Find the Holes in the Graph f(x)=(x^2+6x+5)/(x+5)

$f (x) = \frac{x^{2} + 6 x + 5}{x + 5}$

Step 1

Factor $x^{2} + 6 x + 5$ 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 $5$ and whose sum is $6$ .

$1, 5$

Write the factored form using these integers.

$f (x) = \frac{(x + 1) (x + 5)}{x + 5}$

Step 2

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$f (x) = \frac{(x + 1) (x + 5)}{x + 5}$

Divide $x + 1$ by $1$ .

$f (x) = x + 1$

Step 3

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

$x + 5$

Step 4

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

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $x + 1$ and simplify.

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

$- 5 + 1$

Add $- 5$ and $1$ .

$- 4$

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

$(- 5, - 4)$

Step 5

Find the Holes in the Graph f(x)=(x^2+6x+5)/(x+5)

Related Questions

Support LCMGCF.com by disabling your adblocker.