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

1 Answer
Sep 17, 2015

Answer:

Factorise the denominator and examine the degrees of the numerator and denominator to find vertical asymptotes #x = -1#, #x = 4# and horizontal asymptote #y = 0#.

Explanation:

#f(x) = (2x+4)/(x^2-3x-4) = (2(x+2))/((x-4)(x+1))#

This will have vertical asymptotes #x = -1# and #x = 4#. For both of these values of #x# the denominator is #0# and the numerator is non-zero.

Since the degree #2# of the denominator is greater than the degree #1# of the numerator, we find:

#f(x) -> 0# as #x -> +-oo#

So there is a horizontal asymptote #y = 0#.

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