Find the Holes in the Graph f(x)=(x^2-4)/(x^2-3x+2)

$f (x) = \frac{x^{2} - 4}{x^{2} - 3 x + 2}$

Factor $x^{2} - 4$ .

Rewrite $4$ as $2^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

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.

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 2$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

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

Simplify.

Add $2$ and $2$ .

$\frac{4}{2 - 1}$

Subtract $1$ from $2$ .

$\frac{4}{1}$

Divide $4$ by $1$ .

$4$

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

$(2, 4)$

Find the Holes in the Graph f(x)=(x^2-4)/(x^2-3x+2)

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