What are the asymptotes for #(x^2+4x+3)/(x^2 - 9)#?

2 Answers
Aug 24, 2015

Answer:

The vertical asymptotes is 3.
The horizontal asymptote is 1.

Explanation:

Vertical asymptotes occur at points where the denominator is zero. By factorizing the denominator as a difference of 2 squares, we get (x+3)(x-3), however, factorizing the numerator as a trinomial yields (x+1)(x+3) and so the (x+3) factors cancel, meaning that x=3 is the only vertical asymptote. x=-3 is a point of discontinuity since the function f is undefined at x=-3. One may se the sequential criterion to prove that f(x) is not continuous at x=-3.
For horizontal asymptotes we have to investigate the limits of the function at + and - infinity. In both cases, since the quadratic terms dominate the numerator and denominator, this has value 1 at infinity and so the horizontal asymptote occurs at y=1.

Aug 27, 2015

Answer:

#f(x) = (x^2+4x+3)/(x^2-9)#

has horizontal asymptote #y = 1# and vertical asymptote #x = 3#.

It also has a removable singularity at #(-3, 1/3)#

Explanation:

#f(x) = (x^2+4x+3)/(x^2-9) = ((x+1)(x+3))/((x-3)(x+3)) = (x+1)/(x-3)#

with exclusion #x != -3#

Then

#(x+1)/(x-3) = (x-3+4)/(x-3) = 1+4/(x-3)#

As #x -> +-oo# we have #4/(x-3) -> 0#, so #1+4/(x-3) -> 1#.

That is, #f(x)# has a horizontal asymptote #y = 1#

#f(x)# has a vertical asymptote at #x = 3# where the denominator is #0# (and the numerator is non-zero).

#f(x)# has a removable singularity (not an asymptote) where #x = -3#, since

#lim_(x->-3) f(x) = lim_(x->-3) (x+1)/(x-3) = 1/3#

but #f(-3)# is undefined as both the numerator and denominator are #0#.

Here's a graph of #f(x)# and #y = 1#.

graph{(y - (x^2+4x+3)/(x^2-9))(y - 1) = 0 [-20, 20, -10, 10]}

(The graphing tool does not seem to like graphing the vertical asymptote at the same time as #f(x)#.)