How do you find vertical, horizontal and oblique asymptotes for #(x^3+1)/(x^2+3x)#?

1 Answer
May 6, 2016

Answer:

Vertical asymptotes are #x=0# and #x=-3#.
Oblique asymptote is #y=x#

Explanation:

To find all the asymptotes for function #y=(x^3+1)/(x^2+3x)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x^2+3x=0# or #x(x+3)=0#
i.e. #x=0# and #x+3=0# or #x=-3#.

As the highest degree of numerator is #3# and that of denominator is #2# (higher by just one), we have one oblique asymptote given by #y=x^3/x^2=x# or #y=x#.

Hence, Vertical asymptotes are #x=0# and #x=-3#. Oblique asymptote is given by #y=x#

graph{(x^3+1)/(x^2+3x) [-10, 10, -5, 5]}