Find the Holes in the Graph f(x)=(x+5)/(x^2-25)

$f (x) = \frac{x + 5}{x^{2} - 25}$

Factor $x^{2} - 25$ .

Rewrite $25$ as $5^{2}$ .

$f (x) = \frac{x + 5}{x^{2} - 5^{2}}$

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 $b = 5$ .

$f (x) = \frac{x + 5}{(x + 5) (x - 5)}$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$f (x) = \frac{x + 5}{(x + 5) (x - 5)}$

Rewrite the expression.

$f (x) = \frac{1}{x - 5}$

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

$x + 5$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x - 5}$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $\frac{1}{x - 5}$ and simplify.

Substitute $- 5$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{- 5 - 5}$

Simplify.

Subtract $5$ from $- 5$ .

$\frac{1}{- 10}$

Move the negative in front of the fraction.

$- \frac{1}{10}$

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

$(- 5, - \frac{1}{10})$