Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 44

Answer

a. Find the vertical asymptotes $ x = 0$ and $x = \frac{3}{2}$

Work Step by Step

$y = \frac{x^{2} + 1}{3x - 2x^{2}}$ Take out the common factor "$x$" in the denominator. $y = \frac{x^{2} + 1}{x(3 - 2x)}$ Then set the denominator equal to zero and solve for x. $x = 0$ $3-2x = 0$ $3 = 2x$ $x = \frac{3}{2}$ The vertical asymptotes are $x = 0$ and $x = \frac{3}{2}$.
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