Find the Holes in the Graph f(x)=(x+7)/(x^2+x-42)

$f (x) = \frac{x + 7}{x^{2} + x - 42}$

Factor $x^{2} + x - 42$ 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 $- 42$ and whose sum is $1$ .

$- 6, 7$

Write the factored form using these integers.

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

Cancel the common factor of $x + 7$ .

Cancel the common factor.

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

Rewrite the expression.

$f (x) = \frac{1}{x - 6}$

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

$x + 7$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x - 6}$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Substitute $- 7$ for $x$ in $\frac{1}{x - 6}$ and simplify.

Substitute $- 7$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{- 7 - 6}$

Simplify.

Subtract $6$ from $- 7$ .

$\frac{1}{- 13}$

Move the negative in front of the fraction.

$- \frac{1}{13}$

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

$(- 7, - \frac{1}{13})$

Find the Holes in the Graph f(x)=(x+7)/(x^2+x-42)

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