What are the asymptotes of #f(x)=-x/((x^2-8)(x-2)) #?

1 Answer
Steve M
Oct 25, 2016

the asymptotes are;
Vertical at #x=-sqrt(8),2,sqrt(8)#
Horizontal at # 0# as #x->+-oo #

Explanation:

# f(x)=x/((x^2-8)(x-2)) #
# :. f(x)=x/((x^2-sqrt(8)^2)(x-2)) #
# :. f(x)=x/((x+sqrt(8))(x-sqrt(8))(x-2)) #
So these will be vertical asymptotes.

So we can see that the denominator is zero when #x=-sqrt(8),2,sqrt(8)#

Also We have #f(x)=x/(x^3 + "lower terms") #
So as # x->+-oo => f(x)->0 #

Hence, the asymptotes are;
Vertical at #x=-sqrt(8),2,sqrt(8)#
Horizontal at # 0# as #x->+-oo #

The graph will help to visualise the results;
graph{x/((x^2-8)(x-2)) [-10, 10, -5, 5]}

Related questions
See all questions in Limits at Infinity and Horizontal Asymptotes
Impact of this question
1492 views around the world
You can reuse this answer
Creative Commons License