Chapter 4, Problem 25RE

To determine

To discuss: the given rational function.

Answer to Problem 25RE

  

Explanation of Solution

Given:

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

Explanations:

Step $1$ :

The numerator and the denominator is in the factored form.

Find the domain of the function.

The domain of $H$ is $\left\{x|x\ne 0,x\ne 2\right\}$ . As $x\ne 0$ , the value $0$ does not lies in the domain.

Therefore, there is no $y-\mathrm{int}ercept$ .

Step $2$ :

The function $H$ is in lowest terms. The real zero of the numerator satisfies the equation $x+2=0$ . Here, the only $x-\mathrm{int}ercept$ is $-2$ .

Determine the behavior of the graph near the $x-\mathrm{int}ercept$ by substituting $-2$ for $x$ in the denominator of the rational function.

  $\begin{array}{l}H\left(x\right)=\frac{x+2}{-2\left(-2-2\right)}\\ \approx \frac{x+2}{-2\left(-4\right)}\\ \approx \frac{1}{8}\left(x+2\right)\end{array}$

Plot the point $\left(-2,0\right)$ and indicate a line with slope $\frac{1}{8}$ on the graph.

Step $3$ :

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

By the zero-Product Property, solve the equation and get $X=0$ and $X=2$ . The two lines $X=0$ and $X=2$ are the two vertical asymptotes of the graph.

Graph these asymptotes using a dashes line.

Step $4$ :

Degree of the numerator is less than the degree of the denominator. So, $H$ is proper.

So, the line $y=0$ is a horizontal asymptote of the graph.

Indicate this line on the graph using a dashed line.

Substitute $H\left(X\right)=0$ to determine whether the graph $H$ intersects the horizontal asymptote.

  $\frac{x+2}{x\left(x-2\right)}=0$

Multiply by $x\left(x-2\right)$ on both the sides.

  $\begin{array}{l}\frac{x+2}{x\left(x-2\right)}\left[x\left(x-2\right)\right]=0\left[x\left(x-2\right)\right]\\ x+2=0\end{array}$

Subtract $2$ from both the sides.

  $\begin{array}{l}x+2-2=0-2\\ x=-2\end{array}$

The only solution is $x=-2$ . So, the graph intersects the horizontal asymptote at $\left(-2,0\right)$ .

Step $5$ :

The zero of the numerator is $-2$ and the zeros of the denominator is $0$ and $2$ , divide $x-axis$ into your intervals

The four intervals are $\left(-\infty ,-2\right),\left(-2,0\right),\left(0,2\right),\left(2,\infty \right)$ .

Now, construct the table.

  $-2$ $0$ $2$

Interval	$\left(-\infty ,-2\right)$	$\left(-2,0\right)$	$\left(0,2\right)$	$\left(2,\infty \right)$
Number chosen	$-3$	$-1$	$1$	$3$
Value of $H$	$H\left(-3\right)=-\frac{1}{15}$	$H\left(-1\right)=\frac{1}{3}$	$H\left(1\right)=-3$	$H\left(3\right)=\frac{5}{3}$
Location of graph	Below $x-axis$	Above $x-axis$	Below $x-axis$	Above $x-axis$
Point on graph	$\left(-3,-\frac{1}{15}\right)$	$\left(-1,\frac{1}{3}\right)$	$\left(1,-3\right)$	$\left(3,\frac{5}{3}\right)$

Plot the points from the table obtained.

  

Step $6$ :

Next, the behavior of the graph near the asymptotes is to be determined.

The line $y=0$ or $x-axis$ is a horizontal asymptote and the graph lies below the $x-axis$ for $X<-2$ . Draw a portion of the graph by placing a small horizontal arrow to the far left and under the $x-axis$ . Also note the graph intersects $x-axis$ at the point $\left(-2,0\right)$ .

The vertical asymptote is the line $X=0$ and the graph lies above $x-axis$ for $-2<X<0$ . Draw a portion of the graph by placing a small vertical arrow above the $x-axis$ and approaching the line $X=0$ on the left.

The graph is below $x-axis$ for $0<X<2$ 0 X=0 is a vertical asymptote. So, thegraph will continue at the right of $X=0$ at the bottom.

Also, the graph is below the $x-axis$ when considering the vertical asymptote x = 2. So, the graphwill be at the left of $X=2$ below the $x-axis$ .

The line $X=2$ is a vertical asymptote and the graph lies above $x-axis$ for $X>2$ . Draw thegraph by placing a small vertical arrow to the right of

  $X=2$ and above $x-axis$ .

The line $y=0$ is a horizontal asymptote and the graph lies above $x-axis$ for $X>2$ . Draw thegraph by placing a small horizontal arrow above the $x-axis$ and approaching the line $y=0$ at thetop.

  

Step $7$ ;

The figure shows the final graph.

  

Conclusion:

Therefore,the given rational function is analyzed.