Find the Holes in the Graph f(x)=((x+7)(x-3)(x+1))/((x-3)(x-5))

$f (x) = \frac{(x + 7) (x - 3) (x + 1)}{(x - 3) (x - 5)}$

Step 1

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$f (x) = \frac{(x + 7) (x - 3) (x + 1)}{(x - 3) (x - 5)}$

Rewrite the expression.

$f (x) = \frac{(x + 7) (x + 1)}{x - 5}$

Step 2

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

$x - 3$

Step 3

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{(x + 7) (x + 1)}{x - 5}$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $\frac{(x + 7) (x + 1)}{x - 5}$ and simplify.

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

$\frac{(3 + 7) (3 + 1)}{3 - 5}$

Simplify.

Simplify the numerator.

Add $3$ and $7$ .

$\frac{10 (3 + 1)}{3 - 5}$

Add $3$ and $1$ .

$\frac{10 \cdot 4}{3 - 5}$

Subtract $5$ from $3$ .

$\frac{10 \cdot 4}{- 2}$

Multiply $10$ by $4$ .

$\frac{40}{- 2}$

Divide $40$ by $- 2$ .

$- 20$

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

$(3, - 20)$

Step 4

Find the Holes in the Graph f(x)=((x+7)(x-3)(x+1))/((x-3)(x-5))

Related Questions

Support LCMGCF.com by disabling your adblocker.