Point on the graph of $y=\sqrt{4x+13}$ closest to $(5,0)$? Just did this question on an exam earlier today, I'm curious to see if I'm correct.

What point on the graph of $y=\sqrt{4x+13}$ is closest to $(5,0)$?

My answer: $(-1,3)$

Your method described in the comments seems correct. You start by noting that the distance between $(x,y(x))$ and $(5,0)$ is given by
$$d(x) = \sqrt{(5 - x)^2 + (0 - y(x))^2} = \sqrt{(5 - x)^2 + 4x + 13} = \sqrt{x^2 - 6x + 38}. \tag{1}$$
You want to minimize $d(x)$, so indeed you want to look at the derivative:
$$d'(x) = \frac{1}{2 \sqrt{x^2 - 6x + 38}} \cdot (2x - 6) = \frac{x - 3}{\sqrt{x^2 - 6x + 38}}$$
Solving $d'(x) = 0$ for $x$ we find $x = 3$, and this is indeed a global minimum; this could be proved by showing that $d''(3) > 0$. 
So in the end we find a minimum at $(3, \sqrt{4\cdot3+13}) = (3,5)$, and this point has distance $d(x) = \sqrt{3^2 - 6\cdot 3 + 38} = \sqrt{29}$ from $(5,0)$.

In this case you can take a shortcut from $(1)$ to the solution as noted by Arthur, by noting that 
$$d(x)^2 = x^2 - 6x + 38 = (x - 3)^2 + 29 \geq 29$$
where the inequality follows from $(x-3)^2 \geq 0$, and equality holds only if $x - 3 = 0$, i.e. if $x = 3$.
