Find the Holes in the Graph f(x)=(x-11)/(x^2-121)

$f (x) = \frac{x - 11}{x^{2} - 121}$

Factor $x^{2} - 121$ .

Rewrite $121$ as $11^{2}$ .

$f (x) = \frac{x - 11}{x^{2} - 11^{2}}$

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 $b = 11$ .

$f (x) = \frac{x - 11}{(x + 11) (x - 11)}$

Cancel the common factor of $x - 11$ .

Cancel the common factor.

$f (x) = \frac{x - 11}{(x + 11) (x - 11)}$

Rewrite the expression.

$f (x) = \frac{1}{x + 11}$

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

$x - 11$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x + 11}$ .

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Add $11$ to both sides of the equation.

$x = 11$

Substitute $11$ for $x$ in $\frac{1}{x + 11}$ and simplify.

Substitute $11$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{11 + 11}$

Add $11$ and $11$ .

$\frac{1}{22}$

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

$(11, \frac{1}{22})$