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

2. Label the x – values of these intersections as they appear from left to right on the x – axis as x1, x2, x3, and x4

3. Find the values of x2- x1 and x4 - x3 and name them respectively SL and SR.

1. Since,

2. Through the part one and part two, my conjecture is a D = 1/a.
3. However, there are some possibilities to change my conjecture.

3. Investigating conjecture (a = -1)

1. y = ax2 + bx + c
2. let a = -1, b = 6 and c = -3
3. y = -x2+6x-3

Process

1. x2 - x1 = 0.70 – 1 = -0.30 = SL  
2. x4 - x3 = 3 – 4.3 = - 1.30 = SR
3. So, the ㅣSL - SR ㅣ = ㅣ-0.30 - -1.30ㅣ = ㅣ1.00ㅣ= 1., in other words, D = 1
4. However, there are some problems because my conjecture is D = 1/a but it seems not working in this problem. Since a = -1, if I use my conjecture, then answer is 1 = -1 which is incorrect.
5. Therefore, let’s try another one in order to figure out another conjecture.

4. Investigating Conjecture ( a = -2)

1. let a = -2, b = -6 and c = -5.
2. -2x2-6x-5

Process

i.  x2 - x1 = -3.2 + 2.5 = =- 0.7 = SL

                ii.  x4 - x3 = -1 + 0.80= - 0.20 = SR

                - So, the ㅣSL - SR ㅣ = ㅣ-0.7 - -0.20ㅣ = ㅣ-0.50ㅣ= 0.50, in other words, D = 0.50

                - Even if the graph is on the quadrant 3, the conjecture still works.

                - Again, if I use the conjecture ( D = 1/a, then it gives 0.50 = - 0.50 which is not correct.

3. Investigating Conjecture (a = -3)

1. let a = -3, b = -9 and c =-2
2. -3x2-9x-2

Process

                 i. x2 - x1 = -3.47 +3.12 = - 0.35 = SL

                 ii. x4 - x3 = -0.21 + 0.19 = - 0.02 = SR

Hence, the ㅣSL - SR ㅣ = ㅣ-0.35 - -0.02ㅣ =ㅣ-0.33ㅣ = 0.33, thus, D = 0.33

Again, D = 1/a is not working for this problem.  

f. Patterns (for a<0)

                i. So, I listed all the answer.

        A.  a = -1, D = 1

        B.  a = -2, D = 0.5 or 1/2

        C.  a = -3, D = 0.33 or 1/3

It seems like when a = negative number, then the conjecture ( D = 1./a) doesn’t work. Therefore, what makes the value always positive?

The answer is absolute value

Therefore, it is clear that the conjecture has to be D = 1/ lal

x8 - x7 = 5.0 – 4.9 = 0.1 = Sn

x6 - x5 = 0.9 -0.85 = 0.05 = SL

x4 - x3 = -3.39 +3.29 = -0.1 = SR

x2 - x1 = -3.9 + 3.8 = -0.1 = Sm

Hence, equation will look like

D = ㅣㅣ-0.1 - -0.1ㅣ – ㅣ0.05 – 0.1ㅣㅣ – 0.05

ㅣㅣ0ㅣ – ㅣ-0.05ㅣㅣ- 0.05

0.05-0.05 = 0, in other words, D = 0

Conclusively, through this part six, it is clear that D is always equal to 0. In addition, through this part, I learned that even if the numbers are getting bigger like x5 and x6, there is no change to D. Since the numbers are getting bigger, then the equation is also changing. For instance, when is the cubic polynomials, equation will be D = ㅣ Sm – SR ㅣ- SL. Also, when the equation is quadric polynomials, equation looks like D = ㅣㅣ Sm – SR ㅣ- ㅣSL – Sn ㅣㅣ. Without doubt, D is always equal to 0 because when the numbers are changing, equations are changing also.

Conclusion

Finally, the portfolio ends with enormous amounts of graph and equations. Through theses parts, I found the new concept that can’t be found in class period. I never had time to think seriously about the conjecture so I am actually glad that I experienced and finished this portfolio. First of all, I needed to make the conjecture in order to understand the problems and patterns. Therefore, in part one, two, three and four. I tried to come up with the conjecture that is suitable to problems. Moreover, part five and six were challenging to me. Since, I never thought that there are some problems about cubic polynomials. All in all, this portfolio was hard. However, it was good experiences to me. The Investigation of Parabola helped me a lot to understand the concept of parabolas, lines and intersections. Through this portfolio, I found enormous sources and information that were not in the HL Math book, and solve many problems that I concerned during the class period. Equally important, I am really thanks to Dr. Hall, who is the conductor of our class, for improving the student to learn and fathom the mathematic concepts using this portfolio.