Question 328393
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ ax^2\ +\ bx +\ c]


Is a parabola.  The *[tex \Large x]-coordinate of the vertex is *[tex \Large \frac{-b}{2a}],  The *[tex \Large y]-coordinate is *[tex \Large p\left(\frac{-b}{2a}\right)\ =\ a\left(\frac{-b}{2a}\right)^2\ +\ b\left(\frac{-b}{2a}\right) +\ c]


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


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


The  *[tex \Large y]-intercept is the point *[tex \Large (0,p(0))], which is to say *[tex \Large (0,c)].  By symmetry, there is also a point on the parabola at *[tex \Large \left(\frac{-b}{a},c\right)]


If *[tex \Large b^2\ -\ 4ac\ > 0], then there are two *[tex \Large x]-intercepts at *[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)]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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