Find the Holes in the Graph f(x)=(x+1)/(x^2-16)

$f (x) = \frac{x + 1}{x^{2} - 16}$

Factor $x^{2} - 16$ .

Rewrite $16$ as $4^{2}$ .

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

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 $b = 4$ .

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

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

No holes