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

1 Answer
May 6, 2016

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]}