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

Question

Serenity Jones · Accepted 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)# #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.

Greyson Brown · Answer

undefined To find the asymptotes of $ f(x) = \frac{x+3}{x^2 + 8x + 15} $, follow these steps:

1. Factor the denominator, $ x^2 + 8x + 15 $, to determine any potential vertical asymptotes.
2. Identify any horizontal or slant asymptotes by examining the degrees of the numerator and denominator polynomials.

Applying these steps to the given function, $ f(x) = \frac{x+3}{x^2 + 8x + 15} $:

1. Factor the denominator: $ x^2 + 8x + 15 = (x + 3)(x + 5) $.
2. Determine if there are any vertical asymptotes by setting the denominator equal to zero: $ x^2 + 8x + 15 = 0 $.
   - Solving this equation yields no real solutions, indicating no vertical asymptotes.
3. Identify horizontal or slant asymptotes by comparing the degrees of the numerator and denominator.
   - Since the degree of the numerator (1) is less than the degree of the denominator (2), there is a horizontal asymptote at $ y = 0 $.

Therefore, the function $ f(x) = \frac{x+3}{x^2 + 8x + 15} $ has no vertical asymptotes and a horizontal asymptote at $ y = 0 $.