Find the Holes in the Graph f(x)=(-2x^2+10x)/(4x-20)

$f (x) = \frac{- 2 x^{2} + 10 x}{4 x - 20}$

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

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

$f (x) = \frac{2 x (- x) + 10 x}{4 x - 20}$

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

$f (x) = \frac{2 x (- x) + 2 x (5)}{4 x - 20}$

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

$f (x) = \frac{2 x (- x + 5)}{4 x - 20}$

Factor $4$ out of $4 x - 20$ .

Factor $4$ out of $4 x$ .

$f (x) = \frac{2 x (- x + 5)}{4 (x) - 20}$

Factor $4$ out of $- 20$ .

$f (x) = \frac{2 x (- x + 5)}{4 x + 4 \cdot - 5}$

Factor $4$ out of $4 x + 4 \cdot - 5$ .

$f (x) = \frac{2 x (- x + 5)}{4 (x - 5)}$

Cancel the common factor of $2$ and $4$ .

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

$f (x) = \frac{2 (x (- x + 5))}{4 (x - 5)}$

Cancel the common factors.

Factor $2$ out of $4 (x - 5)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $- x + 5$ and $x - 5$ .

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

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

Rewrite $5$ as $- 1 (- 5)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 5)$ .

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

Rewrite $- (x - 5)$ as $- 1 (x - 5)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move $- 1$ to the left of $x$ .

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

Move the negative in front of the fraction.

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

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

$x - 5$

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

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Substitute $5$ 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 $0$ .

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

Find the Holes in the Graph f(x)=(-2x^2+10x)/(4x-20)

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