Find the Holes in the Graph f(x)=(3x^2+12x)/(-2x^2+2x)

$f (x) = \frac{3 x^{2} + 12 x}{- 2 x^{2} + 2 x}$

Factor $3 x$ out of $3 x^{2} + 12 x$ .

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

$f (x) = \frac{3 x (x) + 12 x}{- 2 x^{2} + 2 x}$

Factor $3 x$ out of $12 x$ .

$f (x) = \frac{3 x (x) + 3 x (4)}{- 2 x^{2} + 2 x}$

Factor $3 x$ out of $3 x (x) + 3 x (4)$ .

$f (x) = \frac{3 x (x + 4)}{- 2 x^{2} + 2 x}$

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

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

$f (x) = \frac{3 x (x + 4)}{2 x (- x) + 2 x}$

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

$f (x) = \frac{3 x (x + 4)}{2 x (- x) + 2 x (1)}$

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

$f (x) = \frac{3 x (x + 4)}{2 x (- x + 1)}$

Cancel the common factor of $x$ .

Cancel the common factor.

$f (x) = \frac{3 x (x + 4)}{2 x (- x + 1)}$

Rewrite the expression.

$f (x) = \frac{3 (x + 4)}{2 (- x + 1)}$

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

$x$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3 (x + 4)}{2 (- x + 1)}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{3 (x + 4)}{2 (- x + 1)}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3 (0 + 4)}{2 (- 0 + 1)}$

Simplify.

Cancel the common factor of $0 + 4$ and $2$ .

Factor $2$ out of $3 (0 + 4)$ .

$\frac{2 (3 (0 + 2))}{2 (- 0 + 1)}$

Cancel the common factors.

Cancel the common factor.

$\frac{2 (3 (0 + 2))}{2 (- 0 + 1)}$

Rewrite the expression.

$\frac{3 (0 + 2)}{- 0 + 1}$

Add $0$ and $2$ .

$\frac{3 \cdot 2}{- 0 + 1}$

Simplify the denominator.

Multiply $- 1$ by $0$ .

$\frac{3 \cdot 2}{0 + 1}$

Add $0$ and $1$ .

$\frac{3 \cdot 2}{1}$

Multiply $3$ by $2$ .

$\frac{6}{1}$

Divide $6$ by $1$ .

$6$

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

$(0, 6)$

Find the Holes in the Graph f(x)=(3x^2+12x)/(-2x^2+2x)

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