# How do you find the domain for f(q)=(q+1)/(q^2+6q-27)?

Jul 5, 2018

The domain is $q \in \left(- \infty , - 9\right) \cup \left(- 9 , 3\right) \cup \left(3 , + \infty\right)$

#### Explanation:

The denominator is

${q}^{2} + 6 q - 27 = \left(q - 3\right) \left(q + 9\right)$

As the denominator must be $\ne 0$

$\left(q - 3\right) \left(q + 9\right) \ne 0$

$\left\{\begin{matrix}q - 3 \ne 0 \\ q + 9 \ne 0\end{matrix}\right.$

$\implies$, $\left\{\begin{matrix}q \ne 3 \\ q \ne - 9\end{matrix}\right.$

The domain is $q \in \left(- \infty , - 9\right) \cup \left(- 9 , 3\right) \cup \left(3 , + \infty\right)$

graph{(x+1)/(x^2+6x-27) [-20.28, 20.26, -10.14, 10.14]}