Question 975177
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ ax^2\ -\ 4x\ +\ 3]


is a parabola with vertex at *[tex \Large \left(x_v,\,y_v\right)] where *[tex \Large x_v\ =\ -\frac{-4}{2a}\ =\ \frac{2}{a}] and *[tex \Large y_v\ =\ f(x_v)].


Evaluate *[tex \Large f\left(\frac{2}{a}\right)], set the resulting polynomial in *[tex \Large a] equal to 12, and then solve for *[tex \Large a]


Hint: Since you are looking for a maximum, the parabola must open downward, which is to say that *[tex \Large a\ <\ 0].  If you don't come up with a negative value for *[tex \Large a], check your arithmetic.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \