Chapter 4, Problem 6CT

In problems $5$ and $6$, find the domain of each function. Find any horizontal, vertical, or oblique asymptotes.

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

To determine

The domain andany 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.