Find the Holes in the Graph f(x)=(4x+16)/(x^2+x-12)

$f (x) = \frac{4 x + 16}{x^{2} + x - 12}$

Factor $4$ out of $4 x + 16$ .

Factor $4$ out of $4 x$ .

$f (x) = \frac{4 (x) + 16}{x^{2} + x - 12}$

Factor $4$ out of $16$ .

$f (x) = \frac{4 x + 4 \cdot 4}{x^{2} + x - 12}$

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

$f (x) = \frac{4 (x + 4)}{x^{2} + x - 12}$

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

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

Cancel the common factor of $x + 4$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{4}{x - 3}$ .

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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

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

Simplify.

Subtract $3$ from $- 4$ .

$\frac{4}{- 7}$

Move the negative in front of the fraction.

$- \frac{4}{7}$

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

$(- 4, - \frac{4}{7})$

Find the Holes in the Graph f(x)=(4x+16)/(x^2+x-12)

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