Question 342647
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Given a quadratic function in standard form, namely:


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


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


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = f(x_v) = f\left(\frac{-b}{2a}\right)]


The axis of symmetry is given by:


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


which is to say:


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


If the lead coefficient is greater than zero, the parabola opens upward and therefore the vertex is a minimum.  Otherwise, the parabola opens downward and the vertex is a maximum.


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