Question 458254
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Given the quadratic function 


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


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


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


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


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


The axis of symmetry is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ x_v]


The lead coefficient is negative, hence the parabola opens downward, hence the vertex is a maximum, and the maximum value is the value of the function (read *[tex \Large y]-coordinate) at the vertex.


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