Question

using interval notation, state the domain of the rational function. $y = \frac{x - 3}{x^{2}-x - 12}$
$(-\infty, 4)\cup(4, \infty)$
$(-\infty, 3)\cup(3, \infty)$
$(-\infty, - 3)\cup(-3, 4)\cup(4, \infty)$
$(-\infty, - 4)\cup(-4, 3)\cup(3, \infty)$

Explanation:

Step1: Find values that make denominator 0

Set $x^{2}-x - 12=0$.
Factor the quadratic: $(x - 4)(x+3)=0$.

Step2: Solve for x

Using the zero - product property, $x - 4=0$ gives $x = 4$ and $x+3=0$ gives $x=-3$.

Step3: Write domain in interval notation

The domain is all real numbers except $x=-3$ and $x = 4$, so it is $(-\infty,-3)\cup(-3,4)\cup(4,\infty)$.

Answer:

$(-\infty,-3)\cup(-3,4)\cup(4,\infty)$