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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