Find the Holes in the Graph f(x)=(x-3)/(x^2-5x+6)

$f (x) = \frac{x - 3}{x^{2} - 5 x + 6}$

Factor $x^{2} - 5 x + 6$ 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 $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$f (x) = \frac{x - 3}{(x - 3) (x - 2)}$

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$f (x) = \frac{x - 3}{(x - 3) (x - 2)}$

Rewrite the expression.

$f (x) = \frac{1}{x - 2}$

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

$x - 3$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x - 2}$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $\frac{1}{x - 2}$ and simplify.

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{3 - 2}$

Simplify.

Subtract $2$ from $3$ .

$\frac{1}{1}$

Divide $1$ by $1$ .

$1$

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

$(3, 1)$

Find the Holes in the Graph f(x)=(x-3)/(x^2-5x+6)

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