Find the Holes in the Graph f(x)=(3x+12)/(x^2+14x+40)

$f (x) = \frac{3 x + 12}{x^{2} + 14 x + 40}$

Factor $3$ out of $3 x + 12$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{3 (x) + 12}{x^{2} + 14 x + 40}$

Factor $3$ out of $12$ .

$f (x) = \frac{3 x + 3 \cdot 4}{x^{2} + 14 x + 40}$

Factor $3$ out of $3 x + 3 \cdot 4$ .

$f (x) = \frac{3 (x + 4)}{x^{2} + 14 x + 40}$

Factor $x^{2} + 14 x + 40$ 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 $40$ and whose sum is $14$ .

$4, 10$

Write the factored form using these integers.

$f (x) = \frac{3 (x + 4)}{(x + 4) (x + 10)}$

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$f (x) = \frac{3 (x + 4)}{(x + 4) (x + 10)}$

Rewrite the expression.

$f (x) = \frac{3}{x + 10}$

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

$x + 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3}{x + 10}$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Substitute $- 4$ for $x$ in $\frac{3}{x + 10}$ and simplify.

Substitute $- 4$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3}{- 4 + 10}$

Simplify.

Add $- 4$ and $10$ .

$\frac{3}{6}$

Cancel the common factor of $3$ and $6$ .

Factor $3$ out of $3$ .

$\frac{3 (1)}{6}$

Cancel the common factors.

Factor $3$ out of $6$ .

$\frac{3 \cdot 1}{3 \cdot 2}$

Cancel the common factor.

$\frac{3 \cdot 1}{3 \cdot 2}$

Rewrite the expression.

$\frac{1}{2}$

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

$(- 4, \frac{1}{2})$

Find the Holes in the Graph f(x)=(3x+12)/(x^2+14x+40)

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