Let f(x) = ax3 + bx2 + cx + 1. Determine the values of a,b, and c so that f(x) has a point of inflection at x = 2, a local minimum at x = -2, and f(1) = 2.

So, f'(x) = 3ax2 + 2bx + c

f"(x) = 6ax + 2b

I know that f'(x) = 0 when x = -2 so: 12a - 4b + c = 0

and f"(x) = 0 when x = 2 so: 6a + b = 0

and a + b + c = 1 since f(1) = 2

I'm not sure if i'm going in the right direction with this question but i'm stuck at this point. Any help would be appreciated.