Find the Holes in the Graph f(x)=(5x-25)/(x^2-6x+5)

$f (x) = \frac{5 x - 25}{x^{2} - 6 x + 5}$

Step 1

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

Factor $5$ out of $5 x$ .

$f (x) = \frac{5 (x) - 25}{x^{2} - 6 x + 5}$

Factor $5$ out of $- 25$ .

$f (x) = \frac{5 x + 5 \cdot - 5}{x^{2} - 6 x + 5}$

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

$f (x) = \frac{5 (x - 5)}{x^{2} - 6 x + 5}$

Step 2

Factor $x^{2} - 6 x + 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$f (x) = \frac{5 (x - 5)}{(x - 5) (x - 1)}$

Step 3

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$f (x) = \frac{5 (x - 5)}{(x - 5) (x - 1)}$

Rewrite the expression.

$f (x) = \frac{5}{x - 1}$

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 $\frac{5}{x - 1}$ .

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Substitute $5$ for $x$ in $\frac{5}{x - 1}$ and simplify.

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

$\frac{5}{5 - 1}$

Subtract $1$ from $5$ .

$\frac{5}{4}$

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

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

Step 6

Find the Holes in the Graph f(x)=(5x-25)/(x^2-6x+5)

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