Find the Holes in the Graph f(x)=(x^2-5x)/(-3x^2-15x)

$f (x) = \frac{x^{2} - 5 x}{- 3 x^{2} - 15 x}$

Step 1

Factor $x$ out of $x^{2} - 5 x$ .

Factor $x$ out of $x^{2}$ .

$f (x) = \frac{x \cdot x - 5 x}{- 3 x^{2} - 15 x}$

Factor $x$ out of $- 5 x$ .

$f (x) = \frac{x \cdot x + x \cdot - 5}{- 3 x^{2} - 15 x}$

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

$f (x) = \frac{x (x - 5)}{- 3 x^{2} - 15 x}$

Step 2

Factor $3 x$ out of $- 3 x^{2} - 15 x$ .

Factor $3 x$ out of $- 3 x^{2}$ .

$f (x) = \frac{x (x - 5)}{3 x (- x) - 15 x}$

Factor $3 x$ out of $- 15 x$ .

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

Factor $3 x$ out of $3 x (- x) + 3 x (- 5)$ .

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

Step 3

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 5}{3 (- x - 5)}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{x - 5}{3 (- x - 5)}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{0 - 5}{3 (- 0 - 5)}$

Simplify.

Cancel the common factor of $0 - 5$ and $- 0 - 5$ .

Factor $5$ out of $0$ .

$\frac{5 (0) - 5}{3 (- 0 - 5)}$

Factor $5$ out of $- 5$ .

$\frac{5 \cdot 0 + 5 \cdot - 1}{3 (- 0 - 5)}$

Factor $5$ out of $5 \cdot 0 + 5 \cdot - 1$ .

$\frac{5 \cdot (0 - 1)}{3 (- 0 - 5)}$

Cancel the common factors.

Factor $5$ out of $3 (- 0 - 5)$ .

$\frac{5 \cdot (0 - 1)}{5 (3 (- 0 - 1))}$

Cancel the common factor.

$\frac{5 \cdot (0 - 1)}{5 (3 (- 0 - 1))}$

Rewrite the expression.

$\frac{0 - 1}{3 (- 0 - 1)}$

Subtract $1$ from $0$ .

$\frac{- 1}{3 (- 0 - 1)}$

Simplify the denominator.

Multiply $- 1$ by $0$ .

$\frac{- 1}{3 (0 - 1)}$

Subtract $1$ from $0$ .

$\frac{- 1}{3 \cdot - 1}$

Multiply $3$ by $- 1$ .

$\frac{- 1}{- 3}$

Dividing two negative values results in a positive value.

$\frac{1}{3}$

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

$(0, \frac{1}{3})$

Step 6

Find the Holes in the Graph f(x)=(x^2-5x)/(-3x^2-15x)

Related Questions

Support LCMGCF.com by disabling your adblocker.