Question 234141
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I think you mean "quadratic function."  That's because *[tex \LARGE  f(x)\ =\ x^2\ -\ 2x\ -\ 35], is, in fact, a quadratic function.  The Quadratic Formula is something else entirely.


Given a quadratic function of the form: *[tex \LARGE  f(x)\ =\ ax^2\ +\ bx\ +\ c]


The *[tex \Large x]-coordinate of the vertex is *[tex \Large \frac{-b}{2a}].  The *[tex \Large y]-coordinate of the vertex is the value of the function at that *[tex \Large x]-value.


So: *[tex \LARGE \frac{-b}{2a}\ =\ \frac{-(-2)}{2(1)}\ =\ 1], and *[tex \LARGE f(1)\ =\ (1)^2\ -\ 2(1)\ -\ 35\ =\ -36].  Hence the vertex is at *[tex \LARGE \left(1,\,-36\right)]



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