Question 287526
Integrate (find the antiderivative) twice:

f'(x)=2x^2+C
f(x)=(2/3)x^3+Cx+D


Now, f has a horizontal tangent at (2,0) IE the function attains a local extrema and f'(2)=0

hence

f'(2)=2*2^2+C=0
implies C=-8.

Now we have f(x)=(2/3)x^3-8x+D.

But we know f(2)=0, so, (2/3)2^3-8(2)+D=0
16/3-48/3=-D
-32/3=-D

implies 32/3=D. 

This gives our final function as

f(x)=(2/3)x^3-8x+32/3.