Find the Holes in the Graph f(x)=(x^2-36)/(x^3-36x)

$f (x) = \frac{x^{2} - 36}{x^{3} - 36 x}$

Factor $x^{2} - 36$ .

Rewrite $36$ as $6^{2}$ .

$f (x) = \frac{x^{2} - 6^{2}}{x^{3} - 36 x}$

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 $b = 6$ .

$f (x) = \frac{(x + 6) (x - 6)}{x^{3} - 36 x}$

Factor $x^{3} - 36 x$ .

Factor $x$ out of $x^{3} - 36 x$ .

Factor $x$ out of $x^{3}$ .

$f (x) = \frac{(x + 6) (x - 6)}{x \cdot x^{2} - 36 x}$

Factor $x$ out of $- 36 x$ .

$f (x) = \frac{(x + 6) (x - 6)}{x \cdot x^{2} + x \cdot - 36}$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 36$ .

$f (x) = \frac{(x + 6) (x - 6)}{x (x^{2} - 36)}$

Rewrite $36$ as $6^{2}$ .

$f (x) = \frac{(x + 6) (x - 6)}{x (x^{2} - 6^{2})}$

Factor.

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 $b = 6$ .

$f (x) = \frac{(x + 6) (x - 6)}{x ((x + 6) (x - 6))}$

Remove unnecessary parentheses.

$f (x) = \frac{(x + 6) (x - 6)}{x (x + 6) (x - 6)}$

Cancel the common factor of $x + 6$ .

Cancel the common factor.

$f (x) = \frac{(x + 6) (x - 6)}{x (x + 6) (x - 6)}$

Rewrite the expression.

$f (x) = \frac{x - 6}{x (x - 6)}$

Cancel the common factor of $x - 6$ .

Cancel the common factor.

$f (x) = \frac{x - 6}{x (x - 6)}$

Rewrite the expression.

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

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

$x + 6, x - 6$

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

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Substitute $- 6$ for $x$ in $\frac{1}{x}$ and simplify.

Substitute $- 6$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{- 6}$

Move the negative in front of the fraction.

$- \frac{1}{6}$

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Substitute $6$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{6}$

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

$(- 6, - \frac{1}{6}), (6, \frac{1}{6})$

Find the Holes in the Graph f(x)=(x^2-36)/(x^3-36x)

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