Question 1038746
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Use the vertex form.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ a(x\ -\ h)^2\ +\ k]


where *[tex \Large (h,k)] is the vertex of the parabola.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ a(x\ -\ 2)^2\ +\ 3]


Substitute the coordinates for (4,4):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4\ =\ a(4\ -\ 2)^2\ +\ 3]


and solve for *[tex \Large a]


Then you can insert the value of *[tex \Large a] and then expand and rearrange to get the standard form of the function if you wish.


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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