Find the Holes in the Graph y=(x^2-3x+2)/(x-2)

$y = \frac{x^{2} - 3 x + 2}{x - 2}$

Step 1

Factor $x^{2} - 3 x + 2$ 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 $2$ and whose sum is $- 3$ .

$- 2, - 1$

Write the factored form using these integers.

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

Step 2

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Divide $x - 1$ by $1$ .

$y = x - 1$

Step 3

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

$x - 2$

Step 4

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

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $x - 1$ and simplify.

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

$2 - 1$

Subtract $1$ from $2$ .

$1$

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

$(2, 1)$

Step 5

Find the Holes in the Graph y=(x^2-3x+2)/(x-2)

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