Find the Holes in the Graph (2x+4)/(-x^2+4)

$\frac{2 x + 4}{- x^{2} + 4}$

Step 1

Factor $2$ out of $2 x + 4$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $4$ .

$\frac{2 x + 2 \cdot 2}{- x^{2} + 4}$

Factor $2$ out of $2 x + 2 \cdot 2$ .

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

Step 2

Factor $- x^{2} + 4$ .

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

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

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

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

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

$\frac{2 (x + 2)}{(2 + x) (2 - x)}$

Step 3

Cancel the common factor of $x + 2$ and $2 + x$ .

Reorder terms.

$\frac{2 (x + 2)}{(x + 2) (2 - x)}$

Cancel the common factor.

$\frac{2 (x + 2)}{(x + 2) (2 - x)}$

Rewrite the expression.

$\frac{2}{2 - x}$

Step 4

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

$2 + x$

Step 5

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

Set $2 + x$ equal to $0$ .

$2 + x = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Substitute $- 2$ for $x$ in $\frac{2}{2 - x}$ and simplify.

Substitute $- 2$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2}{2 + 2}$

Simplify.

Cancel the common factor of $2$ and $2 + 2$ .

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 \cdot 1 + 2 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{1 + 1}$

Add $1$ and $1$ .

$\frac{1}{2}$

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

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

Step 6

Find the Holes in the Graph (2x+4)/(-x^2+4)

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