<s>[INST] 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: $(-1,3)$
 [/INST] Let $P = (x,\sqrt{4x+13})$ be a generic point of your function, and pose $Q=(5,0)$.
We want to find the point $P$ such that the distance between $P$ and $Q$ is minimal. The distance is:
$$d(x) = \sqrt{(x-5)^2 + \left(\sqrt{4x+13} - 0 \right)^2} = $$
$$= \sqrt{(x-5)^2 + |4x+13|} $$
Minimize the distance is equivalent to minimize the square of the distance. That is:
$$f(x) = d^2(x) = (x-5)^2 + |4x+13| $$
Let's take the derivative of $f(x)$ (note that you have to take care about the absolute value!!!):
$$f'(x) = \left\{ \begin{array}{lr}2x - 10 + 4 & x \geq -\frac{13}{4}\\2x - 10 - 4 & x < -\frac{13}{4}\end{array}\right. = \left\{ \begin{array}{lr}2x - 6 & x \geq -\frac{13}{4}\\2x - 14 & x < -\frac{13}{4}\end{array}\right.$$
Now, we must pose $f'(x) = 0$, and then:
$$\left. \begin{array}{lr}2x-6 = 0 & x \geq -\frac{13}{4}\\2x-14 = 0 & x < -\frac{13}{4}\end{array}\right. \Rightarrow \left. \begin{array}{lr}x = 3 & x \geq -\frac{13}{4}\\x = 7 & x < -\frac{13}{4}\end{array}\right.$$
Cleary, the second solution is not admissible since $7 > -\frac{13}{4}$. Hence, the only solution is $x=3$ and then the point you are looking for is: 
$$P=(3, \sqrt{12+13}) = (3, 5)$$
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