How do you find the domain of (3x^2+3x-6)/(x^2-x-12)?

Jan 7, 2018

$x \in \mathbb{R} , x \ne - 3 , 4$

Explanation:

The denominator of the rational expression cannot be zero as this would make it undefined. Equating the denominator to zero and solving gives the values that x cannot be.

$\text{solve } {x}^{2} - x - 12 = 0 \Rightarrow \left(x - 4\right) \left(x + 3\right) = 0$

$\Rightarrow x = - 3 \text{ or "x=4larrcolor(red)"excluded values}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne - 3 , 4$

$\text{in interval notation as}$

$\left(- \infty , - 3\right) \cup \left(- 3 , 4\right) \cup \left(4 , + \infty\right)$
graph{(3x^2+3x-6)/(x^2-x-12) [-10, 10, -5, 5]}