Find the Holes in the Graph (2x-4)/(x^2-4)

$\frac{2 x - 4}{x^{2} - 4}$

Factor $2$ out of $2 x - 4$ .

Factor $2$ out of $2 x$ .

$\frac{2 (x) - 4}{x^{2} - 4}$

Factor $2$ out of $- 4$ .

$\frac{2 x + 2 \cdot - 2}{x^{2} - 4}$

Factor $2$ out of $2 x + 2 \cdot - 2$ .

$\frac{2 (x - 2)}{x^{2} - 4}$

Factor $x^{2} - 4$ .

Rewrite $4$ as $2^{2}$ .

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

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 $b = 2$ .

$\frac{2 (x - 2)}{(x + 2) (x - 2)}$

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{2 (x - 2)}{(x + 2) (x - 2)}$

Rewrite the expression.

$\frac{2}{x + 2}$

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

$x - 2$

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

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2}{2 + 2}$

Simplify.

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

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 \cdot 1 + 2 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{1 + 1}$

Add $1$ and $1$ .

$\frac{1}{2}$

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

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

Find the Holes in the Graph (2x-4)/(x^2-4)

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