Question

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

Answer 1

The vertical asymptotes are `x=0` and `x=4`
The oblique asymptote is `y=x+9`
As `x->+-oo, y->+-oo`

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+5x^2-1=0`
As `x->+-oo, y->+-oo`

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