Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 576: 114


See graph and explanations.

Work Step by Step

Step 1. Given the function $f(x)=\frac{2x^2}{x^2-1}=\frac{2x^2}{(x+1)(x-1)}$, we can identify two vertical asymptotes as $x=\pm1$ Step 2. We can find a horizontal asymptote as $y=2$ Step 3. We can find the x-intercept as $x=0$ and y-intercept as $y=0$ Step 4. The function is even because $f(-x)=f(x)$ Step 5. The signs of the function when crossing the vertical asymptotes can be found as $...(+)...(-1)...(-)...(1)...(+)...$ Step 6. We can graph the function as shown in the figure.
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