How do you find the asymptotes for #f(x) = (x+3 )/(x^2 + 8x + 15)#?

1 Answer
George C.
Nov 13, 2015

Factor the denominator and simplify, finding that #f(x)# has a horizontal asymptote #y=0# and vertical asymptote #x=-5#

Explanation:

#f(x) = (x+3)/(x+8x+15) = (x+3)/((x+3)(x+5))#

#= 1/(x+5)# with exclusion #x != 3#

As #x->+-oo#, #1/(x+5)->0#, so #f(x)# has a horizontal asymptote #y=0#.

When #x = -5#, the denominator is zero and the numerator is non-zero, so #f(x)# has a vertical asymptote #x=-5#

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