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Parabola Investigation

Extracts from this document...

Introduction

Description

In this task, you will investigate the patterns in the intersections of parabolas and the lines y=x and y=2x. Then you will be asked to prove your conjectures and to broaden the scope of the investigation to include other lines and other types of polynomials.

Method

1. Consider the parabola y=(x−3)2+2 = x2−6x+11 and the lines y=x and y=2x.
• The four points of intersections are illustrated in the graph below. Using a graphing program and a quadratic program, one can solve for these points.

The four points of intersection are:

x1≈ (1.764, 5.528)

x2(2.381, 2.381)

x3≈ (4.618, 4.618)

x4≈ (6.236, 12.272)

• The x-values of these intersections are labeled as they appear from left to right on the x-axis as x1, x2, x3, and x4.

• The values of SL and SR are defined and solved.

SL= x2−x1≈ 2.382−1.764≈ 0.618

SR = x4−x3≈ 6.236−4.618≈ 1.618

• Calculate D=|SL− SR|.

D=|SL− SR| ≈ |0.618−1.618|= |−1|= 1

1. Find values of D for other parabolas of the form y= ax2

Middle

 Formula x1 x2 x3 x4 D y= −x2−2x+6 -5.162 -4.372 1.372 1.162 1 y= −x2−5x−8 -5.562 -4 -2 -1.438 1 y=−2x2+3x+5 -1.351 -1.158 2.158 1.851 0.5

The conjecture still holds and the results of D fit the conjecture found in part 2.

Proof:

y= ax2−bx+c

y= x

ax2−(b−1)x+c=0

x= (1−b)±√((b−1)2−4ac)

2a

x2,3=(1−b)±√((b−1)2−4ac)

2a

y= ax2−bx+c

y= 2x

ax2−(b−2)x+c=0

x= (2−b)±√((b−2)2−4ac)

2a

X1,4= (2−b)±√((b−2)2−4ac)

2a

SL= x2−x1

SR= x4−x3

D=|SL−SR| = |(x2−x1) – (x4−x3)|= |(x2+x3)−(x1+x4)|= |(1–b)2/2a – (2–b)2/2a|

= 1/a|1−b−2+b|=1/a

1. To prove that the conjecture will still hold with the lines are changed, I will use the same examples from part 2. Now, the two intersecting lines will be y=1.5x and y=3x.

y= x2−8x+18

y= 2x2−8x+9

y= 4x2−20x+26

 Formula x1 x2 x3 x4 D y= x2−8x+18 2 2.614 6.886 9 1.5 y= 2x2−8x+9 1 1.307 3.443 4.5 0.75 y= 4x2−20x+26 1.546 1.837 3.538 4.204 0.375

Some modifications had to be made to the conjecture. The new conjecture is:

Conclusion

2+x3)x2+(x1x2+x2x3+x1x3)x−(x1x2x3))

= ax3−a(x1+x2+x3)x2+a(x1x2+x2x3+x1x3)x−a(x1x2x3)

From the proof we can see what each of the coefficient equals:

a=a

b= −a(x1+x2+x3)

c= a(x1x2+x2x3+x1x3)

d= a(x1x2x3)

From the expression for b, we can find the sum of the roots:

b= −a(x1+x2+x3)

x1+x2+x3= b/−a = −b/a

From the conjecture and proof for the parabola,

We know that:

D=|(x2+x3)−(x1+x4)|

For a cubic polynomial, we’ve found that the sum of the roots is –b/a so

D= |(−b/a)−( −b/a)|=0

1. The conjecture can be modified to include higher order polynomials and it would very similar to the cubic one. For higher order polynomials, the roots will cancel out so D=0 will always be true.

anxn+an-1xn-1+…+a1x+a0= an(x−xn)(x−xn-1)…(x−x1)

anx3−an(xn+xn-1+…+x1)+ an(xnxn-1+xn-1xn-2+…+xnx1)+an(xnxn-1…x1)

xn+xn-1+…x1=(an−1)/-an

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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