Find the Holes in the Graph f(x)=(x^2+2x-80)/(2x^2-16x)

$f (x) = \frac{x^{2} + 2 x - 80}{2 x^{2} - 16 x}$

Factor $x^{2} + 2 x - 80$ 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 $- 80$ and whose sum is $2$ .

$- 8, 10$

Write the factored form using these integers.

$f (x) = \frac{(x - 8) (x + 10)}{2 x^{2} - 16 x}$

Factor $2 x$ out of $2 x^{2} - 16 x$ .

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

$f (x) = \frac{(x - 8) (x + 10)}{2 x (x) - 16 x}$

Factor $2 x$ out of $- 16 x$ .

$f (x) = \frac{(x - 8) (x + 10)}{2 x (x) + 2 x (- 8)}$

Factor $2 x$ out of $2 x (x) + 2 x (- 8)$ .

$f (x) = \frac{(x - 8) (x + 10)}{2 x (x - 8)}$

Cancel the common factor of $x - 8$ .

Cancel the common factor.

$f (x) = \frac{(x - 8) (x + 10)}{2 x (x - 8)}$

Rewrite the expression.

$f (x) = \frac{x + 10}{2 x}$

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

$x - 8$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 10}{2 x}$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Substitute $8$ for $x$ in $\frac{x + 10}{2 x}$ and simplify.

Substitute $8$ for $x$ to find the $y$ coordinate of the hole.

$\frac{8 + 10}{2 \cdot 8}$

Simplify.

Cancel the common factor of $8 + 10$ and $2$ .

Factor $2$ out of $8$ .

$\frac{2 \cdot 4 + 10}{2 \cdot 8}$

Factor $2$ out of $10$ .

$\frac{2 \cdot 4 + 2 \cdot 5}{2 \cdot 8}$

Factor $2$ out of $2 \cdot 4 + 2 \cdot 5$ .

$\frac{2 \cdot (4 + 5)}{2 \cdot 8}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 8$ .

$\frac{2 \cdot (4 + 5)}{2 (8)}$

Cancel the common factor.

$\frac{2 \cdot (4 + 5)}{2 \cdot 8}$

Rewrite the expression.

$\frac{4 + 5}{8}$

Add $4$ and $5$ .

$\frac{9}{8}$

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

$(8, \frac{9}{8})$

Find the Holes in the Graph f(x)=(x^2+2x-80)/(2x^2-16x)

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