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

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Answer 1 · 2015-07-19T14:20:15

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

with exclusion $x != -3$ .

Hence the asymptotes are $y = 0$ and $x = 3$ .

Use the difference of squares identity: $a^2-b^2 = (a-b)(a+b)$

with $a=x$ and $b=3$ to find:

$(x^2-9) = (x^2-3^2) = (x-3)(x+3)$

So:

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

with exclusion $x != -3$ .

As $x->+-oo$ we find $f(x) -> 0$ , so $y=0$ is a horizontal asymptote.

There is a vertical asymptote at $x=3$ since the denominator becomes zero and the numerator is non-zero.

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