Precalculus Examples

$f (x) = \frac{5 x - 25}{x^{2} - 5 x}$

Step 1

Factor $5$ out of $5 x - 25$ .

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Factor $5$ out of $5 x$ .

$f (x) = \frac{5 (x) - 25}{x^{2} - 5 x}$

Factor $5$ out of $- 25$ .

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

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

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

Step 2

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

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

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

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

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

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

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

Step 3

Cancel the common factor of $x - 5$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x - 5$

Step 5

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

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Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Substitute $5$ for $x$ in $\frac{5}{x}$ and simplify.

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Substitute $5$ for $x$ to find the $y$ coordinate of the hole.

$\frac{5}{5}$

Divide $5$ by $5$ .

$1$

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

$(5, 1)$

Step 6

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