Chapter 4, Problem 6CT

To determine

The domain and any horizontal, vertical or oblique asymptotes of the rational function $r\left(x\right)=\frac{x^2+2x-3}{x+1}$ .

Answer to Problem 6CT

Solution:

The domain of the rational function is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$ and the function has oblique asymptote$y=x+1$ and vertical asymptote is $x=-10$ .

Explanation of Solution

Given information:

The rational function, $r\left(x\right)=\frac{x^2+2x-3}{x+1}$

The domain of a rational function is set of all real numbers excepts those for which the denominator of the function is $0$ .

For the rational function $r\left(x\right)=\frac{x^2+2x-3}{x+1}$ , the denominator is $q\left(x\right)=x+1$ .

Therefore the values of $x$ , for which the denominator of the rational function is $0$ are the zeros of the function $q\left(x\right)$ .

Set $q\left(x\right)=0$

  $\Rightarrow x+1=0$

  $\Rightarrow x=-1$

Hence the denominator of the rational function is $0$ , when the value of $x$ is $-1$ .

Therefore, the domain of the function is the set of all real numbers except $x=-1$ .

Thus, the domain of the rational function is $\left\{x\in ℝ|x\ne -1\right\}$ .

In interval notation domain is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$ .

Next, find the asymptotes of the rational function $r\left(x\right)$ .

Here the rational function $r\left(x\right)=\frac{x^2+2x-3}{x+1}$ is not in the lowest term.

Firstly, convert the rational function in the lowest term.

The numerator $x^2+2x-3$ can factor as $\left(x-1\right)\left(x+3\right)$ and the denominator $x+1$ .

  $\Rightarrow r\left(x\right)=\frac{x^2+2x-3}{x+1}$

  $r\left(x\right)=\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)}$

If $x-a$ is a factor of the denominator of $r\left(x\right)$ , in lowest terms, then $r\left(x\right)$ will have the vertical asymptote $x=a$ .

Here, $x+1$ is a factor of the denominator of $r\left(x\right)$ , then $x=-1$ is a vertical asymptote.

Here the degree of numerator $\left(x-1\right)\left(x+3\right)$ is two and the degree of denominator $x+1$ is one, that means the degree of the numerator is one more than the degree of the denominator therefore the asymptote is oblique asymptote of the form $y=mx+b$ .

The oblique asymptote for the rational function is the quotient of the polynomial division.

To find the oblique asymptote by using the polynomial division,

  $\begin{array}{l}x+1\\ x+1\overline{)x^2+2x-3}\\ \underset{\_}{-x^2+x}\\ x-3\\ \underset{\_}{-x+1}\\ -4\end{array}$

Here the quotient is $x+1$ and remainder is $-4$ .

Therefore, the oblique asymptote for the rational function is $y=x+1$ .

As the rational function is an equation of line, therefore it has no horizontal asymptote.