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Parabola Investigation

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Parabola Investigation

In this task, I will investigate the patterns in the intersections of parabolas and the lines y=x and y=2x, then I will prove my conjectures and to broaden the scope of the investigation to include other lines and other types of polynomials.

1. Consider the parabola y=(x-3)2 +2=x2-6x+11 and the lines y=x and y=2x.

The original graph is shown below (graph 1.1)

Find the intersections of the parabola with y=x and y=2x, Graph 1.2.

Also, label the x-values of the intersections with the line y=x as they appear from left to right on the x-axis as a1 and a2; label the x-values of the intersections with the line y=2x as b1 and b2.

Now, I will using the graph and graph calculator, find the values of a1-b1 and b2-a2 and name them respectively SL and SR.

SL=a1-b1=2.381966-1.763932=0.618034

SR=b2-a2=6.236068-4.618034=1.618034

Now, calculate the quantity D= │SL-SR│

D= │SL-SR│=│0.618034-1.618034│=1

By algebra calculation,

D=│SL-SR│

=│ a1-b1-(b2-a2) │

=│ a1-b1-b2+a2 │

=│ (a1+ a2 )-(b1+b2) │

Now, I will try other parabolas of the form y=ax2+bx+c, a>0, with vertices in quadrant 1, intersected by the lines y=x and y=2x.

y=x2+2x+1 [pic]

From the graph we can see there is no intersection of the parabola and y=x, y=2x.

Using the algebra way:

Solve: (a) x2+2x+1=x

(b) x2+2x+1=2x

(a) x2+2x+1=x

x2+x+1=0

[pic]

X=[pic]

a1+a2= 1

(b) x2+2x+1=2x

x2+1=0

x2=-1

x=±i

b1+b2=0

so D =│ (a1+ a2 )-(b1+b2) │

=│ 1-0 │

=1

y=2x2+3x-1

D=│(a1+a2)-(b1+b2)│

=│(-1.366025+0.366025)+(-1+0.5)│

=│-1+0.5│

=0.5

Base on the example above, I think D is relate with the coefficient of the formula, and I guess it’s

D=[pic]

To prove my guess, I would like to check it in algebra way. Using y1=ax2+bx+c intersected by lines y2=x and y3=2x find the general form of D.

Find the intersected points of y1 and y2, so y1=y2

ax2+bx+c=x

ax2+(b-1)x+c=0

Δ=(b-1)2-4ac

→ x=[pic]

So we can find a1 + a2 in general form:

a1+a2=[pic]

=[pic]

Now, use the same method to find the intersected points of y1 and y3,

ax2+bx+c=2x

ax2+(b-2)x+c=0

Δ=(b-2)2-4ac

→x=[pic]

So

b1+b2=[pic]

=[pic]

Using the equation in part 1,

D=│ (a1+ a2 )-(b1+b2) │

=[pic]

=[pic]

Conclusion:

The general form for D as the D=│SL-SR│ (or =│ (a1+ a2 )-(b1+b2) │,a1a2 represent the x-value of intersection of parabola and y=x, b1b2 represent the x-value of intersection of parabola and y=2x ) is modulo the inverse of the leading coefficient of the parabola, no matter whether the parabola and y=x, y=2x have real intersection or not, because when you add the two roots together, the complex part will cancel.

-----------------------
Graph 1.1

Graph 1.2

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