Find the Holes in the Graph f(x)=(3x^2-15x)/(-2x+10)

$f (x) = \frac{3 x^{2} - 15 x}{- 2 x + 10}$

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

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

$f (x) = \frac{3 x (x) - 15 x}{- 2 x + 10}$

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

$f (x) = \frac{3 x (x) + 3 x (- 5)}{- 2 x + 10}$

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

$f (x) = \frac{3 x (x - 5)}{- 2 x + 10}$

Factor $2$ out of $- 2 x + 10$ .

Factor $2$ out of $- 2 x$ .

$f (x) = \frac{3 x (x - 5)}{2 (- x) + 10}$

Factor $2$ out of $10$ .

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

Factor $2$ out of $2 (- x) + 2 (5)$ .

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

Cancel the common factor of $x - 5$ and $- x + 5$ .

Factor $- 1$ out of $x$ .

$f (x) = \frac{3 x (- 1 (- x) - 5)}{2 (- x + 5)}$

Rewrite $- 5$ as $- 1 (5)$ .

$f (x) = \frac{3 x (- 1 (- x) - 1 \cdot 5)}{2 (- x + 5)}$

Factor $- 1$ out of $- 1 (- x) - 1 (5)$ .

$f (x) = \frac{3 x (- 1 (- x + 5))}{2 (- x + 5)}$

Cancel the common factor.

$f (x) = \frac{3 x (- 1 (- x + 5))}{2 (- x + 5)}$

Rewrite the expression.

$f (x) = \frac{3 x \cdot (- 1)}{2}$

Multiply $- 1$ by $3$ .

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

Move the negative in front of the fraction.

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

Multiply $3$ by $x$ .

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

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 $- \frac{3 x}{2}$ .

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

$- x + 5 = 0$

Solve $- x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$- x = - 5$

Divide each term in $- x = - 5$ by $- 1$ and simplify.

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

Divide $- 5$ by $- 1$ .

$x = 5$

Substitute $5$ for $x$ in $- \frac{3 x}{2}$ and simplify.

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

$- \frac{3 \cdot 5}{2}$

Multiply $3$ by $5$ .

$- \frac{15}{2}$

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

$(5, - \frac{15}{2})$

Find the Holes in the Graph f(x)=(3x^2-15x)/(-2x+10)

Related Questions

Support LCMGCF.com by disabling your adblocker.