- Level: International Baccalaureate
- Subject: Maths
- Word count: 977
Parabola Investigation
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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
- 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
- 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
- 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
= 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
- 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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