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

Apr 7, 2017

see explanation.

#### Explanation:

To find any $\textcolor{b l u e}{\text{excluded values}}$ on the domain.

The denominator of R(x) cannot be zero as this would make R(x) undefined. Evaluating the denominator to zero and solving gives the values that x cannot be.

$\text{solve } {x}^{2} - 4 = 0 \Rightarrow {x}^{2} = 4 \Rightarrow x = \pm 2$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne \pm 2$

To find $\textcolor{b l u e}{\text{excluded values}}$ on the range.

divide all terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$R \left(x\right) = \frac{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2 - \frac{12}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = \frac{1 + \frac{1}{x} - \frac{12}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$

as $x \to \pm \infty , \frac{1}{x} , \frac{12}{x} ^ 2 , \frac{4}{x} ^ 2 \to 0$

${\lim}_{x \to \pm \infty} , R \left(x\right) \to \frac{1 + 0 - 0}{1 - 0}$

$= \frac{1}{1} = 1 \leftarrow \textcolor{red}{\text{ excluded value}}$

$\text{range is } y \in \mathbb{R} , y \ne 1$

$\textcolor{b l u e}{\text{intercepts}}$

• " let x = 0, in equation, for y-intercept"

• " let y = 0, in equation, for x-intercepts"

$x = 0 \to y = \frac{- 12}{- 4} = 3 \leftarrow \textcolor{red}{\text{ y-intercept}}$

The numerator is the only part of R ( x ) that can equal zero when y = 0

$\Rightarrow y = 0 \to {x}^{2} + x - 12 = 0$

$\Rightarrow \left(x + 4\right) \left(x - 3\right) = 0$

$\Rightarrow x = - 4 , x = 3 \leftarrow \textcolor{red}{\text{ x-intercepts}}$
graph{(x^2+x-12)/(x^2-4) [-10, 10, -5, 5]}