Find the Holes in the Graph f(x)=(x^2+2x)/(-2x^2+10x)

$f (x) = \frac{x^{2} + 2 x}{- 2 x^{2} + 10 x}$

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

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

$f (x) = \frac{x \cdot x + 2 x}{- 2 x^{2} + 10 x}$

Factor $x$ out of $2 x$ .

$f (x) = \frac{x \cdot x + x \cdot 2}{- 2 x^{2} + 10 x}$

Factor $x$ out of $x \cdot x + x \cdot 2$ .

$f (x) = \frac{x (x + 2)}{- 2 x^{2} + 10 x}$

Factor $2 x$ out of $- 2 x^{2} + 10 x$ .

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

$f (x) = \frac{x (x + 2)}{2 x (- x) + 10 x}$

Factor $2 x$ out of $10 x$ .

$f (x) = \frac{x (x + 2)}{2 x (- x) + 2 x (5)}$

Factor $2 x$ out of $2 x (- x) + 2 x (5)$ .

$f (x) = \frac{x (x + 2)}{2 x (- x + 5)}$

Cancel the common factor of $x$ .

Cancel the common factor.

$f (x) = \frac{x (x + 2)}{2 x (- x + 5)}$

Rewrite the expression.

$f (x) = \frac{x + 2}{2 (- x + 5)}$

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 $\frac{x + 2}{2 (- x + 5)}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{x + 2}{2 (- x + 5)}$ and simplify.

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

$\frac{0 + 2}{2 (- 0 + 5)}$

Simplify.

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

Factor $2$ out of $0$ .

$\frac{2 \cdot 0 + 2}{2 (- 0 + 5)}$

Factor $2$ out of $2$ .

$\frac{2 \cdot 0 + 2 \cdot 1}{2 (- 0 + 5)}$

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

$\frac{2 \cdot (0 + 1)}{2 (- 0 + 5)}$

Cancel the common factors.

Cancel the common factor.

$\frac{2 \cdot (0 + 1)}{2 (- 0 + 5)}$

Rewrite the expression.

$\frac{0 + 1}{- 0 + 5}$

Add $0$ and $1$ .

$\frac{1}{- 0 + 5}$

Simplify the denominator.

Multiply $- 1$ by $0$ .

$\frac{1}{0 + 5}$

Add $0$ and $5$ .

$\frac{1}{5}$

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

$(0, \frac{1}{5})$

Find the Holes in the Graph f(x)=(x^2+2x)/(-2x^2+10x)

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