Chapter 1, Problem 24RE

Solution

(a).

Find vertical asymptotes of the graph of the function. Be sure to state your answer as equation of line.

  $y=\frac{\left|x\right|}{x+1}$

  $x=-1$

Given:

The given function, $y=\frac{\left|x\right|}{x+1}$

Concept Used:

A vertical asymptote is a vertical line that has the property that either: 1. 2. That is, as approaches from either the positive or negative side, the function approaches infinity. Vertical asymptotes occur at the values where a rational function has a denominator of 0.

Calculation:

The given function, $y=\frac{\left|x\right|}{x+1}$

  $\begin{array}{c}y=\frac{\left|x\right|}{x+1}\\ x+1=0\\ x=-1\end{array}$

The vertical asymptotes of the function $y=\frac{\left|x\right|}{x+1}$ is $x=-1$ .

(b).

Find horizontal asymptotes of the graph of the function. Be sure to state your answer as equation of line.

  $y=\frac{\left|x\right|}{x+1}$

  $y=1,-1$

Given:

The given function, $y=\frac{\left|x\right|}{x+1}$

Concept Used:

• A horizontal asymptote is a horizontal line that tells you how the function will behave at the very edges of a graph.
• If degree of numerator N < degree of denominator D, then the horizontal asymptote is y = 0.
• If degree of numerator N = degree of denominator D, then the horizontal asymptote is y = ratio of the leading coefficients.
• If degree of numerator N > degree of denominator D, then there is no horizontal asymptote.

Calculation:

The given function, $y=\frac{\left|x\right|}{x+1}$

  $\begin{array}{l}y=\frac{\left|x\right|}{x+1}\\ y=\frac{\pm x}{x+1}\\ y=\frac{x}{x+1},\frac{-x}{x+1}\end{array}$

Here degree of numerator (N) = 1

And If degree of denominator (D) = 1

If degree of numerator N =degree of denominator D, then the horizontal asymptote is y = ratio of the leading coefficients.

So N = D

Horizontal asymptote is −

  $\begin{array}{l}y=\frac{x}{x},\frac{-x}{x}\\ y=1,-1\end{array}$

Thus the horizontal asymptote of the function $y=\frac{\left|x\right|}{x+1}$ is $y=1,-1$ .