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

1 Answer
Jul 4, 2016

Answer:

vertical asymptotes x = ± 3
horizontal asymptote y = 0

Explanation:

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]}