Answer
$\frac{x^3}{3}-4x+\frac{16}{3}$
Work Step by Step
$f^{''}(x) = 2x⇒ f^{'}(x) = x^2 + C$, C being the constant of integration
We are given that "f has a horizontal tangent at (2, 0)". This statement reveals two things:
1) $f^{'}(x)=0|_{x=2}$ or simply $f^{'}(2) = 0$ as a horizontal tangent means that the slope of the tangent is zero.
2) The point (2, 0) lies on the curve
By (1), $f'(2)=2^2+C = 0 ⇒ C=-4$
Thus, $f^{'}(x) = x^2 -4$
Also, $f^{'}(x) = x^2 -4 ⇒ f(x) = \frac{x^3}{3} - 4x +C'$
By (2), $ f(2)= \frac{2^3}{3} - 8 +C'=0$ or $C' = \frac{16}{3}$
Putting the value of $C'$,
$f(x)= \frac{x^3}{3}-4x+\frac{16}{3}$
