## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 102

#### Answer

- Vertical asymptote: $x=-2$ - Oblique asymptote: $y=\frac{x}{2}-1$ #### Work Step by Step

$$y=\frac{x^2-1}{2x+4}$$ - As $x\to-2$, $(2x+4)$ approaches $0$ and $x^2-1$ approaches $(-2)^2-1=3\gt0$ , so $(x^2-1)/(2x+4)$ will approach $\infty$. In other words, $$\lim_{x\to-2}\frac{x^2-1}{2x+4}=\infty$$ Therefore, the line $x=-2$ is the vertical asymptote of the graph. - There is one more oblique asymptote, but to find it, we need to change the form of the function: $$y=\frac{(x^2-4)+3}{2x+4}$$ $$y=\frac{x^2-4}{2x+4}+\frac{3}{2x+4}$$ $$y=\frac{(x-2)(x+2)}{2(x+2)}+\frac{3}{2x+4}$$ $$y=\Big(\frac{x}{2}-1\Big)+\frac{3}{2x+4}$$ As $x\to\pm\infty$, $3/(2x+4)$ approaches $0$, making the function turn into $$y=\frac{x}{2}-1$$ which is the oblique asymptote of the graph. The graph, along with 2 asymptotes, is shown below. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.