Question 1038747
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A quadratic function has the form *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c]


So if you let *[tex \Large x] represent the time since the ball was dropped and *[tex \Large y] represent the height of the ball at that time, then we have three ordered pairs that must be in the solution set of the desired function, namely *[tex \Large (0,576)], *[tex \Large (3,432)], and *[tex \Large (6,0)].


Substituting in the general form of the function for the first point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(0)^2\ +\ b(0)\ +\ c\ =\ 576]


which reduces to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ 576]


Then substituting the second and third points as well as the value we now know for the constant term we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(3)^2\ +\ b(3)\ +\ 576\ =\ 432]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(6)^2\ +\ b(6)\ +\ 576\ =\ 0]


Simplifying these two expressions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ =\ -144]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 36a\ +\ 6b\ =\ -576]


Solve the 2X2 system of linear equations to find the coefficients *[tex \Large a] and *[tex \Large b] to put into:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx\ +\ 576]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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