Question

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

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

Short answer

The graph is

Step-by-step solution

Given information

Given function$G\left(x\right)=\frac{x^2-x-12}{x+1}$

Factoring the numerator and denominator

Factoring, we get

$$\begin{gathered} G\left(x\right)=\frac{x^2-x-12}{x+1} \\ G\left(x\right)=\frac{x^2-4x+3x-12}{x+1} \\ G\left(x\right)=\frac{x(x-4)+3(x-4)}{x+1} \\ G\left(x\right)=\frac{(x+3)(x-4)}{x+1} \end{gathered}$$

Calculate the domain

Setting denominator to zero, we get

$$\begin{gathered} x+1=0 \\ x=-1 \end{gathered}$$

The domain is all real numbers except$-1$

Calculate the x intercept

Setting numerator to zero, we get

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

The$y$intercept is$G\left(0\right)=-12$i.e.$(0,-12)$

Calculate the vertical asymptote

The vertical asymptote is zeroes of denominator

$x=-1$

Calculate the horizontal asymptote

The degree of numerator is greater than numerator so there is an oblique asymptote

$G\left(x\right)=x-2+\frac{-10}{x+1}$

$y=x-2$

$$\begin{gathered} G\left(x\right)=\frac{x^2-x-12}{x+1} \\ =x-2 \\ {x}^2-x-12=(x-2)(x+1) \\ 0=10 \end{gathered}$$

This is not true so it doesn't intersect$R$

Sketching the graph

The graph is