Question: I need to define the function $y= |x+2|+|x-3|$ over the relevant intervals,
but I am not entirely sure what this entails. How do I find the needed intervals? Plugging in different values gives me an idea of the shape of the graph, but I don't know how to use that to actually determine the piecewise definition of $y$.

Any guidance would be appreciated.

Answer: 
![enter image description here](https://i.stack.imgur.com/Tgso2.jpg)

Firstly, apologies for the somewhat bad sketch.

If you sketch $y=|x+2|$ and $y=|x-3|$ on the same set of axes, notice the three intervals where the function $y=|x+2|+|x-3|$ will be defined differently. The dotted arrows in the sketch above should help with that. 

Can you see where to go from here?

Edit:
Your function (and what I'm fairly sure the question is asking for) may look something like this when you're done:
$$y=\begin{cases}\text{something}, & x\leq-2\\ \text{something else}, & -2<x<3\\ \text{something else again}, & x\geq3\end{cases}$$
you can use the geometric interpretation of the absolute value function $|a-b|$ as the distance between the points represented $a$ and $b$ on the number line. that is rewrite $y = |x+2| + |x-3|$ as $y=|x-(-2)| + |x-3|.$ now interpret $y$ as the sum of the distances between the point $x$ to the points $-2$ and $3.$ 

with that, it becomes clear now that if $x$ is between $-2$ and $3,$ then $y$ is the constant $5,$ the distance between the points $-2$ and $3.$ if $x > 3,$ then $y$ is twice the distance from the midpoint $\frac 12,$ that is $2(x - \frac12) = 2x - 1.$ similarly if $x < -2, y = 1- 2x.$ in fact the graph of $y$ is symmetric about $x = \frac 12.$

so finally we have $$|x+2| +|x - 3| = \begin{cases} 1 - 2x & \text{ if }x \le -2\\5 & \text{ if } -2 < x < 3\\2x-1 & \text{ if } 3 \le x. \end{cases}$$