Question 622752
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For *[tex \LARGE \ \ \ \ \ \ \ \ \ \ \phi(x)\ =\ ax^2\ +\ bx\ +\ c]


The vertex is located at *[tex \LARGE \left(\frac{-b}{2a},\,\phi\left(\frac{-b}{2a}\right)\right)]


and the axis of symmetry is the vertical line *[tex \LARGE x\ =\ \frac{-b}{2a}]


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