Question 1039113
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The general model for quadratic function is:


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


Given that the point *[tex \LARGE (-2,-20)] is on the graph of *[tex \LARGE  y\ =\ ax^2\ +\ bx\ +\ c], then the following must hold:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-2)^2\ +\ b(-2)\ +\ c\ =\ -20]


which is to say,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ +\ c\ =\ -20]


Similarly, considering the other two given points:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(0)^2\ +\ b(0)\ +\ c\ =\ -4]


which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ -4]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(4)^2\ +\ b(4)\ +\ c\ =\ -20]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16a\ +\ 4b\ +\ c\ =\ -20]


Substituting the now known value of *[tex \LARGE c] and simplifying, we get the following 2X2 system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ =\ -16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16a\ +\ 4b\ =\ -16]


Solve the 2X2 system for a and b, which, along with -4 for c can be substituted into *[tex \LARGE y\ =\ ax^2\ +\ bx\ +\ c] to get the desired function.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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