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

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

Step 1

Factor $x^{2} - x - 6$ 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 $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Step 2

Factor $x^{2} + x - 2$ 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 $1$ .

$- 1, 2$

Write the factored form using these integers.

$f (x) = \frac{(x - 3) (x + 2)}{(x - 1) (x + 2)}$

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

$f (x) = \frac{(x - 3) (x + 2)}{(x - 1) (x + 2)}$

Rewrite the expression.

$f (x) = \frac{x - 3}{x - 1}$

Step 4

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

$x + 2$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 3}{x - 1}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

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

Simplify.

Subtract $3$ from $- 2$ .

$\frac{- 5}{- 2 - 1}$

Subtract $1$ from $- 2$ .

$\frac{- 5}{- 3}$

Dividing two negative values results in a positive value.

$\frac{5}{3}$

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

$(- 2, \frac{5}{3})$

Step 6

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

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