Find the Holes in the Graph f(x)=(3x+12)/(-x^2+16)

$f (x) = \frac{3 x + 12}{- x^{2} + 16}$

Step 1

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

Factor $3$ out of $3 x$ .

$f (x) = \frac{3 (x) + 12}{- x^{2} + 16}$

Factor $3$ out of $12$ .

$f (x) = \frac{3 x + 3 \cdot 4}{- x^{2} + 16}$

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

$f (x) = \frac{3 (x + 4)}{- x^{2} + 16}$

Step 2

Factor $- x^{2} + 16$ .

Rewrite $16$ as $4^{2}$ .

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

Reorder $- x^{2}$ and $4^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = x$ .

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

Step 3

Cancel the common factor of $x + 4$ and $4 + x$ .

Reorder terms.

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$4 + x$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3}{4 - x}$ .

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

$4 + x = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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

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

Add $4$ and $4$ .

$\frac{3}{8}$

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

$(- 4, \frac{3}{8})$

Step 6