- Level: International Baccalaureate
- Subject: Maths
- Word count: 2920
Maths Portfolio Type I Parabola Investigation
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Introduction
Ken Chen
Consider the parabola y = (x – 3)2 + 2 = x2 – 6x + 11, which is drawn together with the lines y = x and y = 2x on the same axes.
Using the program called Qax Grapher, which has functions same as a GDC, the parabola y = (x – 3)2 + 2 and the two lines y = x and y = 2x are graphed as illustrated below.
There are 4 intersections in the axes which have the x-values labeled as x1, x2, x3 and x4 as they appear from left to right on the x-axis.
x1and x4 – the x-values of the intersections between the line y = 2x and the parabola y = x2 – 6 x + 11 on the left and right hand side of the graph respectively.
x2and x3 – the x-values of the intersections between the line y = x and the parabola y = x2 – 6 x + 11 on the left and right hand side of the graph respectively.
Using GDC, these x-values can be found.
x1 = 1.764 x2 = 2.382 x3 = 4.618 x4 = 6.236
Call SL as the difference between x1 and x2 and SR as the difference between x3 and x4.
SL = x2 – x1 = 2.382 – 1.764 = 0.618
and
SR = x4 – x3 = 6.236 – 4.618 = 1.618
D is the difference between the two differences SL and SR.
D = |SL – SR| = 1.618 – 0.618 = 1.
In order to find out the pattern of the D-value, other parabolas which have same characteristics as the parabola above (the parabolas need to be in the form y = ax2 + bx + c (a > 0) with vertices in quadrant 1) are considered.
- Consider the parabolas with a = 1
When the parabola y = (x – 4)2 + 1 = x2 – 8x + 17
Middle
From these examples, the conjecture can be obtained.
“For any four intersections of the two lines y = x, y = 2x and the parabola ax2 + bx + c (a > 0 and its vertex is in quadrant 1); the value of D which is calculated by D = |(x2 – x1) – (x4 – x3)| equals to Error! Reference source not found..”
Testing the Validity of the Statement
To test for the validity of the conjecture above, other parabolas which have a as a real number and have vertices place in any quadrants are investigated.
Firstly, the pattern in intersections of any parabolas that have real value of a and the lines y = x and y = 2x is investigated out to see whether the conjecture above still holds true.
- a = - 4 (a Error! Reference source not found. Z)
When the parabola y = - 4x2 + 32x - 45 and the lines y = x and y = 2x are graphed, four values of x are obtained.
x1 = 1.934 x2 = 2.073 x3 = 5.427 x4 = 5.816
SL = x2 – x1 = 2.073 – 1.934 = 0.139
SR = x4 – x3 = 5.816 – 5.427 = 0.389
D = |SL – SR| = 0.389 – 0.139 = 0.25 = Error! Reference source not found.
- a = Error! Reference source not found.a Error! Reference source not found. Q)
When the parabola y = - ½ x2 + 3x + 5 and the lines y = x and y = 2x are graphed, four values of x are obtained.
x1 = -2.317 x2 = -1.742 x3 = 5.742 x4 = 4.317
SL = x2 – x1 = -1.742 + 2.317 = 0.575
SR = x4
Conclusion
- Quadric polynomial
Consider y = x4 – 6x3 + 9x2 – 2. There are 8 intersections.
x1 = -0.377 x2 = 0.726 x3 = 2.000 x4 = 3.651 x5 = -0.343 x6 = 1.000
x7 = 1.529 x8 = 3.814
x1, x2, x3, x4 are the x-values of the intersections of the quadric polynomial and the y = x
x5, x6, x7, x8 are the x-values of the intersections of the quadric polynomial and the y = 2x
D = (x1 + x2 + x3 + x4) - (x5 + x6 + x7 + x8) = 6 – 6 = 0.
- Polynomial of order 5
Consider y = 2x5 – 6x4 – 2x3 + 7x2 + 2x – 1. There are 10 intersections.
x1 = -0.914 x2 = -0.484 x3 = 0.340 x4 = 1.139 x5 = 2.919 x6 = -1.000
x7 = -0.384 x8 = 0.442 x9 = 1.000 x10 = 2.942
x1, x2, x3, x4, x5 are the x-values of the intersections of the quadric polynomial and the y = x
x6, x7, x8, x9, x10 are the x-values of the intersections of the quadric polynomial and the y = 2x
D = (x1 + x2 + x3 + x4 + x5) - (x6 + x7 + x8 + x9 + x10) = 3 – 3 = 0
The conjecture can be made: “The difference between the sum of all the x-values of the intersections of the polynomial and the line y = x and the sum of all those of the intersections of the polynomial the lines y = 2x equals to 0”
Justification
The justification is same as the justification for cubic polynomial. According to the Viete’s theorem, it can be seen that the sum of all the roots is independent of the coefficient of x. So the sum always equals to Error! Reference source not found. which is a constant for any intersections of the polynomial (with order higher than 3) and any changing lines.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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