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

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

Step 1

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 $10$ .

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

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

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

Step 2

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 $x \cdot x + x \cdot - 2$ .

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

Step 3

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

Factor $- 1$ out of $- x$ .

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

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

$f (x) = \frac{5 (- (x) - 1 \cdot - 2)}{x (x - 2)}$

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

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

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

$f (x) = \frac{5 (- 1 (x - 2))}{x (x - 2)}$

Cancel the common factor.

$f (x) = \frac{5 (- 1 (x - 2))}{x (x - 2)}$

Rewrite the expression.

$f (x) = \frac{5 \cdot (- 1)}{x}$

Step 4

Simplify the numerator.

Multiply $5$ by $- 1$ .

$f (x) = \frac{5 (- 1)}{x}$

Multiply $5$ by $- 1$ .

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

Step 5

Move the negative in front of the fraction.

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

Step 6

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

$x - 2$

Step 7

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 $0$ .

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

Step 8

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

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