Question

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

$G\left(x\right)=\frac{x^3+1}{x^2+2x}$

Short answer

The graph is

Step-by-step solution

Given information

Given the function$G\left(x\right)=\frac{x^3+1}{x^2+2x}$

Factor the numerator and denominator

Factoring , we get

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

Calculate the domain

Substitute , we get

$$\begin{gathered} x^2+2x=0 \\ x(x+2)=0 \\ x=0,-2 \end{gathered}$$

The domain is all real numbers except$0,-2$

Substitute numerator to zero to get x intercept

Substituting, we get

$$\begin{gathered} x^3+1=0 \\ {x}^3=-1 \\ x=-1 \end{gathered}$$

The $x$intercept is$(-1,0)$

Calculate the vertical asymptote

Setting denominator to zero, we get

$$\begin{gathered} x^2+2=0 \\ x(x+2)=0 \\ x=0,-2 \end{gathered}$$

Calculating the oblique asymptote

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

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

So$y=x-2$is the oblique asymptote

Sketching the graph

The graph is