Question 986161
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx\ +\ c]


If *[tex \Large (-1,-4)] is an element of the solution set, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-1)^2\ +\ b(-1)\ +\ c\ =\ -4]


Likewise:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(2)^2\ +\ b(2)\ +\ c\ =\ -7]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(3)^2\ +\ b(3)\ +\ c\ =\ 0]


Simplifying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ +\ c\ =\ -4]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ +\ c\ =\ -7]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ +\ c\ =\ 0]


Solve the 3X3 system of equations for the values of the coefficients of the desired quadratic.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \