Find the Holes in the Graph (2x^2+6x)/(-4x^2+8x)

$\frac{2 x^{2} + 6 x}{- 4 x^{2} + 8 x}$

Factor $2 x$ out of $2 x^{2} + 6 x$ .

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

$\frac{2 x (x) + 6 x}{- 4 x^{2} + 8 x}$

Factor $2 x$ out of $6 x$ .

$\frac{2 x (x) + 2 x (3)}{- 4 x^{2} + 8 x}$

Factor $2 x$ out of $2 x (x) + 2 x (3)$ .

$\frac{2 x (x + 3)}{- 4 x^{2} + 8 x}$

Factor $4 x$ out of $- 4 x^{2} + 8 x$ .

Factor $4 x$ out of $- 4 x^{2}$ .

$\frac{2 x (x + 3)}{4 x (- x) + 8 x}$

Factor $4 x$ out of $8 x$ .

$\frac{2 x (x + 3)}{4 x (- x) + 4 x (2)}$

Factor $4 x$ out of $4 x (- x) + 4 x (2)$ .

$\frac{2 x (x + 3)}{4 x (- x + 2)}$

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2 x (x + 3)$ .

$\frac{2 (x (x + 3))}{4 x (- x + 2)}$

Cancel the common factors.

Factor $2$ out of $4 x (- x + 2)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x$

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

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{x + 3}{2 (- x + 2)}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{0 + 3}{2 (- 0 + 2)}$

Simplify.

Add $0$ and $3$ .

$\frac{3}{2 (- 0 + 2)}$

Simplify the denominator.

Multiply $- 1$ by $0$ .

$\frac{3}{2 (0 + 2)}$

Add $0$ and $2$ .

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

Multiply $2$ by $2$ .

$\frac{3}{4}$

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

$(0, \frac{3}{4})$

Find the Holes in the Graph (2x^2+6x)/(-4x^2+8x)

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