Find the Holes in the Graph f(x)=(x^2+4x)/(2x^2-5x)

$f (x) = \frac{x^{2} + 4 x}{2 x^{2} - 5 x}$

Step 1

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

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

$f (x) = \frac{x \cdot x + 4 x}{2 x^{2} - 5 x}$

Factor $x$ out of $4 x$ .

$f (x) = \frac{x \cdot x + x \cdot 4}{2 x^{2} - 5 x}$

Factor $x$ out of $x \cdot x + x \cdot 4$ .

$f (x) = \frac{x (x + 4)}{2 x^{2} - 5 x}$

Step 2

Factor $x$ out of $2 x^{2} - 5 x$ .

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

$f (x) = \frac{x (x + 4)}{x (2 x) - 5 x}$

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

$f (x) = \frac{x (x + 4)}{x (2 x) + x \cdot - 5}$

Factor $x$ out of $x (2 x) + x \cdot - 5$ .

$f (x) = \frac{x (x + 4)}{x (2 x - 5)}$

Step 3

Cancel the common factor of $x$ .

Cancel the common factor.

$f (x) = \frac{x (x + 4)}{x (2 x - 5)}$

Rewrite the expression.

$f (x) = \frac{x + 4}{2 x - 5}$

Step 4

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

$x$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 4}{2 x - 5}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{x + 4}{2 x - 5}$ and simplify.

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

$\frac{0 + 4}{2 \cdot 0 - 5}$

Simplify.

Add $0$ and $4$ .

$\frac{4}{2 \cdot 0 - 5}$

Simplify the denominator.

Multiply $2$ by $0$ .

$\frac{4}{0 - 5}$

Subtract $5$ from $0$ .

$\frac{4}{- 5}$

Move the negative in front of the fraction.

$- \frac{4}{5}$

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

$(0, - \frac{4}{5})$

Step 6

Find the Holes in the Graph f(x)=(x^2+4x)/(2x^2-5x)

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