Question 638748
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A quadratic polynomial function of the form *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c] has a graph that is a parabola.  Any point on the parabola has coordinates *[tex \LARGE \left{x,\rho(x)\right)]


Therefore if *[tex \LARGE (-180,40)] is a point on the parabola, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-180)^2\ +\ b(-180)\ +\  c\ =\ 40]


likewise


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(180)^2\ +\ b(180)\ +\ c\ =\ 40]


and


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


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 32400a\ -\ 180b\ =\ 40]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 32400a\ +\ 180b\ =\ 40]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0a\ +\ 0b\ +\ c\ =\ 0]


Solve the 3X3 system for *[tex \LARGE a,\,b,\,c] and substitute into *[tex \LARGE y\ =\ ax^2\ +\ bx\ +\ c] 


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