Question 394799
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Given


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


The *[tex \Large x]-coordinate of the vertex is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v\ =\ -\frac{b}{2a}]


The *[tex \Large y]-coordinate of the vertex is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v\ =\ \rho(-\frac{b}{2a})\ =\ a\left(-\frac{b}{2a}\right)^2\ +\ b\left(-\frac{b}{2a}\right)\ +\ c]


Hence the vertex is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x_v,y_v\right)]


The line of symmetry is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ x_v]


The maximum or minimum value is *[tex \Large y_v\ =\ \rho\left(-\frac{b}{2a}\right)]


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


{{{drawing(
500, 500, -12,12,-12,12,
grid(1),
graph(
500, 500, -12,12,-12,12,
3x^2-12x+16))}}}


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