Question 366636
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For the quadratic function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ ax^2\ +\ bx +\ c]


The *[tex \Large x]-intercepts are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-b\ +\ \sqrt{b^2\ -\ 4ac}}{2a},0\right)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-b\ -\ \sqrt{b^2\ -\ 4ac}}{2a},0\right)]


The *[tex \Large y]-intercept is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (0,p(0))]


The vertex of the parabola is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-b}{2a},p\left(\frac{-b}{2a}\right)\right)]


If *[tex \Large a\ >\ 0], then the parabola opens upward and the vertex is a minimum.  If *[tex \Large a\ <\ 0], then the parabola opens downward and the vertex is a maximum.


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