Question

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

Answer 1

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]}