Posted by Jacob on Wednesday, February 27, 2013 at 7:08pm.
f'(x) = -2x+a
if f'=0 at x=2, then a=4
f(x) = -x^2 + 4x + b
f(2) = 13 = -4+8+b
b=5
f(x) = -x^2 + 4x + 5 = -(x+1)(x-5)
vertex is at x=(5-1)/2 = 2 as desired
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