Find the Holes in the Graph f(x)=(-3x^2+3x)/(5x^2-30x+25)

$f (x) = \frac{- 3 x^{2} + 3 x}{5 x^{2} - 30 x + 25}$

Factor $3 x$ out of $- 3 x^{2} + 3 x$ .

Factor $3 x$ out of $- 3 x^{2}$ .

$f (x) = \frac{3 x (- x) + 3 x}{5 x^{2} - 30 x + 25}$

Factor $3 x$ out of $3 x$ .

$f (x) = \frac{3 x (- x) + 3 x (1)}{5 x^{2} - 30 x + 25}$

Factor $3 x$ out of $3 x (- x) + 3 x (1)$ .

$f (x) = \frac{3 x (- x + 1)}{5 x^{2} - 30 x + 25}$

Factor $5 x^{2} - 30 x + 25$ .

Factor $5$ out of $5 x^{2} - 30 x + 25$ .

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

$f (x) = \frac{3 x (- x + 1)}{5 (x^{2}) - 30 x + 25}$

Factor $5$ out of $- 30 x$ .

$f (x) = \frac{3 x (- x + 1)}{5 (x^{2}) + 5 (- 6 x) + 25}$

Factor $5$ out of $25$ .

$f (x) = \frac{3 x (- x + 1)}{5 x^{2} + 5 (- 6 x) + 5 \cdot 5}$

Factor $5$ out of $5 x^{2} + 5 (- 6 x)$ .

$f (x) = \frac{3 x (- x + 1)}{5 (x^{2} - 6 x) + 5 \cdot 5}$

Factor $5$ out of $5 (x^{2} - 6 x) + 5 \cdot 5$ .

$f (x) = \frac{3 x (- x + 1)}{5 (x^{2} - 6 x + 5)}$

Factor.

Factor $x^{2} - 6 x + 5$ 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 $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$f (x) = \frac{3 x (- x + 1)}{5 ((x - 5) (x - 1))}$

Remove unnecessary parentheses.

$f (x) = \frac{3 x (- x + 1)}{5 (x - 5) (x - 1)}$

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

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

$f (x) = \frac{3 x (- (x) + 1)}{5 (x - 5) (x - 1)}$

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

$f (x) = \frac{3 x (- (x) - 1 \cdot - 1)}{5 (x - 5) (x - 1)}$

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

$f (x) = \frac{3 x (- (x - 1))}{5 (x - 5) (x - 1)}$

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

$f (x) = \frac{3 x (- 1 (x - 1))}{5 (x - 5) (x - 1)}$

Cancel the common factor.

$f (x) = \frac{3 x (- 1 (x - 1))}{5 (x - 5) (x - 1)}$

Rewrite the expression.

$f (x) = \frac{3 x \cdot (- 1)}{5 (x - 5)}$

Multiply $- 1$ by $3$ .

$f (x) = \frac{- 3 x}{5 (x - 5)}$

Move the negative in front of the fraction.

$f (x) = - \frac{(3) x}{5 (x - 5)}$

Multiply $3$ by $x$ .

$f (x) = - \frac{3 x}{5 (x - 5)}$

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 $- \frac{3 x}{5 (x - 5)}$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $- \frac{3 x}{5 (x - 5)}$ and simplify.

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

$- \frac{3 \cdot 1}{5 (1 - 5)}$

Simplify.

Multiply $3$ by $1$ .

$- \frac{3}{5 (1 - 5)}$

Subtract $5$ from $1$ .

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

Multiply $5$ by $- 4$ .

$- \frac{3}{- 20}$

Move the negative in front of the fraction.

$- - \frac{3}{20}$

Multiply $- - \frac{3}{20}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{3}{20})$

Multiply $\frac{3}{20}$ by $1$ .

$\frac{3}{20}$

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

$(1, \frac{3}{20})$

Find the Holes in the Graph f(x)=(-3x^2+3x)/(5x^2-30x+25)

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