For which values of $k$ is $f(x) = \begin{cases} k^{2}x, & x < 2 \\ x + 3k, & x \ge 2 \end{cases}$ continuous at $x = 2$?

Explanation

1. 01

   The pieces must meet at $x = 2$: $2k^{2} = 2 + 3k$.

2. 02

   $2k^{2} - 3k - 2 = (2k + 1)(k - 2) = 0$, so $k = 2$ or $k = -\frac{1}{2}$.