Before starting the solution recall that in order for a function to be continuous at $x = a$ both $f\left( a \right)$ and $\mathop {\lim }\limits_{x \to a} f\left( x \right)$ must exist and we must have,

$$\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)$$

Using this idea it should be fairly clear where the function is not continuous.

First notice that at $x = - 8$ we have,

$$\mathop {\lim }\limits_{x \to - {8^ - }} f\left( x \right) = - 6 = \mathop {\lim }\limits_{x \to - {8^ + }} f\left( x \right)$$

and therefore, we also know that $\mathop {\lim }\limits_{x \to - 8} f\left( x \right) = - 6$. We can also see that $f\left( { - 8} \right) = - 3$ and so we have,

$$- 6 = \mathop {\lim }\limits_{x \to - 8} f\left( x \right) \ne f\left( { - 8} \right) = - 3$$

Because the function and limit have different values we can conclude that $f\left( x \right)$ is discontinuous at $x = - 8$.

Next let’s take a look at $x = - 2$ we have,

$$\mathop {\lim }\limits_{x \to - {2^ - }} f\left( x \right) = 3 \ne \infty = \mathop {\lim }\limits_{x \to - {2^ + }} f\left( x \right)$$

and therefore, we also know that $\mathop {\lim }\limits_{x \to \, - 2} f\left( x \right)$ doesn’t exist. We can therefore conclude that $f\left( x \right)$ is discontinuous at $x = - 2$ because the limit does not exist.

Finally let’s take a look at $x = 6$. Here we can see we have,

$$\mathop {\lim }\limits_{x \to {6^ - }} f\left( x \right) = 2 \ne 5 = \mathop {\lim }\limits_{x \to {6^ + }} f\left( x \right)$$

and therefore, we also know that $$\mathop {\lim }\limits_{x \to \,6} f\left( x \right)$$ doesn’t exist. So, once again, because the limit does not exist, we can conclude that $f\left( x \right)$ is discontinuous at $x = 6$.

All other points on this graph will have both the function and limit exist and we’ll have $\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)$ and so will be continuous.

In summary then the points of discontinuity for this graph are : $x = - 8$, $x = - 2$ and $x = 6$.