Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.6 - Rational Expressions - 3.6 Exercises - Page 309: 37

Answer

Vertical asymptotes: $x=2$ and $x=-\dfrac{7}{4}$ Horizontal asymptotes: $y=\dfrac{1}{2}$

Work Step by Step

$r(x)=\dfrac{(x+1)(2x-3)}{(x-2)(4x+7)}$ Vertical asymptotes A rational function has vertical asymptotes where the function is undefined, that is, where the denominator is zero. Set the denominator equal to 0 and solve for x to find the vertical asymptotes of this function: $(x-2)(4x+7)=0$ $x-2=0$ $x=2$ $4x+7=0$ $4x=-7$ $x=-\dfrac{7}{4}$ Horizontal asymptote Rewrite the function by evaluating the products in the numerator and in the denominator: $r(x)=\dfrac{2x^{2}-3x+2x-3}{4x^{2}+7x-8x-14}=\dfrac{2x^{2}-x-3}{4x^{2}-x-14}$ The degrees of the numerator and the denominator are the same. Divide the leading coefficient of the numerator and the leading coefficient of the denominator to obtain the horizontal asymptote: $y=\dfrac{2}{4}=\dfrac{1}{2}$
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