Question 347826
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The vertex of a parabola described by the function 


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


has an *[tex \Large x]-coordinate of *[tex \Large \frac{-b}{2a}] and a *[tex \Large y]-coordinate of *[tex \Large f\left(\frac{-b}{2a}\right)].  The axis of symmetry is the vertical line *[tex \Large x\ =\ \frac{-b}{2a}].  If the lead coefficient is positive, the parabola opens upward.  Therefore the value of the function at the vertex (i.e. the *[tex \Large y]-coordinate as discussed above) is the minimum value of the function.  If the lead coordinate is less than zero, then the parabola opens downward and the vertex is a maximum.


For your function, the *[tex \Large x]-coordinate of the vertex is *[tex \Large \frac{-2}{2(-2)}\ =\ \frac{1}{2}].  You can calculate *[tex \Large f\left(\frac{1}{2}\right)] to get the *[tex \Large y]-coordinate of the vertex.  The rest of your questions fall right in line once you have the coordinates of the vertex.



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