Chapter 4, Problem 31RE

To determine

To discuss: the given polynomial function.

Answer to Problem 31RE

  

Explanation of Solution

Given:

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

Calculation:

Step $1$ :

The numerator and the denominator are in the factored form.

Find the domain of the function.

The domain of $R$ is $\left\{x|x\ne 0\right\}$ .

As $0$ is in the domain, $y-\mathrm{int}ercept$ is found out by substituting $0$ for $X$

  $R\left(X\right)$ .

  $\begin{array}{l}R\left(0\right)=\frac{2\left(0^4\right)}{{\left(0-1\right)}^2}\\ =\frac{2\left(0\right)}{-1^2}\\ =\frac{0}{1}\\ =0\end{array}$

So, the $y-\mathrm{int}ercept$ is $0$ .

Plot the point $\left(0,0\right)$ .

Step $2$ :

The function $R$ is in lowest terms. The real zero of the numerator satisfies the equation $x^4=0$ . Here, the only $x-\mathrm{int}ercept$ is $0$ .

Determine the behavior of the graph near the $x-\mathrm{int}ercept$ . For this, substitute $0$ for $x$ in the denominator.

  $\begin{array}{l}R\left(x\right)=\frac{x^4}{{\left(x-1\right)}^2}\\ \approx \frac{x^4}{{\left(0-1\right)}^2}\\ \approx \frac{x^4}{1}\\ \approx x^4\end{array}$

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

Step $3$ :

  $R$ is in lowest terms. The real zeros of the denominator are the real solutions of the equation ${\left(X-1\right)}^2=0$ .

Solve the equation, $X=1$ is obtained.

The line $X=1$ is the vertical asymptote of the graph.

Graph this asymptotes using a dashes line.

Step $4$ :

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

  $x^2-2x+1\begin{array}{c}\hfill 2x^2+4x+6\\ \hfill \overline{)\begin{array}{l}2x^4\\ \frac{2x^4-4x^3+2x^2}{\begin{array}{l}4x^3-2x^2\\ \frac{4x^3-8x^2+4x}{\begin{array}{l}6x^2-4x\\ \frac{6x^2-12x+6}{8x-6}\end{array}}\end{array}}\end{array}}\end{array}$

Therefore, $R\left(x\right)=2x^2+4x+6+\frac{8x-6}{{\left(x-1\right)}^2}$

The quotient obtained is $2x^2+4x+6$ . So, the graph does not have any horizontal or oblique asymptotes.

Step $5$ :

The zero of the numerator, $0$ , and the zero of the denominator, $1$ divides the $x-axis$ into three intervals.

The three intervals are $\left(-\infty ,0\right),\left(0,1\right),\left(1,\infty \right)$ .

Construct the table.

  $0$ $1$

Interval	$\left(-\infty ,0\right)$	$\left(0,1\right)$	$\left(1,\infty \right)$
Number chosen	$-2$	$0.5$	$2$
Value of $R$	$R\left(-2\right)=\frac{32}{9}$	$R\left(0.5\right)=0.5$	$R\left(2\right)=32$
Location of graph	Above $x-axis$	Above $x-axis$	Above $x-axis$
Point on graph	$\left(-2,\frac{32}{9}\right)$	$\left(0.5,0.5\right)$	$\left(2,32\right)$

Plot the points from the table obtained.

  

Step $6$ :

Next, the behavior of the graphs is to be determined.

The graph of $R$ is above the $x-axis$ for $x<0$ and does not intersect with $y=x^2+4x+6$ . So, place arrows above $y=x^2+4x+6$ on the region $x<0$ .

The graph is above the $x-axis$ for $0<x<1$

  $x<0$ and the line $x=1$ is a vertical asymptote. So , the graph will continue at the left of $x=1$ at the top of the $x-axis$ and also it starts from the right of $x=1$ at the top.

The graph of $R$ is above the $x-axis$ for $x>1$ and does not intersect with $y=x^2+4x+6$ . So, place arrows above $y=x^2+4x+6$ on the region $x>1$ .

  

Step $7$ ;

The figure shows the final graph.

  

Conclusion:

Therefore, the given rational function is analyzed