Find the Holes in the Graph f(x)=(x^2-3x)/(2x^2-3x-9)

$f (x) = \frac{x^{2} - 3 x}{2 x^{2} - 3 x - 9}$

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

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

$f (x) = \frac{x \cdot x - 3 x}{2 x^{2} - 3 x - 9}$

Factor $x$ out of $- 3 x$ .

$f (x) = \frac{x \cdot x + x \cdot - 3}{2 x^{2} - 3 x - 9}$

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

$f (x) = \frac{x (x - 3)}{2 x^{2} - 3 x - 9}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 9 = - 18$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

$f (x) = \frac{x (x - 3)}{2 x^{2} - 3 x - 9}$

Rewrite $- 3$ as $3$ plus $- 6$

$f (x) = \frac{x (x - 3)}{2 x^{2} + (3 - 6) x - 9}$

Apply the distributive property.

$f (x) = \frac{x (x - 3)}{2 x^{2} + 3 x - 6 x - 9}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{x (x - 3)}{(2 x^{2} + 3 x) - 6 x - 9}$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

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

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 3$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$\frac{3}{2 \cdot 3 + 3}$

Simplify.

Cancel the common factor of $3$ and $2 \cdot 3 + 3$ .

Factor $3$ out of $3$ .

$\frac{3 \cdot 1}{2 \cdot 3 + 3}$

Cancel the common factors.

Factor $3$ out of $2 \cdot 3$ .

$\frac{3 \cdot 1}{3 \cdot 2 + 3}$

Factor $3$ out of $3$ .

$\frac{3 \cdot 1}{3 \cdot 2 + 3 (1)}$

Factor $3$ out of $3 \cdot 2 + 3 (1)$ .

$\frac{3 \cdot 1}{3 \cdot (2 + 1)}$

Cancel the common factor.

$\frac{3 \cdot 1}{3 \cdot (2 + 1)}$

Rewrite the expression.

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

Add $2$ and $1$ .

$\frac{1}{3}$

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

$(3, \frac{1}{3})$

Find the Holes in the Graph f(x)=(x^2-3x)/(2x^2-3x-9)

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