How to find the asymptotes of #f(x) = (x+3) /( x^2 + 8x + 15)# ?

1 Answer
Feb 28, 2016

Answer:

Simplify #f(x)# to see that there is a horizontal asymptote #y=0#, vertical asymptote #x=-5# and hole (removable singularity) at #(-3, 1/2)#

Explanation:

#f(x) = (x+3)/(x^2+8x+15)#

#=(x+3)/((x+3)(x+5))#

#=1/(x+5)#

with exclusion #x != -3#

This function has horizontal asymptote #y=0# since #1/(x+5)->0# as x->+-oo#

It has a vertical asymptote #x=-5#, where the denominator is zero, but the numerator non-zero.

It has a hole (removable singularity) at #(-3, 1/2)# since the left and right limits exist at #(-3, 1/2)# and are equal.