Find the Holes in the Graph f(x)=(-4x^2+4x)/(5x^2-25x+20)

$f (x) = \frac{- 4 x^{2} + 4 x}{5 x^{2} - 25 x + 20}$

Factor $4 x$ out of $- 4 x^{2} + 4 x$ .

Factor $4 x$ out of $- 4 x^{2}$ .

$f (x) = \frac{4 x (- x) + 4 x}{5 x^{2} - 25 x + 20}$

Factor $4 x$ out of $4 x$ .

$f (x) = \frac{4 x (- x) + 4 x (1)}{5 x^{2} - 25 x + 20}$

Factor $4 x$ out of $4 x (- x) + 4 x (1)$ .

$f (x) = \frac{4 x (- x + 1)}{5 x^{2} - 25 x + 20}$

Factor $5 x^{2} - 25 x + 20$ .

Factor $5$ out of $5 x^{2} - 25 x + 20$ .

Factor $5$ out of $5 x^{2}$ .

$f (x) = \frac{4 x (- x + 1)}{5 (x^{2}) - 25 x + 20}$

Factor $5$ out of $- 25 x$ .

$f (x) = \frac{4 x (- x + 1)}{5 (x^{2}) + 5 (- 5 x) + 20}$

Factor $5$ out of $20$ .

$f (x) = \frac{4 x (- x + 1)}{5 x^{2} + 5 (- 5 x) + 5 \cdot 4}$

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

$f (x) = \frac{4 x (- x + 1)}{5 (x^{2} - 5 x) + 5 \cdot 4}$

Factor $5$ out of $5 (x^{2} - 5 x) + 5 \cdot 4$ .

$f (x) = \frac{4 x (- x + 1)}{5 (x^{2} - 5 x + 4)}$

Factor.

Factor $x^{2} - 5 x + 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$f (x) = \frac{4 x (- x + 1)}{5 ((x - 4) (x - 1))}$

Remove unnecessary parentheses.

$f (x) = \frac{4 x (- x + 1)}{5 (x - 4) (x - 1)}$

Cancel the common factor of $- x + 1$ and $x - 1$ .

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

$f (x) = \frac{4 x (- (x) + 1)}{5 (x - 4) (x - 1)}$

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

$f (x) = \frac{4 x (- (x) - 1 \cdot - 1)}{5 (x - 4) (x - 1)}$

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

$f (x) = \frac{4 x (- (x - 1))}{5 (x - 4) (x - 1)}$

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

$f (x) = \frac{4 x (- 1 (x - 1))}{5 (x - 4) (x - 1)}$

Cancel the common factor.

$f (x) = \frac{4 x (- 1 (x - 1))}{5 (x - 4) (x - 1)}$

Rewrite the expression.

$f (x) = \frac{4 x \cdot (- 1)}{5 (x - 4)}$

Multiply $- 1$ by $4$ .

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

Move the negative in front of the fraction.

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

Multiply $4$ by $x$ .

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

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

$x - 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{4 x}{5 (x - 4)}$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $- \frac{4 x}{5 (x - 4)}$ and simplify.

Substitute $1$ for $x$ to find the $y$ coordinate of the hole.

$- \frac{4 \cdot 1}{5 (1 - 4)}$

Simplify.

Multiply $4$ by $1$ .

$- \frac{4}{5 (1 - 4)}$

Subtract $4$ from $1$ .

$- \frac{4}{5 \cdot - 3}$

Multiply $5$ by $- 3$ .

$- \frac{4}{- 15}$

Move the negative in front of the fraction.

$- - \frac{4}{15}$

Multiply $- - \frac{4}{15}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{4}{15})$

Multiply $\frac{4}{15}$ by $1$ .

$\frac{4}{15}$

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

$(1, \frac{4}{15})$

Find the Holes in the Graph f(x)=(-4x^2+4x)/(5x^2-25x+20)

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