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

Question

Theodore Amidon · Accepted Answer

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

Lincoln Anderson · Answer

undefined To find the asymptotes for $ f(x) = \frac{x + 3}{x^2 + 8x + 15} $:

1. Determine if there are any vertical asymptotes by identifying any values of $ x $ that make the denominator equal to zero.
2. Determine if there are any horizontal or oblique asymptotes.

Vertical asymptotes occur where the denominator of the function becomes zero. Solve $ x^2 + 8x + 15 = 0 $ to find these values.

Factorizing the quadratic equation gives $ (x + 5)(x + 3) = 0 $. Thus, $ x = -5 $ and $ x = -3 $.

For horizontal or oblique asymptotes, compare the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, there are no horizontal asymptotes.

Therefore, the vertical asymptotes for the function $ f(x) = \frac{x + 3}{x^2 + 8x + 15} $ are at $ x = -5 $ and $ x = -3 $.