Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Review Exercises - Page 431: 75

Answer

vertical asymptote: $x=-1$ no horizontal asymptote. slant asymptote: $y=x-1$ See graph.

Work Step by Step

Step 1. Rewrite the function as $y(x)=\frac{x^2}{x+1}=\frac{x^2-1+1}{x+1}=x-1+\frac{1}{x+1}$ Step 2. The vertical asymptote can be found as $x=-1$ Step 3. There is no horizontal asymptote. Step 4. A slant asymptote can be found as $y=x-1$ Step 5. See graph.
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