Find the Holes in the Graph f(x)=(x+1)/(x^2-1)

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

Step 1

Factor $x^{2} - 1$ .

Rewrite $1$ as $1^{2}$ .

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

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 $b = 1$ .

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

Step 2

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

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

$x + 1$

Step 4

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

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

Substitute $- 1$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Subtract $1$ from $- 1$ .

$\frac{1}{- 2}$

Move the negative in front of the fraction.

$- \frac{1}{2}$

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

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

Step 5