Chapter 4, Problem 32RE

To determine

To calculate:The domain of rational function. Also, to find out the horizontal, vertical , or oblique asymptote.

Answer to Problem 32RE

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

The vertical asymptote is $x=-2$ .

The horizontal asymptote is $y=1$.

Explanation of Solution

Given information:

The rational function $R(x)=\frac{x^2+3x+2}{{\left(x+2\right)}^2}$ .

Formula used:

Vertical Asymptote: A rational function $R(x)=\frac{p(x)}{q(x)}$ , in lowest terms, will have a

asymptote $x=r$ if $r$ is a real zero of the denominator q .

Horizontal Asymptote: If a rational function is proper , the line $y=0$ is a asymptote of the graph.

Oblique Asymptote :it occur when the degree of the denominator of a rational

function is one less than the degree of the numerator. The asymptote will be along line

  $y=mx+b$ where $m\ne 0$

Calculation:

Consider ,

  $R(x)=\frac{x^2+3x+2}{{(x+2)}^2}$

The domain of R is the set of all real numbers except $x=-2$ .

Now, Ris not in the lowest terms.

Numerator:-

  $\begin{array}{l}=x^2+3x+2\\ =x^2+2x+x+2\\ =x(x+2)+1(x+2)\\ =(x+1)(x+2)\end{array}$

Therefore,

  $R(x)=\frac{(x+1)(x+2)}{{(x+2)}^2}$

  $=\frac{(x+1)}{(x+2)}$

Denominator:-

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

The line $x=-2$ are the vertical asymptote of the graph of R .

Since the degree of the numerator ,1,is equal to the degree of denominator 1.The rational function is improper.

Using long division method we get,

  $1$

  $x+2\overline{)\begin{array}{l}x+1\\ x+2\end{array}}$

  $x+2\overline{)\begin{array}{l}x+1\\ -x-2\end{array}}$

  $x+2\overline{)-1}$

As a result,

  $R(x)=\frac{x^2+3x+2}{{(x+2)}^2}$

  $=1+\frac{(-1)}{x+2}$

As $x\to -\infty$ or $x\to \infty$ ,

  $\frac{(-1)}{x+2}$

  $=\frac{-1}{x}\to 0$

and $R(x)\to 1$ .

This implies $y=1$ is a horizontal asymptote.

Hence, the domain of R is $\left\{x|x\ne -2\right\}$ .

The function has vertical and horizontal asymptote.

Vertical Asymptote at $x=-2$ and Horizontal Asymptote at $y=1$ .