Question

Discuss each rational function following Steps 1–7 given on page 228.

$F\left(x\right)=\frac{x^3}{x^2-4}$

Short answer

The domain is $\left\{x|x\ne -2,2\right\}$and graph of the function is

Step-by-step solution

Step 1. Given Information 

The given function is$F\left(x\right)=\frac{x^3}{x^2-4}$

Step 2. Explanation 

Factor the numerator and denominator.

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

The domain is$\left\{x|x\ne -2,2\right\}$

Since 0 is in the domain, find the y-intercept.

$$\begin{gathered} F\left(0\right)=\frac{0^3}{(0+2)(0-2)} \\ =\frac{0}{-4} \\ =0 \end{gathered}$$

Thus, the y-intercept is 0

The real zero of the numerator satisfies the equation$x^3=0$

Hence, the only x-intercept is 0.

Step 3. Explanation  

Determine the behavior of the graph near the x-intercept by substituting 0 in the denominator.

$$\begin{gathered} F\left(x\right)=\frac{x^3}{(0+2)(0-2)} \\ =\frac{x^3}{2(-2)} \\ \approx -\frac{1}{4}{x}^3 \end{gathered}$$

Plot the point $(0,0)$and indicate a line with slope $-\frac{1}{4}$on the graph.

The real zeroes of the denominator are the real solutions of the equation $(x+2)(x-2)=0$

The two lines $x=-2,x=2$are the two vertical asymptotes of the graph.

The degree of numerator is greater than the degree of denominator. So, the function is improper. Use the long division to find the horizontal asymptote.

Thus, $F\left(x\right)=x+\frac{4x}{(x+2)(x-2)}$

As the quotient is $x$. So, the line $y=x$is an oblique asymptote of the graph.

Step 4. Calculation  

Substitute $F\left(x\right)=x$to determine whether the graph intersects the asymptote $y=x$

$$\begin{gathered} \frac{x^3}{x^2-4}=x \\ {x}^3=x^3-4x \\ 0=-4x \\ 0=x \end{gathered}$$

Thus, the point of intersection is $(0,0)$

Construct the table and determine the values in the four intervals.

Step 5. Graph  

Plot the points and lines on the graph.