## Precalculus (6th Edition)

$(-\infty, 8) \cup (8, +\infty)$
The denominator is not allowed to be equal to zero since dividing zero to ny number is undefined. Since $8$ will make the denominator equal to zero, then the value of $x$ is not allowed to be $8$. The function is defined for all real numbers except $8$. Thus, the function's domain is: $(-\infty, 8) \cup (8, +\infty)$.