<PRE><B>Question 362161</B>
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You have a quadratic equation.  Quadratic equations do not have a vertex.  The corresponding quadratic &lt;i&gt;&lt;b&gt;function&lt;/b&gt;&lt;/i&gt; does have a vertex.  Is that what you meant?


&lt;u&gt;&lt;b&gt;Quadratic Function&lt;/b&gt;&lt;/u&gt;


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ ax^2\ +\ bx +\ c]


The *[tex \Large x]-coordinate of the vertex of the general quadratic function is given by *[tex \Large \frac{-b}{2a}].  The *[tex \Large y]-coordinate is simply the value of the function at that *[tex \Large x]-value, that is to say: *[tex \Large \rho\left(\frac{-b}{2a}\right)]


&lt;u&gt;&lt;b&gt;Quadratic Equation&lt;/b&gt;&lt;/u&gt;


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx +\ c\ =\ 0]


To find the roots, either factor the trinomial and use the Zero Product Rule, or use the Quadratic Formula.  Since your given quadratic equation does not factor over the rationals, the quadratic equation is your only choice.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


Just plug in the coefficients and do the arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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