Find the Holes in the Graph f(x)=(x^2-100)/(x-10)

$f (x) = \frac{x^{2} - 100}{x - 10}$

Factor $x^{2} - 100$ .

Rewrite $100$ as $10^{2}$ .

$f (x) = \frac{x^{2} - 10^{2}}{x - 10}$

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 $b = 10$ .

$f (x) = \frac{(x + 10) (x - 10)}{x - 10}$

Cancel the common factor of $x - 10$ .

Cancel the common factor.

$f (x) = \frac{(x + 10) (x - 10)}{x - 10}$

Divide $x + 10$ by $1$ .

$f (x) = x + 10$

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

$x - 10$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x + 10$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Substitute $10$ for $x$ in $x + 10$ and simplify.

Substitute $10$ for $x$ to find the $y$ coordinate of the hole.

$10 + 10$

Add $10$ and $10$ .

$20$

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

$(10, 20)$