Question 519031
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For any quadratic function of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx +\ c]


The coordinates of the vertex are found by:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = f(x_v) = f\left(\frac{-b}{2a}\right)]


And the equation of the axis of symmetry is:


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


Which is to say:


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


Just plug in the numbers 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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