Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.6 - Rational Functions and Their Graphs - Exercise Set - Page 402: 135

Answer

$f(x)=\frac{2(x^2-9)}{x^2-4}$

Work Step by Step

Step 1. With the two vertical asymptotes, we can set the denominator as $(x+2)(x-2)=x^2-4$ Step 2. With a horizontal asymptote of $y=2$, we can write $f(x)=\frac{2(x^2+a)}{x^2-4}$ Step 3. The x-intercepts happens when $y=0$; we have $x^2+a=0$ when $x=\pm3$, which gives $a=-9$ Step 4 Check $f(x)=\frac{2(x^2-9)}{x^2-4}$ for the y-intercept by letting $x=0$, which gives $y=\frac{9}{2}$ Step 5. Check for y-symmetry: $f(-x)=\frac{2((-x)^2-9)}{(-x)^2-4}=\frac{2(x^2-9)}{x^2-4}=f(x)$ Step 6. We conclude that one of the rational functions is $f(x)=\frac{2(x^2-9)}{x^2-4}$
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