<PRE><B>Question 753633</B>
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The vertex of *[tex \LARGE f(x)\ =\ ax^2\ +\ bx\ +\ c] is at the point *[tex \LARGE \left(\frac{-b}{2a},\,f\left(\frac{-b}{2a}\right)\right)] and the equation of the axis of symmetry is *[tex \LARGE x\ =\frac{-b}{2a}].  Re-write your equation into *[tex \LARGE f(x)\ =\ ax^2\ +\ bx\ +\ c] form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{1}{8}x^2\ +\ 0\,\cdot\,x\ +\ 0]


and then calculate *[tex \LARGE \frac{-b}{2a}\ \ ] and *[tex \LARGE \ \ f\left(\frac{-b}{2a}\right)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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