<PRE><B>Question 788407</B>
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You could make a table of *[tex \LARGE x] values and then calculate the *[tex \LARGE y]-values that correspond.  But it would probably be helpful to compare this equation to the equation of a parabola with vertex at *[tex \LARGE  (h,\,k)], namely


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a(x\ -\ h)^2\ +\ k]


That gives you the vertex directly and allows you to plot two points for every one you calculate after that because of symmetry.


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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