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

1 Answer
Oct 22, 2016

Answer:

There is only a slant asymptote #y=x#

Explanation:

The domain of the function is #RR# as the denominator is always #>0#
So there is no vertical asymptote

The degree of the numerator is #># degree of denominator, so we expect a slant asymptote

A long division gives

#f(x)=(x^3-2x-3)/(x^2+2)=x-(4x+3)/(x^2+2)#

So the slant asymptote is #y=x#
For horizontal asymptotes, we look at the limit of #f(x)# as #x->+-oo#

limit #f(x)# as #x->+oo# is #+oo#

limit #f(x)# as #x->-oo# is #-oo#
So there is no horizontal asymptote

graph{(x^3-2*x-3)/(x^2+2) [-10, 10, -5, 5]}