Find the Holes in the Graph f(x)=(5x-10)/(x^2-2x)

$f (x) = \frac{5 x - 10}{x^{2} - 2 x}$

Factor $5$ out of $5 x - 10$ .

Factor $5$ out of $5 x$ .

$f (x) = \frac{5 (x) - 10}{x^{2} - 2 x}$

Factor $5$ out of $- 10$ .

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

Factor $5$ out of $5 x + 5 \cdot - 2$ .

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

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

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

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

Factor $x$ out of $- 2 x$ .

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

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

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

$f (x) = \frac{5}{x}$

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

$x - 2$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{5}{x}$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

$\frac{5}{2}$

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

$(2, \frac{5}{2})$