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

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

Step 1

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

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

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

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

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

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

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

Step 2

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

Since no factors can be removed from the denominator, there are no holes in the graph.

No holes

Step 4