How do you find the asymptotes for #(x-4)/(x^2-3x-4)#?

1 Answer
Jan 16, 2016

Answer:

You start by factoring the denominator, which can be written as #(x+1)(x-4)#

Explanation:

Now the vertical discontinuities are clear, because that's when the denominator becomes zero:
#x=-1andx=+4#

If we near #x=4# from both sides, it will be seen that:
#lim_(x->4+) y = lim_(x->4-) y=1/5#
So this is a 'repairable' discontinuity, while #x=-1# is not.

We can now do the following:
#(cancel(x-4))/((x+1)(cancel(x-4))) =1/(x+1)#

The horizontal asymptote is when #x# becomes very large. In this case the whole thing goes to zero.

Answer:
#x=-1#
#y=0#
graph{(x-4)/(x^2-3x-4) [-10, 10, -5, 5]}