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

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

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

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

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

Raise $x$ to the power of $1$ .

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

Factor $x$ out of $x^{1}$ .

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

Factor $x$ out of $x \cdot x + x \cdot 1$ .

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

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

Factor $4$ out of $4 x$ .

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

Factor $4$ out of $4$ .

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

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

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x}{4}$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ in $\frac{x}{4}$ and simplify.

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

$\frac{- 1}{4}$

Move the negative in front of the fraction.

$- \frac{1}{4}$

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

$(- 1, - \frac{1}{4})$