# How do you find vertical, horizontal and oblique asymptotes for y = (4x^3 + x^2 + x + 5 )/( x^2 + 3x)?

Jun 15, 2016

Vertical asymptotes are $x = 0$ and $x = - 3$ and oblique asymptote is $y = 4 x$.

#### Explanation:

To find the asymptotes for function $\frac{4 {x}^{3} + {x}^{2} + x + 5}{{x}^{2} + 3 x}$, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or ${x}^{2} + 3 x = 0$ i.e. $x \left(x + 3\right) = 0$ and hence $x = - 3$ and $x = 0$ are two vertical asymptotes.

As the highest degree of numerator is $3$ and is just one higher than that of denominator i.e. $2$, we would have obique asymptote given by $y = 4 {x}^{3} / {x}^{2} = 4 x$ and there is no horizontal asymptote.

Hence, Vertical asymptotes are $x = 0$ and $x = - 3$ and oblique asymptote is $y = 4 x$.

The graph is as shown below.
graph{(4x^3+x^2+x+5)/(x^2+3x) [-40, 40, -20, 20]}