Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.2 One-to-One Functions; Inverse Functions - 4.2 Assess Your Understanding - Page 294: 108

Answer

domain $\{x|x\ne-\frac{3}{2},2 \}$. H.A. $y=3$, V.A. $x=-\frac{3}{2}$.

Work Step by Step

Step 1. $R(x)=\frac{6x^2-11x-2}{2x^2-x-6}=\frac{(6x+1)(x-2)}{(2x+3)(x-2)}=\frac{6x+1}{2x+3}, x\ne2$, thus we have the domain $\{x|x\ne-\frac{3}{2},2 \}$. Step 2. We can identify the horizontal asymptote H.A. $y=3$, vertical asymptote V.A. $x=-\frac{3}{2}$, oblique asymptote O.A. $none$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.