Find the Holes in the Graph f(x)=(x^2+7x)/(x^3-5x^2-14x)

$f (x) = \frac{x^{2} + 7 x}{x^{3} - 5 x^{2} - 14 x}$

Factor $x$ out of $x^{2} + 7 x$ .

Factor $x$ out of $x^{2}$ .

$f (x) = \frac{x \cdot x + 7 x}{x^{3} - 5 x^{2} - 14 x}$

Factor $x$ out of $7 x$ .

$f (x) = \frac{x \cdot x + x \cdot 7}{x^{3} - 5 x^{2} - 14 x}$

Factor $x$ out of $x \cdot x + x \cdot 7$ .

$f (x) = \frac{x (x + 7)}{x^{3} - 5 x^{2} - 14 x}$

Factor $x^{3} - 5 x^{2} - 14 x$ .

Factor $x$ out of $x^{3} - 5 x^{2} - 14 x$ .

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

$f (x) = \frac{x (x + 7)}{x \cdot x^{2} - 5 x^{2} - 14 x}$

Factor $x$ out of $- 5 x^{2}$ .

$f (x) = \frac{x (x + 7)}{x \cdot x^{2} + x (- 5 x) - 14 x}$

Factor $x$ out of $- 14 x$ .

$f (x) = \frac{x (x + 7)}{x \cdot x^{2} + x (- 5 x) + x \cdot - 14}$

Factor $x$ out of $x \cdot x^{2} + x (- 5 x)$ .

$f (x) = \frac{x (x + 7)}{x (x^{2} - 5 x) + x \cdot - 14}$

Factor $x$ out of $x (x^{2} - 5 x) + x \cdot - 14$ .

$f (x) = \frac{x (x + 7)}{x (x^{2} - 5 x - 14)}$

Factor.

Factor $x^{2} - 5 x - 14$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 14$ and whose sum is $- 5$ .

$- 7, 2$

Write the factored form using these integers.

$f (x) = \frac{x (x + 7)}{x ((x - 7) (x + 2))}$

Remove unnecessary parentheses.

$f (x) = \frac{x (x + 7)}{x (x - 7) (x + 2)}$

Cancel the common factor of $x$ .

Cancel the common factor.

$f (x) = \frac{x (x + 7)}{x (x - 7) (x + 2)}$

Rewrite the expression.

$f (x) = \frac{x + 7}{(x - 7) (x + 2)}$

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

$x$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 7}{(x - 7) (x + 2)}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{x + 7}{(x - 7) (x + 2)}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{0 + 7}{(0 - 7) (0 + 2)}$

Simplify.

Cancel the common factor of $0 + 7$ and $0 - 7$ .

Factor $7$ out of $0$ .

$\frac{7 (0) + 7}{(0 - 7) (0 + 2)}$

Factor $7$ out of $7$ .

$\frac{7 \cdot 0 + 7 \cdot 1}{(0 - 7) (0 + 2)}$

Factor $7$ out of $7 \cdot 0 + 7 \cdot 1$ .

$\frac{7 \cdot (0 + 1)}{(0 - 7) (0 + 2)}$

Cancel the common factors.

Factor $7$ out of $(0 - 7) (0 + 2)$ .

$\frac{7 \cdot (0 + 1)}{7 ((0 - 1) (0 + 2))}$

Cancel the common factor.

$\frac{7 \cdot (0 + 1)}{7 ((0 - 1) (0 + 2))}$

Rewrite the expression.

$\frac{0 + 1}{(0 - 1) (0 + 2)}$

Cancel the common factor of $0 + 1$ and $0 - 1$ .

Rewrite $0$ as $- 1 (0)$ .

$\frac{- 1 (0) + 1}{(0 - 1) (0 + 2)}$

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{- 1 \cdot 0 - 1 \cdot - 1}{(0 - 1) (0 + 2)}$

Factor $- 1$ out of $- 1 \cdot 0 - 1 \cdot - 1$ .

$\frac{- 1 \cdot (0 - 1)}{(0 - 1) (0 + 2)}$

Cancel the common factor.

$\frac{- 1 \cdot (0 - 1)}{(0 - 1) (0 + 2)}$

Rewrite the expression.

$\frac{- 1}{0 + 2}$

Add $0$ and $2$ .

$\frac{- 1}{2}$

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 $0$ .

$(0, - \frac{1}{2})$

Find the Holes in the Graph f(x)=(x^2+7x)/(x^3-5x^2-14x)

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