The Parabola Project Report
By: Juan Estrada
Introduction
In This project we were asked to investigate the patterns in the intersections of parabolas and the lines y=x and y=2x.I was then asked to prove my conjectures and to broaden the scope of the investigation to include other lines and other types of polynomials. They gave methods to do. I was given 4 steps in completing this project and each of these steps had more steps.
Steps
For the first step they asked me to consider this parabola y=(x-3)^2+2=x^2-6x+11 and the lines y=x and y=2x. I then we use technology to find the four intersections illustrated on the picture that they provided us with that included the parabola and the lines y=x and y=2x. We were asked to label the values of from left to right on the x-axis as Xsub1, Xsub2, Xsub3, and Xsub4. Were then asked to find the values of Xsub2-Xsub1 and Xsub4-Xsub3 and to name them respectively SL and SR. For the last part of the first step I was asked to calculate D=|SL-SR|, here is what I got. Xsub1:1.76393203, Xsub2: 2.2381966011, Xsub3: 4.618033989 and Xsub4: 6.236067978. After I finished finding these points I went ahead and continued and found the answers for Xsub2-Xsub1 (SL) and Xsub4-Xsub3 (SR). SL= 0.618033988 and SR=1.6180339887. I was then asked to subtract these as I said earlier and got that D=|SL-SR| came out to be a +1.
For the second step I was asked to find the values of D for other parabola of the form Y=ax^2+bx+c, a>0, with vertices in quadrant 1, intersected by the lines y=x and y=2x. I was also told to consider various values of a, beginning with a=1. Lastly I was asked to make a conjecture about the values of D for these parabolas. So I started to play around with parabolas in quadrant 1. I started to manipulate the value of “a” starting at 1. I didn’t get too complex because it would involve a lot more work.
For the third step I was asked to investigate my conjecture for any real values of “a” and any placement of the vertex. I was also asked to refine my conjecture as necessary and to prove it. Lastly, I was asked to maintain the labeling convention used in parts 1 and 2 by having the intersections of the first line to be Xsub2 and Xsub3 and the intersections with the second line to be Xsub1 and Xsub4. So as I started to play around with the equation of the parabola, I started to notice a weird pattern as I looked closely. When I decided to put 3x^2 my answer was 1/3 and if I put 5 instead I would get 1/5, and if you notice, when I did 1 the answer was 1/1, or just 1. So I came up with a hypothesis that the answer to these equations will absolutely always be the inverse of my value for the parabola equation. The vertex of the parabola didn’t affect the outcome, and as long as the parabola had four intersections then it should work with the lines y=x and y=2x. An example is the line f(x)=(x^2-6)+11 intersecting lines y=x and y=2x. I started to enter the parabola and the two intersecting lines along with the four values of the intersections. The intersections given to me by the calculator were Xsub1= 1.763932023, Xsub2= 2.381966011, Xsub3= 4.618033989 and Xsub4= 6.236067978. After this, start going through the same process as the first part of the process, you have to enter the exact same numbers as appeared or else the final answer will be completely inaccurate. As I finished the process for this equation, I found the answer to be 1 and also for the problem the value of “a” was also 1. Based on this data along with other tries I found my conjecture to be correct.
For the last part of this project, I was asked if my conjecture hold if the intersecting lines changed. I had already done this above.Conjecture
I Conjecture that the “a” value has basically everything to do with the outcome of the answers. “a” will be the inverse of the final answer