# How do you find the intervals on which the function is continuous given y = (2)/((x + 4)^2) + 8?

Feb 23, 2018

The function is continuous at all points except where x=-4. The domain of the function can be given by.
$\left(- \infty , - 4\right) \cup \left(- 4 , \infty\right)$

#### Explanation:

The given function is defined only for the points where denominator i.e. ${\left(x + 4\right)}^{2}$ is not equal to zero.

the only point on the real number line where the given function is not defined is at $x = - 4$.

Hence, its interval is given by $\left(- \infty , - 4\right) \cup \left(- 4 , \infty\right)$. graph{(2)/((x+4)^(2))+8 [-10, 10, -5, 5]} graph{(2)/((x+4)^(2))+8 [-24.85, 16.47, 1.22, 21.87]}

Edit: Move the graph to view the horizontal asymptote at $y = {8}^{+}$.