Question

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

$H\left(x\right)=\frac{x^3-1}{x^2-9}$

Short answer

The graph of the function $H\left(x\right)=\frac{x^3-1}{x^2-9}$is,

Step-by-step solution

Step 1 Given function is,

$H\left(x\right)=\frac{x^3-1}{x^2-9}$

which may also written as,

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

The domain of $H$is $\left\{x|x\ne -3,x\ne 3\right\}$

So, there will be domain in $0$.

Substitute $x=0$in the given function.

$$\begin{gathered} H\left(0\right)=\frac{0-1}{0-9} \\ H\left(0\right)=\frac{1}{9} \end{gathered}$$

So, the obtained point is,$\left(0,\frac{1}{9}\right)$.

Step 2 The function H is in lower terms.

The real zeros of the denominator are the real solutions of the equation.

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

The two lines $x=3,x=-3$are two vertical asymptotes of the graph.

Graph these asymptotes using the dashed lines.

Step 3 In the given function degree of the numerator is greater than the degree of the denominator.

Use the method of long division to find the horizontal asymptote.

Therefore,$H\left(x\right)=x+\frac{9x-1}{\left(x+3\right)\left(x-3\right)}$.

The quotient is$x$. So, the line$y=x$is an oblique asymptote of the graph. Indicate the line$y=x$on the graph using a dashed line.

Step 4 Substitute Hx=x to determine whether the graph intersects at the asymptote y=x.

$$\begin{gathered} x=\frac{x^3-1}{x^2-9} \\ x(x^2-9)=x^3-1 \\ {x}^3-9x=x^3-1 \\ -9x=-1 \\ x=\frac{1}{9} \end{gathered}$$

As $y=x$,$y=\frac{1}{9}$.

So, the point of intersection is,$\left(\frac{1}{9},\frac{1}{9}\right)$.

Step 5 Now plot a graph by using the known values.