Find the Holes in the Graph f(x)=(x-6)/(x^2-7x+6)

$f (x) = \frac{x - 6}{x^{2} - 7 x + 6}$

Step 1

Factor $x^{2} - 7 x + 6$ 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 $6$ and whose sum is $- 7$ .

$- 6, - 1$

Write the factored form using these integers.

$f (x) = \frac{x - 6}{(x - 6) (x - 1)}$

Step 2

Cancel the common factor of $x - 6$ .

Cancel the common factor.

$f (x) = \frac{x - 6}{(x - 6) (x - 1)}$

Rewrite the expression.

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

Step 3

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

$x - 6$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x - 1}$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Substitute $6$ for $x$ in $\frac{1}{x - 1}$ and simplify.

Substitute $6$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{6 - 1}$

Subtract $1$ from $6$ .

$\frac{1}{5}$

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

$(6, \frac{1}{5})$

Step 5

Find the Holes in the Graph f(x)=(x-6)/(x^2-7x+6)

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