# How do you find the domain for R(x) = (x^2 + x - 12)/(x^2 - 4)?

##### 1 Answer
Jan 13, 2016

Find that $R \left(x\right)$ is well defined for all $x$ except $\pm 2$, so the domain of $R \left(x\right)$ is $\left(- \infty , - 2\right) \cup \left(- 2 , 2\right) \cup \left(2 , \infty\right)$.

#### Explanation:

Both the numerator and denominator are well defined for any Real value of $x$. So the quotient $R \left(x\right)$ will be well defined except when the denominator is zero.

${x}^{2} - 4 = 0$ when $x = \pm \sqrt{4} = \pm 2$

So the domain of $R \left(x\right)$ is $\left(- \infty , - 2\right) \cup \left(- 2 , 2\right) \cup \left(2 , \infty\right)$

Note that ${x}^{2} + x - 12 = \left(x + 4\right) \left(x - 3\right)$, so the numerator is non-zero when $x = \pm 2$. As a result, $R \left(x\right)$ has vertical asymptotes at $x = \pm 2$ ...

graph{(x^2+x-12)/(x^2-4) [-20, 20, -10, 10]}