Find the Holes in the Graph f(x)=(-5x+15)/(2x^2-16x+30)

$f (x) = \frac{- 5 x + 15}{2 x^{2} - 16 x + 30}$

Step 1

Factor $5$ out of $- 5 x + 15$ .

Factor $5$ out of $- 5 x$ .

$f (x) = \frac{5 (- x) + 15}{2 x^{2} - 16 x + 30}$

Factor $5$ out of $15$ .

$f (x) = \frac{5 (- x) + 5 (3)}{2 x^{2} - 16 x + 30}$

Factor $5$ out of $5 (- x) + 5 (3)$ .

$f (x) = \frac{5 (- x + 3)}{2 x^{2} - 16 x + 30}$

Step 2

Factor $2 x^{2} - 16 x + 30$ .

Factor $2$ out of $2 x^{2} - 16 x + 30$ .

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

$f (x) = \frac{5 (- x + 3)}{2 (x^{2}) - 16 x + 30}$

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

$f (x) = \frac{5 (- x + 3)}{2 (x^{2}) + 2 (- 8 x) + 30}$

Factor $2$ out of $30$ .

$f (x) = \frac{5 (- x + 3)}{2 x^{2} + 2 (- 8 x) + 2 \cdot 15}$

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

$f (x) = \frac{5 (- x + 3)}{2 (x^{2} - 8 x) + 2 \cdot 15}$

Factor $2$ out of $2 (x^{2} - 8 x) + 2 \cdot 15$ .

$f (x) = \frac{5 (- x + 3)}{2 (x^{2} - 8 x + 15)}$

Factor.

Factor $x^{2} - 8 x + 15$ 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 $15$ and whose sum is $- 8$ .

$- 5, - 3$

Write the factored form using these integers.

$f (x) = \frac{5 (- x + 3)}{2 ((x - 5) (x - 3))}$

Remove unnecessary parentheses.

$f (x) = \frac{5 (- x + 3)}{2 (x - 5) (x - 3)}$

Step 3

Cancel the common factor of $- x + 3$ and $x - 3$ .

Factor $- 1$ out of $- x$ .

$f (x) = \frac{5 (- (x) + 3)}{2 (x - 5) (x - 3)}$

Rewrite $3$ as $- 1 (- 3)$ .

$f (x) = \frac{5 (- (x) - 1 \cdot - 3)}{2 (x - 5) (x - 3)}$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

$f (x) = \frac{5 (- (x - 3))}{2 (x - 5) (x - 3)}$

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

$f (x) = \frac{5 (- 1 (x - 3))}{2 (x - 5) (x - 3)}$

Cancel the common factor.

$f (x) = \frac{5 (- 1 (x - 3))}{2 (x - 5) (x - 3)}$

Rewrite the expression.

$f (x) = \frac{5 \cdot (- 1)}{2 (x - 5)}$

Step 4

Simplify the numerator.

Multiply $5$ by $- 1$ .

$f (x) = \frac{5 (- 1)}{2 (x - 5)}$

Multiply $5$ by $- 1$ .

$f (x) = \frac{- 5}{2 (x - 5)}$

Step 5

Move the negative in front of the fraction.

$f (x) = - \frac{5}{2 (x - 5)}$

Step 6

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

$x - 3$

Step 7

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{5}{2 (x - 5)}$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $- \frac{5}{2 (x - 5)}$ and simplify.

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

$- \frac{5}{2 (3 - 5)}$

Simplify.

Subtract $5$ from $3$ .

$- \frac{5}{2 \cdot - 2}$

Multiply $2$ by $- 2$ .

$- \frac{5}{- 4}$

Move the negative in front of the fraction.

$- - \frac{5}{4}$

Multiply $- - \frac{5}{4}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{5}{4})$

Multiply $\frac{5}{4}$ by $1$ .

$\frac{5}{4}$

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

$(3, \frac{5}{4})$

Step 8

Find the Holes in the Graph f(x)=(-5x+15)/(2x^2-16x+30)

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