Question

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

Answer 1

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`

Answer 2

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]}