problem: If we express $3x^2 + x - 4$ in the form $a(x - h)^2 + k$, then what is $k$?
solution: We complete the square.  First, we factor 3 out of the terms $3x^2 + x$ to get $3 \left( x^2 + \frac{x}{3} \right)$.  We can square $x + \frac{1}{6}$ to get $x^2 + \frac{x}{3} + \frac{1}{36}$, so \begin{align*}
  3 \left( x^2 + \frac{x}{3} \right) &= 3 \left[ \left( x + \frac{1}{6} \right)^2 - \frac{1}{36} \right]\\
  &= 3 \left( x + \frac{1}{6} \right)^2 - \frac{3}{36}\\
  & = 3 \left( x + \frac{1}{6} \right)^2 - \frac{1}{12},\end{align*}and \begin{align*}3 \left( x^2 + \frac{x}{3} \right) - 4 &= 3 \left( x + \frac{1}{6} \right)^2 - \frac{1}{12} - 4\\
  & = 3 \left( x + \frac{1}{6} \right)^2 - \frac{49}{12}.\end{align*}We see that $k = \boxed{-\frac{49}{12}}$.