Question

In Problems $7-44$, follow Steps 1 through $7$on page $228$to analyze the graph of each function.

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

Short answer

The graph is

Step-by-step solution

Given information

Given the function$R\left(x\right)=\frac{x^2+x-12}{x^2-4}$

Factor the numerator and denominator

Factoring, we get

$$\begin{gathered} R\left(x\right)=\frac{x^2+x-12}{x^2-4} \\ R\left(x\right)=\frac{x^2+4x-3x-12}{(x-2)(x+2)} \\ R\left(x\right)=\frac{x(x+4)-3(x+4)}{(x-2)(x+2)} \\ R\left(x\right)=\frac{(x-3)(x+4)}{(x-2)(x+2)} \end{gathered}$$

Calculate the domain

On setting the denominator to zero, we get

$x=-2,2$

Substitute numerator equal to zero to get the x intercept

Substituting, we get

$$\begin{gathered} (x-3)(x+4)=0 \\ x=3,-4 \end{gathered}$$

The $x$intercept is $(3,0),(-4,0)$

The$y$intercept is$R\left(0\right)=3$

Calculate the vertical asymptote

Substituting the denominator to zero, we get

$$\begin{gathered} (x-2)(x+2)=0 \\ x=-2,2 \end{gathered}$$is the vertical asymptote

Calculation of horizontal asymptote

Here the degree of numerator and denominator is same so the horizontal asymptote is

$$\begin{gathered} y=\frac{a}{b} \\ y=\frac{1}{1} \\ y=1 \end{gathered}$$

Sketching the graph

The graph is