Find the Holes in the Graph f(x)=(x-6)/(x^2-6x)

$f (x) = \frac{x - 6}{x^{2} - 6 x}$

Factor $x$ out of $x^{2} - 6 x$ .

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

$f (x) = \frac{x - 6}{x \cdot x - 6 x}$

Factor $x$ out of $- 6 x$ .

$f (x) = \frac{x - 6}{x \cdot x + x \cdot - 6}$

Factor $x$ out of $x \cdot x + x \cdot - 6$ .

$f (x) = \frac{x - 6}{x (x - 6)}$

Cancel the common factor of $x - 6$ .

Cancel the common factor.

$f (x) = \frac{x - 6}{x (x - 6)}$

Rewrite the expression.

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

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

$x - 6$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x}$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Substitute $6$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{6}$

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

$(6, \frac{1}{6})$