How do you find the vertical, horizontal and slant asymptotes of: #f(x)= x^3 / (x^2-1)#?

Question

1221 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-21T21:12:19

The vertical asymptotes are #x=1# and #x=-1#
The slant asymptote is #y=x#

Firstly the domain of #f(x)#

is #RR-(1,-1)#
since we cannot divide by zero
#x-1!=0# and #x+1!=0#
so #x!=-1# and #x!=-1#

So #x=1# and #x=-1# are vertical asymptotes

A clant asymptote exists only if the degree of the numerator is greater than that of the numerator

We can do a long division

#f(x)=x^3/(x^2-1)=x+x/(x^2-1)#

So the slant asymptote is #y=x#