## Precalculus (10th Edition)

$\{z|z \ge- 3, z\ne 2\}$ $\:$ 
$h(z)=\frac{\sqrt{z+3}}{z-2}$ We don't take square root of negative values. This means the value inside the square root must be positive. Exclude the negative values from square root. This can be described by the mathematical expression as: $z+3\ge0$ $z\ge-3$ Secondly, for a rational function to be defined, the denominator part of it must not be equal to zero. This can be described mathematically as: $z-2\ne0$ $z\ne2$ So the domain of the given function must only include the values $z\ge-3$ and $z\ne2$ for which it is defined.