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Mathematics 15 Online
OpenStudy (anonymous):

Find a such that f(x)=ax2-4x+3 has a max. value of 12.

OpenStudy (anonymous):

max is at the vertex, first coordinate of the vertex is \(-\frac{b}{2a}\)

OpenStudy (anonymous):

in your case it is \(\frac{2}{a}\) so i guess you evaluate at \(x=\frac{2}{a}\) set the result equal to 12 and solve for \(a\)

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