The function $h(x) = x^3 + bx^2 + d$ has a critical point at $(2, -4)$. Determine the constants $b$ and $d$ and find the equation of $h(x)$. The function $h(x) = x^3 + bx^2 + d$ has a critical point at $(2, -4)$. Determine the constants $b$ and $d$ and find the equation of $h(x)$. Please help me answer this question.

The critical points occur where the derivative is equal to zero.
\begin{align*}
h(x) & = x^3 + bx^2 + d\\
h'(x) & = 3x^2 + 2bx\\
      & = x(3x + 2b)
\end{align*}
Setting the derivative equal to zero yields
\begin{align*}
x & = 0 & 3x + 2b & = 0\\
  & & 3x & = -2b\\
  & & x & = -\frac{2b}{3}
\end{align*}
We know that a critical point occurs when $x = 2$.  Hence, 
$$2 = -\frac{2b}{3} \implies b = -3$$
Thus, 
$$h(x) = x^3 - 3x^2 + d$$
We also know that $h$ has the critical point $(2, -4)$.  Hence, 
$$h(2) = 2^3 - 3 \cdot 2^2 + d = 8 - 12 + d = -4 + d = -4$$
Thus, $d = 0$. 
As a check, you should verify that $h(x) = x^3 - 3x^2$ has a critical point at $(2, -4)$.
