# What is the range of the function  h(x) = 1 /(x^2 - 9)?

$y \setminus \ge q - \frac{1}{9}$, $y \ne 0$
$y$ is in the range if and only if the polynomial $p \left(x\right) = y {x}^{2} - y 9 - 1 = 0$ has a root, and it has a root if and only if $9 + \frac{1}{y} \ge q 0$ and $y \ne 0$, so
$y \setminus \ge q - \frac{1}{9}$, $y \ne 0$