Chapter 4, Problem 32RE

To determine

To discuss: the given rational function.

Answer to Problem 32RE

  

Explanation of Solution

Given:

  $R\left(x\right)=\frac{x^4}{x^2-9}$

Explanations:

Factor the denominator of $R$ .

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

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

As $0$ is in the domain, the $y-\mathrm{int}er\sec t$ can be determined by substituting $0$ for $x$ .

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

The $y-\mathrm{int}er\sec t$ is $0$ .

The function $F$ is in the lowest terms. The real zeros of the numerator is $X=0$ . The $X-\mathrm{int}er\sec t$ is $0$ .

Determine the behavior of the graph near the $x-\mathrm{int}ercept$ .

  $\begin{array}{l}F\left(x\right)=\frac{x^4}{\left(0+3\right)\left(0-3\right)}\\ \approx \frac{x^4}{-9}\\ \approx -\frac{x^4}{9}\end{array}$

Plot the point $\left(0,0\right)$ and indicate a line with slope $-\frac{1}{9}$ on the graph.

The real zeros of the denominator are the real solutions of the equation $X+3=0$ and $X+3=0$

Thus, $X=3$ and $-3$ .

The lines $X=-3$ and $X=3$ are the vertical asymptotes of the graph. The rational function is not defined at $X=-3$ and $3$ . Thus, there appears a hole in the graph at the point where $X=-3$ and $3$ .

Graph the asymptote $X=-3$ and $X=3$ using a dashed lines.

The degree of the numerator is greater than the degree of the denominator. The function $R$ is improper. Use the method of long division to find the horizontal asymptote.

  $x^2-9\begin{array}{c}\hfill x^2-9\\ \hfill \overline{)\begin{array}{l}{x}^4\\ \frac{x^4-9x^2}{\begin{array}{l}-9x^2\\ \frac{-9x^2+81}{-81}\end{array}}\end{array}}\end{array}$

The quotient is $x^2-9$ . The parabola $y=x^2-9$ is an oblique asymptote of the graph. Indicate the line $y=x^2-9$ on the graph using a dashed line.

The zero of the numerator, $0$ , and the zeros of the denominator, $-3$ and $3$ , divide the $x-axis$ into four intervals. The four intervals are $\left(-\infty ,-3\right),\left(-3,0\right),\left(-3,0\right)$ , and $\left(3,\infty \right)$

Interval	$\left(-\infty ,-3\right)$	$\left(-3,0\right)$	$\left(0,3\right)$	$\left(3,\infty \right)$
Number chosen	$-4$	$-1$	$1$	$4$
Value of $R$	$36.57$	$-0.125$	$-0.125$	$36.57$
Location of graph	Above $x-axis$	Below $x-axis$	Below $x-axis$	Above $x-axis$
Point on graph	$\left(-4,36.57\right)$	$\left(-1,-0.125\right)$	$\left(1,-0.125\right)$	$\left(4,36.57\right)$

Plot the $y-\mathrm{int}ercept$ and the points obtained from the table on a coordinate plane. Draw the vertical and horizontal asymptotes using dashed lines.

  

Since the line $x=-3$ is a horizontal asymptote and the graph Iles above the $x-axis$ for $x<-3$ ,

sketch a portion of the graph by placing a small arrow to the far left and above the $x-axis$ .

Since the line $y=x^2-9$ is an oblique asymptote and the graph lies below $x-axis$ for $x<-2$ , continue placing an arrow well below the $x-axis$ and approaching the line $y=x^2$ on the left.

Similarly, the other arrows in the graph can be accounted.

  

The figure shows the final graph.

  

Conclusion:

Therefore, the given rational function is analyzed