When given the graph of a function, the domain would include all the points that there is a graph. The strategy is to find what <em>is not</em> included.
What we are looking for are points of discontinuity. Think of it as when you remove your pencil from the paper.
From left to right, the graph stops at x = -3. So anything less than -3 is in the domain. Next, the graph starts up again at x =-1 after an asymptote (the vertical dashed lines). This piece goes to x = 4. So our domain is from -1 to 4.
Lastly, there's a jump from 4 to 5 and the graph goes on again. After 5, we take all the stuff more than it. So x > 5 is in the domain.
So x < -3, - 1 < x < 4, and x > 5 appears to be our domain. However, end points needed to be checked to see if we include them or not. Again we go left to right.
At x = -3 there is a filled (or closed) circle and that means we include -3.
At x = -1 there is an asymptote. Asymptotes are things you get close to but don't get to. (Think of it as the "I'm Not Touching" game you play on car trips.) So we exclude -1.
At x = 4 there is an unfilled (or open) circle and that means we exclude 4.
At x = 5 there is a filled circle so we include 5.
Now we refine our domain for the endpoints.
x ≤ -3, -1 < x < 4, x ≥ 5 is our domain.
The problem gives us intervals, and we gave it in inequalities. When we include an endpoint we use brackets - [ and } and when we exclude and endpoint we use parentheses - ( and ). Let's go back to x ≤ -3. Anything less works, and -3 is included (closed circle). That interval is (-∞, -3]. Next is the piece between -1 and 4. Since both are excluded, (-1,4) is our interval. We include 5 to write x ≥5 as the interval [5,∞).
Put the bolded ones all together and use the union, ∪, symbol to connect them, since something on the graph could be in any piece.
Our domain is (-∞, -3] ∪(-1,4) ∪ [5,∞).
where s is the side.