How do you find the asymptotes for #R(x) = (3x) / (x^2 - 9)#?

Question

Sadie Alexander · Accepted Answer

vertical asymptotes x = ± 3
horizontal asymptote y = 0

The denominator of R(x) cannot be zero.This would give division by zero which is undefined. Setting the denominator equal to zero and solving for x gives the values that x cannot be and if the numerator is non-zero for these values of x then they are the asymptotes. solve: #x^2-9=0rArr(x-3)(x+3)=0rArrx=±3# #rArrx=-3,x=3" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),R(x)toc" (a constant)"# divide terms on numerator/denominator by highest exponent of x, that is #x^2# #((3x)/x^2)/(x^2/x^2-9/x^2)=(3/x)/(1-9/x^2)# as #xto+-oo,R(x)to0/(1-0)# #rArry=0" is the asymptote"# graph{(3x)/(x^2-9) [-10, 10, -5, 5]}

Xavier Alonso · Answer

undefined To find the asymptotes of the rational function $ R(x) = \frac{3x}{x^2 - 9} $, follow these steps:

1. Identify the vertical asymptotes by finding the values of $ x $ that make the denominator zero. These values are excluded from the domain of the function because division by zero is undefined.

For $ x^2 - 9 = 0 $:
$ x^2 = 9 $
$ x = \pm 3 $

So, the vertical asymptotes are $ x = 3 $ and $ x = -3 $.

2. Determine the horizontal asymptote, if it exists, by comparing the degrees of the numerator and denominator.

The degree of the numerator is 1, and the degree of the denominator is 2. Since the degree of the denominator is greater, the horizontal asymptote is at $ y = 0 $.

Therefore, the asymptotes for the function $ R(x) = \frac{3x}{x^2 - 9} $ are:
- Vertical asymptotes: $ x = 3 $ and $ x = -3 $
- Horizontal asymptote: $ y = 0 $