Find the Holes in the Graph f(x)=(x^2+9x+20)/(4x+20)

$f (x) = \frac{x^{2} + 9 x + 20}{4 x + 20}$

Factor $x^{2} + 9 x + 20$ 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 $20$ and whose sum is $9$ .

$4, 5$

Write the factored form using these integers.

$f (x) = \frac{(x + 4) (x + 5)}{4 x + 20}$

Factor $4$ out of $4 x + 20$ .

Factor $4$ out of $4 x$ .

$f (x) = \frac{(x + 4) (x + 5)}{4 (x) + 20}$

Factor $4$ out of $20$ .

$f (x) = \frac{(x + 4) (x + 5)}{4 x + 4 \cdot 5}$

Factor $4$ out of $4 x + 4 \cdot 5$ .

$f (x) = \frac{(x + 4) (x + 5)}{4 (x + 5)}$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$f (x) = \frac{(x + 4) (x + 5)}{4 (x + 5)}$

Rewrite the expression.

$f (x) = \frac{x + 4}{4}$

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

$x + 5$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 4}{4}$ .

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $\frac{x + 4}{4}$ and simplify.

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

$\frac{- 5 + 4}{4}$

Simplify.

Add $- 5$ and $4$ .

$\frac{- 1}{4}$

Move the negative in front of the fraction.

$- \frac{1}{4}$

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

$(- 5, - \frac{1}{4})$

Find the Holes in the Graph f(x)=(x^2+9x+20)/(4x+20)

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