Question 1128654
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General Quadratic Function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ ax^2\ +\ bx\ +\ c]


Stop thinking about a point on the graph of a function as *[tex \LARGE (x,\,y)], rather consider it as *[tex \LARGE (x,\,f(x))]. 


So, if *[tex \LARGE f(1)\ =\ 0], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ +\ c\ =\ 0]


And if *[tex \LARGE f(2)\ =\ -7], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ +\ c\ =\ -7]


And if *[tex \LARGE f(5)\ =\ -16], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25a\ +\ 5b\ +\ c\ =\ -16]


Solve the 3X3 system for a, b, and c.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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