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

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

Factor $x^{2} - 4$ .

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

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

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 $b = 2$ .

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

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

Add $2$ and $2$ .

$\frac{1}{4}$

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

$(2, \frac{1}{4})$