Precalculus (6th Edition) Blitzer

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

Chapter 11 - Cumulative Review Exercises - Page 1180: 7


See graph and explanations.

Work Step by Step

Step 1. Rewrite the function as $f(x)=\frac{(2x-1)(x-2)}{(x+2)(x-2)}=\frac{2x-1}{x+2}$ with a hole at $(2,\frac{3}{4})$ Step 2. We have $f(-x)=\frac{-2x-1}{-x+2}=\frac{2x+1}{x-2}$ Thus, there is no symmetry. Step 3. We can find a vertical asymptote as $x=-2$ Step 4. We can find a horizontal asymptote as $y=2$ Step 5. Find the x-intercepts as $x=\frac{1}{2}$ Step 6. Find the y-intercept as $f(0)=-\frac{1}{2}$ Step 7. Using test points and the information above, we can graph the function as shown.
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