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f'(x)=2ax+b
We have f'(2)=0 so 4a+b=0 and b=-4a
f(x)=ax^2-4ax+c
or f(2)=4 and f(3)=7
4a-8a+c=4 and 9a-12a+c=7
-4a+c=4 and -3a+c=7
a=3 and c=16
Finally f(x)=3x^2-12x+16
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