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

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

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

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

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

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

Factor $2$ out of $- 2$ .

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

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

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

Rewrite $1$ as $1^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Remove unnecessary parentheses.

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

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

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

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

Factor $x$ out of $- x$ .

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

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{2 (x + 1)}{x}$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

$\frac{2 (1 + 1)}{1}$

Simplify.

Divide $2 (1 + 1)$ by $1$ .

$2 (1 + 1)$

Add $1$ and $1$ .

$2 \cdot 2$

Multiply $2$ by $2$ .

$4$

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

$(1, 4)$

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

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