Question 370633
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The vertex of *[tex \Large p(x)\ =\ ax^2\ +\ bx +\ c] is at the point *[tex \Large (x_v,y_v)] where *[tex \Large x_v\ =\ \frac{-b}{2a}] and *[tex \Large y_v\ =\ p(x_v)\ =\ a(x_v)^2\ +\ b(x_v) +\ c]



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