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

1 Answer
Feb 25, 2016

Answer:

Only one vertical asymptote is #x+5=0#. There is no horizontal or slanting asymptote.

Explanation:

To find all the asymptotes for function #y=(x+3)/(x^2+8x+15)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x^2+8x+15=0#, which when factorized becomes #(x+3)(x+5)=0# i.e. #x+3=0# and #x+5=0#.

But we have #(x+3)# in numerator too. Hence, function becomes #y=1/((x+5)# and hence only one vertical asymptote #y=x+5#.

There is no horizontal (which requires degree of numerator and denominator to be equal) and slanting asymptote (which requires degree of numerator to be just greater than that of denominator by #1#).

graph{(x+3)/(x^2+8x+15) [-10, 10, -5, 5]}