# What are the vertical and horizontal asymptotes of y = ((x-3)(x+3))/(x^2-9)?

Oct 21, 2015

The function is a constant line, so its only asymptote are horizontal, and they are the line itself, i.e. $y = 1$.

#### Explanation:

Unless you misspelled something, this was a tricky exercise: expanding the numerator, you get $\left(x - 3\right) \left(x + 3\right) = {x}^{2} - 9$, and so the function is identically equal to $1$.

This means that your function is this horizontal line:

graph{((x-3)(x+3))/(x^2-9) [-20.56, 19.99, -11.12, 9.15]}

As every line, it is defined for every real number $x$, and so it has no vertical asymptotes. And in a sense, the line is its own vertical asymptote, since

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right) = {\lim}_{x \setminus \to \setminus \pm \setminus \infty} 1 = 1$.