Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.3 Polynomial and Rational Functions - 2.3 Exercises - Page 75: 35

Answer

$$ y=\frac{6-3x}{4x+12} $$ We obtain that : Asymptotes: $y=-\frac{3}{4} $ and $ x=-3$ $x$-intercept: $2$ the value when $y= 0$. $y$-intercept: $\frac{1}{2}$ the value when $x= 0$
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Work Step by Step

$$ y=\frac{6-3x}{4x+12} $$ The value $ x=-3$makes the denominator 0, but not the numerator, so the line $ x=-3$ is a vertical asymptote. To find a horizontal asymptote, let $x$ get larger and we obtain : $$ y=\lim\limits_{x \to \infty}\frac{6-3x}{4x+12}=-\frac{3}{4} $$ This means that the line $y=-\frac{3}{4} $ is a horizontal asymptote. So, we have: Asymptotes: $y=-\frac{3}{4} $ and $ x=-3$ $x$-intercept: $2$ the value when $y= 0$. $y$-intercept: $\frac{1}{2}$ the value when $x= 0$
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