Find the Holes in the Graph f(x)=(x^2-6x-7)/(3x+3)

$f (x) = \frac{x^{2} - 6 x - 7}{3 x + 3}$

Step 1

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

$- 7, 1$

Write the factored form using these integers.

$f (x) = \frac{(x - 7) (x + 1)}{3 x + 3}$

Step 2

Factor $3$ out of $3 x + 3$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{(x - 7) (x + 1)}{3 (x) + 3}$

Factor $3$ out of $3$ .

$f (x) = \frac{(x - 7) (x + 1)}{3 (x) + 3 (1)}$

Factor $3$ out of $3 (x) + 3 (1)$ .

$f (x) = \frac{(x - 7) (x + 1)}{3 (x + 1)}$

Step 3

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$f (x) = \frac{(x - 7) (x + 1)}{3 (x + 1)}$

Rewrite the expression.

$f (x) = \frac{x - 7}{3}$

Step 4

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

$x + 1$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 7}{3}$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ in $\frac{x - 7}{3}$ and simplify.

Substitute $- 1$ for $x$ to find the $y$ coordinate of the hole.

$\frac{- 1 - 7}{3}$

Simplify.

Subtract $7$ from $- 1$ .

$\frac{- 8}{3}$

Move the negative in front of the fraction.

$- \frac{8}{3}$

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

$(- 1, - \frac{8}{3})$

Step 6

Find the Holes in the Graph f(x)=(x^2-6x-7)/(3x+3)

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