Find the Holes in the Graph y=(x+3)/(x^2+8x+15)

$y = \frac{x + 3}{x^{2} + 8 x + 15}$

Step 1

Factor $x^{2} + 8 x + 15$ 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 $15$ and whose sum is $8$ .

$3, 5$

Write the factored form using these integers.

$y = \frac{x + 3}{(x + 3) (x + 5)}$

Step 2

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$y = \frac{x + 3}{(x + 3) (x + 5)}$

Rewrite the expression.

$y = \frac{1}{x + 5}$

Step 3

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

$x + 3$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x + 5}$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $\frac{1}{x + 5}$ and simplify.

Substitute $- 3$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{- 3 + 5}$

Add $- 3$ and $5$ .

$\frac{1}{2}$

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

$(- 3, \frac{1}{2})$

Step 5

Find the Holes in the Graph y=(x+3)/(x^2+8x+15)

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