Question 320535
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You did absolutely everything correctly except for one thing.


You began with the function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -x^2\ +\ 6x]


You calculated the *[tex \Large x]-coordinate of the vertex by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-b}{2a}\ =\ \frac{-6}{2(-1)}\ =\ 3]


And then you proceeded to evaluate the function at this value of *[tex \Large x].  Here is where you made your error.


The function you started with was:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -x^2\ +\ 6x]


But you evaluated:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -2x^2\ +\ 6x\ =\ -2(3)^2\ +\ 6(3)\ =\ 0]


Your evaluation calculation is spot on, but where did you get that extra factor of 2 in the lead coefficient?


What you should have done:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -(3)^2\ +\ 6(3)\ =\ -9\ +\ 18\ =\ 9]


and therefore the vertex is at the point *[tex \Large (3,\,9)]


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