How do you find the vertical, horizontal or slant asymptotes for #(x-4)/(x^2-3x-4)#?

1 Answer
Nov 6, 2016

Answer:

The vertical asymptote is #x=-1#
The horizontal asymptote is #y=0#
There is no slant asymptote

Explanation:

We can factorise the denominator #x^2-3x-4=(x+1)(x-4)#
#:.(x-4)/(x^2-3x-4)=cancel(x-4)/((x+1)cancel(x-4))=1/(x+1)#

So the vertical asymptote is #x=-1# as we cannot divide by #0#
Limit #1/(x+1)=0^-#
#x->-oo#

Limit #1/(x+1)=0^+#
#x->+oo#

So the horizontal asymptote is #y=0#
And the intercept with the y axis is #(0,1)#
As the degree of the numerator is #<# the degree of the denominator, there is no slant asymptote
graph{1/(x+1) [-12.66, 12.65, -6.33, 6.33]}