Question 1032297
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A quadratic function has the form:


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


If *[tex \Large \left(x,\,\rho(x)\right)\ =\ (-2,-20)] then


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


Likewise, for *[tex \Large \left(x,\,\rho(x)\right)\ =\ (0,-4)] and *[tex \Large \left(x,\,\rho(x)\right)\ =\ (4,-20)]


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


and


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



Which yields the 3X3 system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ +\ c\ =\ -20]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0a\ +\ 0b\ +\ c\ =\ -4]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16a\ +\ 4b\ + c\ =\ -20]


Which can be simplified to the 2X2 system:


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


Solve the 2X2 system for *[tex \Large a] and *[tex \Large b]. You already know that *[tex \Large c\ =\ -4]



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

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

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