Ib math HL portfolio parabola investigation

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Math HL Portfolio

Parabola Investigation


In this portfolio I am going to investigate the patterns formed when a parabola,

y=ax2+bx+c intersects with the lines y=x and y=2x. Further I would be broadening the scope of this investigation to other lines and other orders of polynomials and observing patterns in their respective intersections.  

The first part of my portfolio would include parabolas with their turning points located in the 1st quadrant of a graph. I would be investigating the patterns formed when the two lines y=x and y=2x intersect with parabolas that have different coefficients of x2 but lie in the first quadrant and form a conjecture. Then I would prove it and test its limitations.

The second part would include testing and modifying my conjecture for the turning point of the parabola situated in any quadrant and testing for all real co-efficient of x2.

The third part would include modifications to the conjecture if the intersecting lines are changed and hence finding its limitations.

In the final part of my portfolio I will also try and make a conjecture for polynomials of higher order and derive any patterns or observations plausible.


A polynomial is a mathematical expression involving a sum of powers in one or more  multiplied by coefficients. A polynomial in one variable (i.e., a univariate polynomial) with constant coefficients is given by:

A parabola is the name given to the shape of the graph formed with any polynomial of degree 2. This polynomial can be expressed in three forms:

  • f(x) = ax2+bx+c , a≠0  This is the general form.
  • f(x)=a(x-h)2 + k, a≠0  This is the vertex form where the point (h,k) is its turning point.
  • f(x)=a(x-m)(x-n), a≠0  This is the factored form where m and n are the two roots of the equation.

In the 1st part of my portfolio I will only consider a>o, giving me a minimum point for each parabola rather than a maximum. The 2nd would include values of a < 0.

We know that for a graph to have its turning point in the first quadrant and a>0, it should have no real roots. This means that the parabola would not cut with the x-axis and its determinant would be greater than zero. Hence: ,where a >0 .

We also know that for the turning point (h,k) to be in the first quadrant both h and k must be greater than zero .

We want to find four intersection points. This means that the parabola MUST intersect with y=x. If a parabola cuts y=x, then it also cuts y=2x. For a parabola to cut y=x at two points, its turning point should be below the line. For the turning point to be below the line the x-coordinate of the vertex should be greater than the y-coordinate. This implies h>k. I am going to keep the values of h at 3 and k at 2 constant in this part, and change the values of ‘a’ to see any emerging pattern.

Thus our equation is :

y = a(x-3)2 + 2

  1. Now let us consider the parabola y= (x-3)2+2 = x2-6x+11 and the lines y=x and y=2x.
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The vertex of this equation is (3, 2) and the value, a = 1. Thus, this parabola has its turning point in the first quadrant.

  1. Using the software  ‘Fx Graph4’ I have plotted these 3 functions on a graph.
  2. Then I have found the four intersection points of these three functions and labeled them as X1, X2, X3, X4 respectively from left to right. (see Graph 1)

Graph 1: Value of a=1

  1. Then I found out the values of (X2 – X1) and (X4 – X3) and called them SL and ...

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