1. Prove that (f(x+h)-f(x-h))/2=f'(x)

2. Prove that any parabola sastifies the equation (f(x+h)-f(x-h))/2=f'(x)
For the first question, I tried to solve it but there is an extra h tacked on to one side. I have no clue what to do for the second question.

First off, the equation should be

(f(x+h)-f(x-h))/(2h)

if f(x) = a(x-b)^2+c, then
f(x+h) = a(x-b+h)^2+c
f(x-h) = a(x-b-h)^2+c
f(x+h)-f(x-h) = a((x-b +h)^2 - (a-b -h)^2
= 4ah(x-b)

divide that by 2h and you have 2a(x-b)

Now, f'(x) = 2a(x-b)

ta-daa