Find the Holes in the Graph f(x)=(-3x^2+15x)/(x^2-7x+10)

$f (x) = \frac{- 3 x^{2} + 15 x}{x^{2} - 7 x + 10}$

Step 1

Factor $3 x$ out of $- 3 x^{2} + 15 x$ .

Factor $3 x$ out of $- 3 x^{2}$ .

$f (x) = \frac{3 x (- x) + 15 x}{x^{2} - 7 x + 10}$

Factor $3 x$ out of $15 x$ .

$f (x) = \frac{3 x (- x) + 3 x (5)}{x^{2} - 7 x + 10}$

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

$f (x) = \frac{3 x (- x + 5)}{x^{2} - 7 x + 10}$

Step 2

Factor $x^{2} - 7 x + 10$ 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 $10$ and whose sum is $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

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

Step 3

Cancel the common factor of $- x + 5$ and $x - 5$ .

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

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

Rewrite $5$ as $- 1 (- 5)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 5)$ .

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

Rewrite $- (x - 5)$ as $- 1 (x - 5)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Multiply $- 1$ by $3$ .

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

Step 5

Move the negative in front of the fraction.

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

Step 6

Multiply $3$ by $x$ .

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

Step 7

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

$x - 5$

Step 8

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{3 x}{x - 2}$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

Substitute $5$ for $x$ to find the $y$ coordinate of the hole.

$- \frac{3 \cdot 5}{5 - 2}$

Simplify.

Multiply $3$ by $5$ .

$- \frac{15}{5 - 2}$

Subtract $2$ from $5$ .

$- \frac{15}{3}$

Divide $15$ by $3$ .

$- 1 \cdot 5$

Multiply $- 1$ by $5$ .

$- 5$

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

$(5, - 5)$

Step 9

Find the Holes in the Graph f(x)=(-3x^2+15x)/(x^2-7x+10)

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