Question 328147
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You didn't bother to share the function, but there are a few things that I can share that might help anyway.


For any quadratic polynomial function of the form:


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


If *[tex \Large a\ >\ 0] the graph opens upward, that is, the vertex is a minimum.  If *[tex \Large a\ <\ 0] the graph opens downward.


The *[tex \Large x]-coordinate of the vertex is given by *[tex \Large \frac{-b}{2a}],  and the *[tex \Large y]-coordinate of the vertex is given by *[tex \Large p\left(\frac{-b}{2a}\right)\ =\ a\left(\frac{-b}{2a}\right)^2\ +\ b\left(\frac{-b}{2a}\right)\ +\ c]


So from what you have told me:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ <\ 0]


The *[tex \Large x]-coordinate of the vertex is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-b}{2a}\ =\ -1]


The *[tex \Large y]-coordinate of the vertex is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(-1)\ =\ a(-1)^2\ +\ b(-1) +\ c\ =\ 8]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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