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

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

Factor $x$ out of $x^{2} + 4 x$ .

Factor $x$ out of $x^{2}$ .

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

Factor $x$ out of $4 x$ .

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

Factor $x$ out of $x \cdot x + x \cdot 4$ .

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

Cancel the common factor of $x + 4$ and $- x - 4$ .

Factor $- 1$ out of $x$ .

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

Rewrite $4$ as $- 1 (- 4)$ .

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

Factor $- 1$ out of $- 1 (- x) - 1 (- 4)$ .

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

Cancel the common factor.

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

Divide $x \cdot (- 1)$ by $1$ .

$f (x) = x \cdot (- 1)$

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

$- x - 4$

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

Set $- x - 4$ equal to $0$ .

$- x - 4 = 0$

Solve $- x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$- x = 4$

Divide each term in $- x = 4$ by $- 1$ and simplify.

Divide each term in $- x = 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{4}{- 1}$

Divide $x$ by $1$ .

$x = \frac{4}{- 1}$

Simplify the right side.

Divide $4$ by $- 1$ .

$x = - 4$

Substitute $- 4$ for $x$ in $x \cdot (- 1)$ and simplify.

Substitute $- 4$ for $x$ to find the $y$ coordinate of the hole.

$- 4 \cdot - 1$

Multiply $- 4$ by $- 1$ .

$4$

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

$(- 4, 4)$