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

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

Cancel the common factor of $x + 1$ and $- x - 1$ .

Factor $- 1$ out of $x$ .

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

Rewrite $1$ as $- 1 (- 1)$ .

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

Factor $- 1$ out of $- 1 (- x) - 1 (- 1)$ .

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

Cancel the common factor.

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

Divide $- 1$ by $1$ .

$f (x) = - 1$

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 $- 1$ .

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

$- x - 1 = 0$

Solve $- x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$- x = 1$

Divide each term in $- x = 1$ by $- 1$ and simplify.

Divide each term in $- x = 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{1}{- 1}$

Simplify the right side.

Divide $1$ by $- 1$ .

$x = - 1$

Because the cancelled expression is constant with respect to $x$ , it is the $y$ coordinate for every hole.

$- 1$

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

$(- 1, - 1)$