Let \[f(x) =
\begin{cases}
2x^2 - 3&\text{if } x\le 2, \\
ax + 4 &\text{if } x>2.
\end{cases}
\]Find $a$ if the graph of $y=f(x)$ is continuous (which means the graph can be drawn without lifting your pencil from the paper).
So, for the graph of $y=f(x)$ to be continuous, the function values at $x=2$ must match for both pieces. That is, I need to find $a$ such that $2(2)^2-3 = a(2) + 4$. Simplifying, I get $2(4)-3 = 2a + 4$, or $5 = 2a + 4$. Subtracting $4$ from both sides, I get $1 = 2a$. Dividing both sides by $2$, I get $a = \frac{1}{2}$. This is the value of $a$ that makes the graph of $y=f(x)$ continuous. 