Find the Holes in the Graph f(x)=(x^2-5x)/(2x^2+7x)

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

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

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

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

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

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

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

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

Factor $x$ out of $2 x^{2} + 7 x$ .

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

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

Factor $x$ out of $7 x$ .

$f (x) = \frac{x (x - 5)}{x (2 x) + x \cdot 7}$

Factor $x$ out of $x (2 x) + x \cdot 7$ .

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x - 5}{2 x + 7}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{x - 5}{2 x + 7}$ and simplify.

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

$\frac{0 - 5}{2 \cdot 0 + 7}$

Simplify.

Subtract $5$ from $0$ .

$\frac{- 5}{2 \cdot 0 + 7}$

Simplify the denominator.

Multiply $2$ by $0$ .

$\frac{- 5}{0 + 7}$

Add $0$ and $7$ .

$\frac{- 5}{7}$

Move the negative in front of the fraction.

$- \frac{5}{7}$

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

$(0, - \frac{5}{7})$