How do you identify all asymptotes or holes and intercepts for #f(x)=(x^2-2x)/(x^3+1)#?

1 Answer
Oct 24, 2016

Answer:

The vertical asymptote is #x=-1#
The intercepts with the x axis are #(0,0)# and #(2,0)#

Explanation:

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

As we cannot divide by #0#, the denominator #!=0#
#x^3+1!=0# #=># #x^3!=-1# #=># #x!=-1#
So x=-1 is a vertical asymptote

To see the intercept with the x axis,
#f(x)=0#
#(x^2-2x)/(x^3+1)=0#
#x^2-2x=0#
#x(x-2)=0#
so #x=0# and #x=2#
The points are #(0,0)# and #(2,0)#

We can find the limits at #+-oo#
limit f(x) = #x^2/x^3=1/x=0#
#xrarr-oo#

limit f(x) = #x^2/x^3=1/x=0#
#xrarr+oo#