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

May 6, 2016

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

#### Explanation:

To find all the asymptotes for function $y = \frac{{x}^{3} + 1}{{x}^{2} + 3 x}$, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or ${x}^{2} + 3 x = 0$ or $x \left(x + 3\right) = 0$
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]}