# How do you graph y=(x^3+5x^2-1)/(x^2-4x) using asymptotes, intercepts, end behavior?

Oct 18, 2016

The vertical asymptotes are $x = 0$ and $x = 4$
The oblique asymptote is $y = x + 9$
As $x \to \pm \infty , y \to \pm \infty$

#### Explanation:

To determine the oblique asymptote, do a long division
(x^3+5x^2-1)/(x^2-4x)=x+9+(36x)/(x^2-4x
So the oblique asymptote is $y = x + 9$
To determine the intercepts put $y = 0$But you cannot solve the equation ${x}^{3} + 5 {x}^{2} - 1 = 0$
As $x \to \pm \infty , y \to \pm \infty$

graph{(x^3+5x^2-1)/(x^2-4x) [-20, 20, -10, 10]}