Question 332331
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I'll do you one better.  I'll tell you how to find the *[tex \Large y]-coordinate of the vertex of any parabola with equation in standard form.


Given 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx +\ c]


first find the *[tex \Large x]-coordinate of the vertex using:


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


then evaluate your function for that value of the independent variable:


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



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