,
5
25

Ϫ

25 110 ϩ 1210
,
10

Ϫ

25 110 Ϫ 1210
10

24. Ϫ

Ϫ ,

22 ϩ 13
2
22 ϩ 12
2

26.

22 Ϫ 12
2

27. Ϫ

23 210
,
25
5

24 7 3 210
, ,
,
25 25
10

18. Ϫ

23 16 ϩ 4 13
6

4 25
,
9

Ϫ

31.

23
3

14. Ϫ

4 22 7 23 16 Ϫ 4 13
, ,
9
9
6

21. Ϫ

29.

7 26
,
9 3

Ϫ ,

28 ϩ 155
4

235
,
18

Ϫ

23.

4 26
,
25

?

cos2 2x ϩ sin2 2x ϭ 1
1ϭ1

3 255 23 28 Ϫ 155
, ,
,
32
32
4

17. Ϫ

19.

12. 1.64

22 Ϫ 12
2

28. Ϫ
30. Ϫ
?

sin 2x ϭ 2 cot x sin2 x cos x
?
2 sin x cos x ϭ 2
ؒ sin2 x sin x
2 sin x cos x ϭ 2 sin x cos x

22 Ϫ 13
2
22 Ϫ 13
2

32.
2aϮ

3

2 cos2

x
2

?

ϭ 1 ϩ cos x

2

1 ϩ cos x b 2

1 ϩ cos x b 2

2a

?

ϭ 1 ϩ cos x
?

ϭ 1 ϩ cos x

1 ϩ cos x ϭ 1 ϩ cos x

403

Algebra 2

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?

1
2

?

1
2

?

1
2

34. sin2 x ϭ (1 Ϫ cos2 x)

33.
?

sin4 x Ϫ cos4 x ϭ 2 sin2 x Ϫ 1
(sin2 x Ϫ cos2 x)(sin2 x ϩ cos2 x)

sin2 x ϭ [1 Ϫ (1 Ϫ 2 sin2 x)]

?

ϭ 2 sin2 x Ϫ 1
(sin2 x Ϫ cos2 x) ؒ 1

sin2 x ϭ (2 sin2 x) sin2 x ϭ sin2 x

?

ϭ 2 sin2 x Ϫ 1
[sin2 x Ϫ (1 Ϫ sin2 x)] ؒ 1
?

ϭ 2 sin2 x Ϫ 1 sin2 x Ϫ 1 ϩ sin2 x
?

ϭ 2 sin2 x Ϫ 1
2 sin2 x Ϫ 1 ϭ 2 sin2 x Ϫ 1
35.

3

3

tan2

x
2

?

1 Ϫ cos x
1 ϩ cos x

?

1 Ϫ cos x
1 ϩ cos x

ϭ

x

sin2 2 cos 2 x

ϭ

36.

2

1 Ϫ cos x 2 b 2
? 1 Ϫ cos x
1 ϩ cos x 2 b 2

1 Ϫ cos x
1 ϩ cos x

39. 2 ϩ 23

ϭ

ϭ

1 cos x
Ϫ
sin x cos x sin x
1 Ϫ cos2 x sin x cos x sin2 x sin x cos x sin x cos x

1 ϩ cos x
1 Ϫ cos x
1 ϩ cos x

37. 46.3Њ

38.

40.

2 g 3
3

?

ϭ tan x
?

ϭ tan x
?

ϭ tan x
?

ϭ tan x

tan x ϭ tan x

1 Ϫ cos L
1 ϩ cos L
1 Ϫ cos L
1 ϩ cos L

v 2 (tan ␪ Ϫ tan ␪ sin2 ␪)
?

2 g v 2 tan ␪(1 Ϫ sin2 ␪)

?

2 g v 2 tan ␪ cos2 ␪

?

2 2 v sin g v 2 sin 2␪ g ϭ ϭ ϭ ϭ ©Glencoe/McGraw-Hill

404

␪ cos ␪

Algebra 2

Chapter 14

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41.

1 tan 4

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y

42.

Ϫ

2.5
2
1.5
1
0.5
3␲ Ϫ␲
2

Ϫ2

y ϭ sin2 x x O

2

Ϫ1

1
Ϫ1.5
y ϭ Ϫ 2 cos 2x
Ϫ2

3␲
2

y ϭ Ϫcos2 x

Ϫ2.5

Sample answer: They all have the same shape and are vertical translations of each other.

2

43. The maxima occur at x ϭ Ϯ and 3␲
Ϯ . The
2

44.

y
2.5
2
1.5
1
0.5

minima occur

at x ϭ 0, Ϯ␲ and Ϯ2␲.

Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ1
Ϫ1.5
Ϫ2
Ϫ2.5

y ϭ sin 2x x O

90˚ 180˚ 270˚

45. The graph of f(x) crosses the x-axis at the points specified in Exercise 41.

46. c ϭ 1 and d ϭ 0.5

47. Sample answer: The sound waves associated with music can be modeled using trigonometric functions.
Answers should include the following information.
• In moving from one harmonic to the next, the number of vibrations that appear as sine waves increase by 1.

48. D

405

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• The period of the function as you move from the nth harmonic to the (n ϩ 1)th harmonic decreases from
2␲
2␲
.
to
26 Ϫ 22
4

n

49. B
51.

53. Ϫ
55.

26 ϩ 22
4

nϩ1

50.

23
2

54.

1
2

22
2

26 ϩ 22
4

52. Ϫ

56.
?

cot2 ␪ Ϫ sin2 ␪ ϭ

cos2 ␪ csc2 ␪ Ϫ sin2 ␪ sin2 ␪ csc2 ␪
1

2

2

?

cot ␪ Ϫ sin ␪ ϭ

cos2 ␪ sin2 ␪ Ϫ sin2 ␪
1

sin2 ␪ sin2 ␪

cot2 ␪ Ϫ sin2 ␪
1
cot2 ␪ Ϫ sin2 ␪ ϭ cot2 ␪ Ϫ sin2 ␪
?

cot2 ␪ Ϫ sin2 ␪ ϭ

58. 101 or 10

57. cos ␪(cos ␪ ϩ cot ␪)
?

ϭ cot ␪ cos ␪1sin ␪ ϩ 12
?

ϭ

cos ␪ cos ␪ sin ␪ ϩ cot ␪ cos ␪ sin ␪

?

ϭ cos2 ␪ ϩ cot ␪ cos ␪ ϭ cos ␪(cos ␪ ϩ cot ␪)
59. 102.5 or about 316 times greater 60. Ϫ6, 5

61. 1, Ϫ1

62. 0, Ϫ2

63.

5
,
2

1 1
2 2

64. Ϫ ,

Ϫ2
1
2

65. 0, Ϫ

406

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Chapter 14
Practice Quiz 2
Page 797
?

1. sin ␪ sec ␪ ϭ tan ␪ sin ␪ ؒ

1

cos sin ␪ cos ␪

?

2.

sec ␪ Ϫ cos ␪ ϭ sin ␪ tan ␪ cos ␪ ?
1
Ϫ cos ␪ ؒ ϭ sin ␪ tan ␪ cos ␪ cos ␪ cos2 ␪ ?
1
Ϫ ϭ sin ␪ tan ␪ cos ␪ cos ␪
1 Ϫ cos2 ␪ ? ϭ sin ␪ tan ␪ cos ␪ sin2 ␪ ? ϭ sin ␪ tan ␪ cos ␪ sin ␪ ? ϭ sin ␪ tan ␪ sin ␪ cos ␪ sin ␪ tan ␪ ϭ sin ␪ tan ␪

4.

sin (90Њ ϩ ␪) ϭ cos ␪

?

ϭ tan ␪
?

ϭ tan ␪
?

tan ␪ ϭ tan ␪

sin ␪(cos ␪ ϩ 1) cos ␪ sin ␪ cos ␪ ϩ sin ␪
?
sin ␪ ϩ tan ␪ ϭ cos ␪ sin ␪
? sin ␪ cos ␪ sin ␪ ϩ tan ␪ ϭ ϩ cos cos ␪ sin ␪ ϩ tan ␪ ϭ sin ␪ ϩ tan ␪

?
3. sin ␪ ϩ tan ␪ ϭ

?

sin 90Њ cos ␪ ϩ cos 90Њ sin ␪ ϭ cos ␪
?

cos ␪ ϩ 0 ϭ cos ␪ cos ␪ ϭ cos ␪

3␲
?
Ϫ ␪b ϭ Ϫsin ␪
2
3␲
3␲
? cos cos ␪ ϩ sin sin ␪ ϭ Ϫsin ␪
2
2 cos a

5.

6. sin (␪ ϩ 30Њ) ϩ cos (␪ ϩ 60Њ)
?

?

ϭa
?

Ϫsin ␪ ϭ Ϫsin ␪

9.

sin ␪ ϩ

1 a cos
2

23
2

?

1
2

␪Ϫ

9 282
82

13
2
1
2

cos ␪b ϩ sin ␪b

1
2

ϭ cos ␪ ϩ cos ␪

22 Ϫ 13
2

13
2

ϭ (sin ␪ cos 30Њ ϩ cos ␪ sin 30Њ) ϩ
(cos ␪ cos 60Њ Ϫ sin ␪ sin 60Њ)

0 ϩ (Ϫ1 ؒ sin ␪) ϭ Ϫsin ␪

7.

?

22 Ϫ 12
2

ϭ cos ␪
8. Ϫ
10.

407

Algebra 2

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Lesson 14-7

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Solving Trigonometric Equations
Pages 802–804
2. Sample answer: The function is periodic with two solutions in each of its infinite number of periods.

1. Sample answer: If sec ␪ ϭ 0
1
then ϭ 0. Since no value cos ␪

of ␪ makes

1 cos ␪

ϭ 0. there

are no solutions.
3. Sample answer: sin ␪ ϭ 2

4. 60Њ, 120Њ, 240Њ, 300Њ

5. 135Њ, 225Њ

6.

7.

6

␲ ␲ 5␲ 3␲
, , ,
6 2 6 2

8. 0 ϩ

9. 0 ϩ k␲

2k␲
3

10. 90Њ ϩ k ؒ 360Њ, 180Њ ϩ k ؒ 360Њ
12.

11. 60Њ ϩ k ؒ 360Њ, 300Њ ϩ k ؒ 360Њ

7␲
6

11␲
6

ϩ 2k␲,

ϩ 2k␲ or

210Њ ϩ k ؒ 360Њ, 330Њ ϩ k ؒ 360Њ
13.

6

ϩ 2k␲,

5␲
6

ϩ 2k␲,

2

ϩ 2k␲

14. 31.3Њ

or 30Њ ϩ k ؒ 360Њ, 150Њ ϩ
360Њ, 90Њ ϩ k ؒ 360Њ
15. 60Њ, 300Њ

16. 240Њ, 300Њ

17. 210Њ, 330Њ

18. 30Њ, 150Њ, 210Њ, 330Њ

19.

␲ 5␲ 3␲
, ,
6 6 2

20.

2

21.

7␲ 11␲
,
6
6

22.

␲ 3␲ 2␲ 4␲
, , ,
2 2 3 3

23.

3

ϩ 2k␲,

5␲
3

ϩ 2k␲

24. ␲ ϩ 2k␲,
5␲
3

25.

2␲
3

27.

3

ϩ 2k␲,

ϩ 2k␲,

4␲
3

5␲
3

ϩ 2k␲

3

ϩ 2k␲,

ϩ 2k␲

26. 0 ϩ 2k␲

ϩ 2k␲

28. 0 ϩ k␲,
408

6

ϩ 2k␲,

5␲
6

ϩ 2k␲

Algebra 2

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29. 45Њ ϩ k ؒ 180Њ

30. 0Њ ϩ k ؒ 180Њ

31. 270Њ ϩ k ؒ 360Њ

32. 30Њ ϩ k ؒ 360Њ,
150Њ ϩ k ؒ 360Њ

33. 0Њ ϩ k ؒ 180Њ, 60Њ ϩ k ؒ 180Њ

34. 120Њ ϩ k ؒ 360Њ,
240Њ ϩ k ؒ 360Њ

3␲
ϩ 2k␲, ϩ 2k␲
2
2

35. 0 ϩ 2k␲,

36.

7␲
6

ϩ 2k␲,

11␲
6

ϩ 2k␲ or

210Њ ϩ k ؒ 360Њ,
330Њ ϩ k ؒ 360Њ

or 0Њ ϩ k ؒ 360Њ, 90Њ ϩ k ؒ 360Њ,
270Њ ϩ k ؒ 360Њ
38.

37. 0 ϩ k␲ or 0Њ ϩ k ؒ 180Њ

2

ϩ k␲,

2␲
3

ϩ 2k␲,

4␲
3

ϩ 2k␲

or 90Њ ϩ k ؒ 180Њ, 120Њ ϩ k ؒ
360Њ, 240Њ ϩ k ؒ 360Њ
39. 0 ϩ 2k␲,

3

ϩ 2k␲,

5␲
3

40.

ϩ 2␲,

2

ϩ 4k␲ or 90Њ ϩ k ؒ 720Њ

or 0Њ ϩ k ؒ 360Њ, 60Њ ϩ k ؒ
360Њ, 300Њ ϩ k ؒ 360Њ
41. S ϭ

352 tan ␪

43. y ϭ

3
2

y
4
3.5
3
2.5
2
1.5
1
0.5
O
Ϫ1
Ϫ1

ϩ
3

or S ϭ 352 cot ␪

3
2

44. 10

sin (␲t)
3

y ϭ 2 ϩ 2 sin (␲ t )

1 2 3 4 5 6 7 8 9

t

Temperatures are cyclic and can be modeled by trigonometric functions.
Answers should include the following information.

45. (4.964, Ϫ0.598)

409

Algebra 2

Chapter 14

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• A temperature could occur twice in a given period such as when the temperature rises in the spring and falls in autumn.

24 7 210 3 210
, ,
,
25 25 10
10

5 211 7 23 233
, ,
,
18
18 6
6

1 1 23
23
,Ϫ , ,
2
2 2 2

7 25 2 25
24
,Ϫ ,
,
25
25 5
5

47. D

48. B

49.

50.

51.

53. Ϫ

23
2

52.
54.

22
2

55. b ϭ 11.0, c ϭ 12.2, mЄC ϭ 78