Question 443064: I do not want an answer to the following problem. I did the work, and I am not getting the right answer. I would like to know where I made my mistake so that I can get the correct answer. Thank you.
Month Share Value (in dollars)
0 55
3 16
5 0
7 -8
9 -8
11 0
13 16
15 40
1. Using the table above, find the quadratic equation that represents the monthly value of a share of ACME Corporation since January 2009. Use the variable x to represent the number of months after January 2009 and the variable y to represent the monthly value.
I used the points (0,55),(5,0), and (11,0) to come up with the quadratic equation. I used (0,55) to solve for c in the equation ax^2+bx+c=y. I got 55 for c. I plugged that into the equation with (5,0) for x and y. I got 25a+5b+55=0, and (11,0) to get 121a+11b+55=y. I multiplied the first equation by -11, and the second one was multiplied by 5 to cancel out b and solve for a. The equation I ended up with is x^2-5x+55, but I do not get a true equation when I plug in the points. Please help me understand what I did wrong. This is due tomorrow.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I do not want an answer to the following problem. I did the work, and I am not getting the right answer. I would like to know where I made my mistake so that I can get the correct answer. Thank you.
Month Share Value (in dollars)
0 55
3 16
5 0
7 -8
9 -8
11 0
13 16
15 40
1. Using the table above, find the quadratic equation that represents the monthly value of a share of ACME Corporation since January 2009. Use the variable x to represent the number of months after January 2009 and the variable y to represent the monthly value.
----
I ran a Quadratic Regression program against the data and got
the following: y = x^2-16x+55, which is the correct equation.
------------------------------------------
I used the points (0,55),(5,0), and (11,0) to come up with the quadratic equation. I used (0,55) to solve for c in the equation ax^2+bx+c=y. I got 55 for c. I plugged that into the equation with (5,0) for x and y. I got 25a+5b+55=0, and (11,0) to get 121a+11b+55=y. I multiplied the first equation by -11, and the second one was multiplied by 5 to cancel out b and solve for a. The equation I ended up with is x^2-5x+55, but I do not get a true equation when I plug in the points. Please help me understand what I did wrong. This is due tomorrow.
---
Using your method with (0,55),(5,0),(11,0) on ax^2+bx+c = y, I get:
(0,55): 0 + 0 + 1 = 55
(5,0):: 25 + 5 + 1 = 0
(11,0): 121+ 11 + 1 = 0
-----
Using a matrix method on that I get:
a =1; b = -16 ; c = 55
Which gives me y = x^2-16x+55 as the proper answer.
======================================================
Cheers,
Stan H.
|
|
|
