Mark M. answered 04/16/20
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
f'(x) = ∫f"(x)dx = (1/2)x2 - 2x + C1
Since f'(2) = -5, we have -2 + C1 = -5. So, C1 = -3
f'(x) = (1/2)x2 - 2x - 3
f(x) = ∫f'(x) dx = (1/6)x3 - x2 - 3x + C2
Since f(0) = -4, C2 = -4
So, f(x) = (1/6)x3 - x2 - 3x - 4.
Therefore, f(6) = 36 - 36 - 18 - 4 = -22