How do you find the asymptotes for #f(x)= (x^2+4x+3)/(x^2 - 9)#?

2 Answers
Mar 7, 2016

Answer:

vertical asymptote=#-3,+3#
horizontal asymptote = #1#

Explanation:

The vertical asymptote is equal to denominator set to zero.Given that your denominator has a x squared the result will be x=#+- 3#. This is because #(-3)^2=9#and #(3)^2=9# and when you subtract both with 9, both are equal zero.

The horizontal asymptote in this case is equal to the division of the x with largest exponent in both numerator and denominator .
#x^2/x^2= 1#

Mar 7, 2016

Answer:

vertical asymptote x = 3
horizontal asymptote y = 1

Explanation:

First step is to factor the function:

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

and simplifying : #(cancel(x+3)(x+1))/(cancel(x+3)(x-3))#

left with # (x+1)/(x-3)#

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let denominator equal zero.

solve: x - 3 = 0 → x = 3 is equation of asymptote.

Horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0#

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

as x approaches ∞ , # 1/x " and " -3/x → 0 #

hence y = 1 is the asymptote

here is the graph of the function.
graph{(x^2+4x+3)/(x^2-9) [-10, 10, -5, 5]}