## College Algebra (6th Edition)

Published by Pearson

# Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.4 - Page 387: 42

#### Answer

Horizontal asymptote: none Slant asymptote: $y = 5x$

#### Work Step by Step

There are 3 scenarios for finding horizontal asymptotes in rational functions: 1) If the degree of the numerator's function is less than the degree of the denominator's function, then the horizontal asymptote is equal to zero. 2) If the degrees of both the numerator and the denominator's functions are equal, then the horizontal asymptote is given by the ratio between the numerator and denominator's leading coefficients. 3) If the degree of the numerator's function is greater than the degree of the denominator's function, then there is no horizontal asymptote. A SLANT asymptote might exist, however, if the numerator's degree is greater than the denominator's degree by exactly 1 degree. In this case, the slant asymptote is calculated by using long division and using the resulting linear equation without the remainder as the asymptote (See pages 402-403 for an in-depth explanation of slant asymptotes). With this in mind, we see that the numerator's leading coefficient has exactly 1 degree higher than the denominator's. Therefore, there is no horizontal asymptote, but we can calculate a slant asymptote instead using long division: ......................$5x + \frac{-5x}{3x^{2} + 1}$ ...................__________________ $(3x^{2} + 1)|| 15x^{3} + 0x^{2} + 0x + 0$ ................$-(15x^{3} + 0x^{2} +5x)$ .............................................$-5x$ Therefore, the slant asymptote is given by the equation $y = 5x$

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