Question

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

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

Short answer

The graph is

Step-by-step solution

Given information

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

Factoring the numerator and denominator

Factoring, we get

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

Calculate the domain

Setting the denominator to zero, we get

$$\begin{gathered} x+2=0 \\ x=-2 \end{gathered}$$

The domain is all real numbers except$-2$

Calculate the x intercept 

Setting the numerator to zero, we get

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

The$y$intercept is$F\left(0\right)=-6$i.e.$(0,-6)$

Calculate the vertical asymptote

The vertical asymptote is zeroes of denominator

$x=-2$

Calculate the horizontal asymptote

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

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

The oblique asymptote is $y=x-1$

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

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

Sketching the graph

The graph is