Question 318297
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You have a quadratic function in the form *[tex \LARGE y\ =\ f(x)\ =\ ax^2\ +\ bx\ +\ c]


The *[tex \Large x]-coordinate of the vertex is given by *[tex \Large \frac{-b}{2a}]


The *[tex \Large y]-coordinate of the vertex is given by *[tex \Large f\left(\frac{-b}{2a}\right)]


For your problem:


The *[tex \Large x]-coordinate of the vertex is *[tex \Large \frac{-(2)}{2(1)}\ =\ -1]


The *[tex \Large y]-coordinate of the vertex is *[tex \Large f\left(-1\right)\ =\ (-1)^2\ +\ 2(-1)\ -\ 3\ =\ -4]


Hence the vertex is at:  *[tex \Large \left(-1,-4\right)]



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