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

1 Answer
Oct 18, 2016

Answer:

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]}