Question 312543
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Re-write your function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -\frac{10}{3}\left(x\ -\ (-5)\right)^2\ +\ (-5)]


Since the lead coefficient is negative, this parabola opens downward making the vertex a maximum.


The vertex form equation of a parabola with vertical axis symmetry like this one is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ f(x)\ =\ y\ =\ 4p(x\ -\ h)^2\ +\ k]


Your vertex is therefore at *[tex \Large (-5,-5)]


The *[tex \Large y]-coordinate of the vertex is the value of the function at the vertex, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ f(-5)\ =\ -5]


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