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

1 Answer
Aug 11, 2015

Answer:

# x = 4, x = -1 " and " f(x) = 0 #

Explanation:

To find the vertical asymptotes:
# lim_{x rarr a^+} f(x) = pm oo " and " lim_{x rarr a^-} f(x) = pm oo #
To find the horizontal asymptotes:
# lim_{x rarr pm oo} f(x) = b #

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

In this case the vertical asymptotes are when the denominator of #f(x)# in its simplest form is equal to zero. Hence:
# (x-4)(x+1) = 0 #
# x = 4 " or " x = -1 #

Note: # lim_{x rarr 4^+-} f(x) = +- oo " and " lim_{x rarr -1^+-} f(x) = ""_+^(-) oo #

To find the horizontal asymptotes:
# lim_(x rarr pm oo) f(x) = lim_(x rarr pm oo) (2(x+2))/(x^2-3x-4) = 0^pm #

Hence the asymptotes for #f(x)# are:
# x = 4, x = -1 " and " f(x) = 0 #

graph{(2x+4)/(x^2-3x-4) [-5, 8, -10, 10]}