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

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

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

Factor $x$ out of $2 x$ .

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

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

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

Factor $x$ out of $x \cdot 2 + x (- x)$ .

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

Multiply $x$ by $2 - x$ .

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{2 + x}{2 - x}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{2 + x}{2 - x}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2 + 0}{2 - 0}$

Simplify.

Cancel the common factor of $2 + 0$ and $2 - 0$ .

Rewrite $2$ as $- 1 (- 2)$ .

$\frac{2 + 0}{- 1 (- 2) - 0}$

Factor $- 1$ out of $- 0$ .

$\frac{2 + 0}{- 1 (- 2) - (0)}$

Factor $- 1$ out of $- 1 (- 2) - (0)$ .

$\frac{2 + 0}{- 1 (- 2 + 0)}$

Factor $2$ out of $2$ .

$\frac{2 (1) + 0}{- 1 (- 2 + 0)}$

Factor $2$ out of $0$ .

$\frac{2 \cdot 1 + 2 \cdot 0}{- 1 (- 2 + 0)}$

Factor $2$ out of $2 \cdot 1 + 2 \cdot 0$ .

$\frac{2 \cdot (1 + 0)}{- 1 (- 2 + 0)}$

Cancel the common factors.

Factor $2$ out of $- 1 (- 2 + 0)$ .

$\frac{2 \cdot (1 + 0)}{2 (- 1 (- 1 + 0))}$

Cancel the common factor.

$\frac{2 \cdot (1 + 0)}{2 (- 1 (- 1 + 0))}$

Rewrite the expression.

$\frac{1 + 0}{- 1 (- 1 + 0)}$

Cancel the common factor of $1 + 0$ and $- 1 + 0$ .

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

$\frac{- 1 (- 1) + 0}{- 1 (- 1 + 0)}$

Rewrite $0$ as $- 1 (0)$ .

$\frac{- 1 \cdot - 1 - 1 \cdot 0}{- 1 (- 1 + 0)}$

Factor $- 1$ out of $- 1 \cdot - 1 - 1 \cdot 0$ .

$\frac{- 1 \cdot (- 1 + 0)}{- 1 (- 1 + 0)}$

Cancel the common factor.

$\frac{- 1 \cdot (- 1 + 0)}{- 1 (- 1 + 0)}$

Rewrite the expression.

$\frac{- 1}{- 1}$

Move the negative one from the denominator of $\frac{- 1}{- 1}$ .

$- 1 \cdot - 1$

Rewrite $- 1 \cdot - 1$ as $- - 1$ .

$- - 1$

Multiply $- 1$ by $- 1$ .

$1$

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

$(0, 1)$

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

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