Find the Holes in the Graph f(x)=(x+4)/(x^2-1)

$f (x) = \frac{x + 4}{x^{2} - 1}$

Factor $x^{2} - 1$ .

Rewrite $1$ as $1^{2}$ .

$f (x) = \frac{x + 4}{x^{2} - 1^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$f (x) = \frac{x + 4}{(x + 1) (x - 1)}$

Since no factors can be removed from the denominator, there are no holes in the graph.

No holes