Find the Holes in the Graph f(x)=(x^2-4x-5)/(x+1)

$f (x) = \frac{x^{2} - 4 x - 5}{x + 1}$

Factor $x^{2} - 4 x - 5$ 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 $- 4$ .

$- 5, 1$

Write the factored form using these integers.

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Divide $x - 5$ by $1$ .

$f (x) = x - 5$

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

$x + 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x - 5$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$- 1 - 5$

Subtract $5$ from $- 1$ .

$- 6$

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

$(- 1, - 6)$

Find the Holes in the Graph f(x)=(x^2-4x-5)/(x+1)

Related Questions

Support LCMGCF.com by disabling your adblocker.