Question 486903
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There is another way to do this.


When your parabola is in *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c] form, find the *[tex \Large x] coordinate of the vertex with *[tex \Large \frac{-b}{2a}].  Then find the *[tex \Large y] coordinate of the vertex by *[tex \LARGE \rho\left(\frac{-b}{2a}\right)\ =\ a\left(\frac{-b}{2a}\right)^2\ +\ b\left(\frac{-b}{2a}\right)\ +\ c]


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