IB HL Math

Block 2

12/09/08

Bonmin Koo

Parabola Investigation: IB HL Math Type I Portfolio

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.

Background

Functions: 1)ax + b, a ≠ 0                              - Linear Equation

         2) ax2+ bx+c                                 - Quadratic

3) ax3+ bx2+cx+d                             - Cubic

         4) ax4+ bx3+cx2+dx+e                         - Quartic

        

Introduction

The investigation of parabola brings me to think and promote my knowledge about quadratic functions. Also, it helps me to find the ‘patterns’ and ‘rules’ of parabola which was another great source to improve my mathematic skills. In the HL type I, it asks about thoughtful questions such as relationship between parabolas and lines. There are six questions in this portfolio and each question requires rumination. I will try to focus on the member of the family of polynomials precisely since they are the basic functions of these problems and it will help me to ‘link’ some ideas. Equally important, interestingly, this portfolio wants the student to find their ‘own’ conjecture. Thus, I will try to focus on the patterns the questions have in order to find accuracy conjectures.

Part One

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

  1. Using the technology find the four intersections illustrated on the right
  1. Intersection points #1: (1.76, 3.53), (6.24, 12.47) – these points are intersection points with y = 2x
  2. Intersection points #2: (2.38, 2.38), (4.62, 4.62) – these points are intersection points with y = x
  1. Label the x – values of these intersections as they appear from left to right on the x – axis as x1, x2, x3, and x4

  1. Find the values of x2- x1 and x4 - x3 and name them respectively SL and SR.
  1. Since, x2- x1 = 2.38 – 1.76 = 0.62. Hence, SL = 0.62
  2. In addition, x4 - x3 = 6.24 – 4.62 = 1.62. Thus, SR. = 1.62
  1. Finally, calculate D = ㅣSL - SR 
  1. So, ㅣSL - SR = ㅣ0.62 – 1.62ㅣ = ㅣ-1ㅣ
  2. Therefore, D equals to 1 because of the absolute value, -1 becomes to 1.
  3. Equally important, it is clear that when the number is greater than 0. In other words, when D > 0, it intersects with two different points.

Part Two

  1. Find the values of D for 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. Consider various values of a, beginning with a =1. Make a conjecture about the value of D for these parabolas.

  1. To find the conjecture (a = 1)
  1. Equation: y = ax2 + bx + c
  2. So let say a=1, b = -4 and c = 5
  3. y = x2-4x+5

Process

  • x2 - x1 = 1.39 – 1 = 0.39 = SL
  • x4 - x3 = 5 – 3.61 = 1.39 = SR
  • So, the ㅣSL - SR ㅣ = ㅣ0.39 – 1.39ㅣ = ㅣ-1ㅣ= 1, in other words, D = 1. 

  1. To find the conjecture (a = 2)
  1. Equation: y = ax2 + bx + c
  2. let a =2, b = -10, and c = 14
  3. y = 2x2-10x+14

Process

  • x2 - x1 = 2 – 1.58 = 0.41 = SL
  • x4 - x3 = 4.41 – 3.5 = 0.91 = SR
  • So, the ㅣSL - SR ㅣ = ㅣ0.41 – 0.91ㅣ= ㅣ-0.5ㅣ= 0.5, in other words, D = 0.5

  1. To find the conjecture (a = 3)
  1. let a = 3, b = -11 and c = 11
  2. y = 3x2-11x+11

Process

  • x2 - x1 = 1.42 – 1.14 = 0.28 = SL
  • x4 - x3 = 3.18 – 2.57 = 0.61 = SR
  • So, the ㅣSL - SR ㅣ = ㅣ0.28 – 0.61ㅣ = ㅣ-0.33ㅣ= 0.33, in other words, D = 0.33

  1. To find the conjectures ( a = 4)
  1. let a = 4, b = -9 and c = 6
  2. y = 4x2-9x+6

Process

  • x2 - x1 = 1 – 0.75 = 0.25 = SL
  • x4 - x3 = 2 – 1.5 = 0.5 = SR
  • So, the ㅣSL - SR ㅣ = ㅣ0.25 – 0.5ㅣ = ㅣ-0.25ㅣ = 0.25, in other words, D =0.25

  1. To find the conjectures (a =5)
  1. let a = 5, b =-14 and c = 11
  2. 5x2-14x+11

        Process

  • x2 - x1 = 1.27 – 1 = 0.27 = SL
  • x4 - x3 = 2.2 – 1.72 = 0.47 = SR
  • So, the ㅣSL - SR ㅣ = ㅣ0.27 – 0.47ㅣ = ㅣ-0.20ㅣ = 0.20, in other words, D = 0.20

  1. Patterns

  1. a = 1, D = 1
  2. a = 2, D = 0.5 or 1/2
  3. a = 3, D = 0.33 or 1/3
  4. a = 4, D = 0.25 or 1/4
  5. a = 5, D = 0.20 or 1/5
Join now!

Clearly, there is pattern. According to the results that I have, I can make the conjecture.

Since (D)(a) = 1, D has to be equal to 1/a.

Consequently, the conjecture is D = 1/a

It is not about the conjecture but I am introducing this rule in order to demonstrate the exceptions.

  1. Basic rule

i. Relationship

  1. According to the answer from part 1, when D = 1, it clearly depicts that graph intersects with two different points.
  2. In fact, the vertex of graph is equal to ( -b/2a, (-(b2- 4ac)/ 4a) ) ...

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