## Calculus with Applications (10th Edition)

Published by Pearson

# 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 $x= 0$

#### 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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