Find the Holes in the Graph f(x)=(x^2-6x+5)/(x^2+4x-5)

$f (x) = \frac{x^{2} - 6 x + 5}{x^{2} + 4 x - 5}$

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 $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

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

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 $4$ .

$- 1, 5$

Write the factored form using these integers.

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x - 5}{x + 5}$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

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

Simplify.

Subtract $5$ from $1$ .

$\frac{- 4}{1 + 5}$

Add $1$ and $5$ .

$\frac{- 4}{6}$

Cancel the common factor of $- 4$ and $6$ .

Factor $2$ out of $- 4$ .

$\frac{2 (- 2)}{6}$

Cancel the common factors.

Factor $2$ out of $6$ .

$\frac{2 \cdot - 2}{2 \cdot 3}$

Cancel the common factor.

$\frac{2 \cdot - 2}{2 \cdot 3}$

Rewrite the expression.

$\frac{- 2}{3}$

Move the negative in front of the fraction.

$- \frac{2}{3}$

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

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

Find the Holes in the Graph f(x)=(x^2-6x+5)/(x^2+4x-5)

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