Find the Holes in the Graph f(x)=(x^2+2x-35)/(3x-15)

$f (x) = \frac{x^{2} + 2 x - 35}{3 x - 15}$

Step 1

Factor $x^{2} + 2 x - 35$ 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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Write the factored form using these integers.

$f (x) = \frac{(x - 5) (x + 7)}{3 x - 15}$

Step 2

Factor $3$ out of $3 x - 15$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{(x - 5) (x + 7)}{3 (x) - 15}$

Factor $3$ out of $- 15$ .

$f (x) = \frac{(x - 5) (x + 7)}{3 x + 3 \cdot - 5}$

Factor $3$ out of $3 x + 3 \cdot - 5$ .

$f (x) = \frac{(x - 5) (x + 7)}{3 (x - 5)}$

Step 3

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$f (x) = \frac{(x - 5) (x + 7)}{3 (x - 5)}$

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.

$x - 5$

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 $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Substitute $5$ for $x$ in $\frac{x + 7}{3}$ and simplify.

Substitute $5$ for $x$ to find the $y$ coordinate of the hole.

$\frac{5 + 7}{3}$

Simplify.

Add $5$ and $7$ .

$\frac{12}{3}$

Divide $12$ by $3$ .

$4$

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

$(5, 4)$

Step 6

Find the Holes in the Graph f(x)=(x^2+2x-35)/(3x-15)

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