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

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

Factor $x^{2} - 1$ .

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

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

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 $b = 1$ .

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Divide $x - 1$ by $1$ .

$f (x) = x - 1$

To find the holes in the graph, look at the denominator factors that were cancelled.

$x + 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x - 1$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ in $x - 1$ and simplify.

Substitute $- 1$ for $x$ to find the $y$ coordinate of the hole.

$- 1 - 1$

Subtract $1$ from $- 1$ .

$- 2$

The holes in the graph are the points where any of the cancelled factors are equal to $0$ .

$(- 1, - 2)$