Find the Holes in the Graph f(x)=(3x^2-15x)/(4x^2+4x)

$f (x) = \frac{3 x^{2} - 15 x}{4 x^{2} + 4 x}$

Step 1

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

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

$f (x) = \frac{3 x (x) - 15 x}{4 x^{2} + 4 x}$

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

$f (x) = \frac{3 x (x) + 3 x (- 5)}{4 x^{2} + 4 x}$

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

$f (x) = \frac{3 x (x - 5)}{4 x^{2} + 4 x}$

Step 2

Factor $4 x$ out of $4 x^{2} + 4 x$ .

Factor $4 x$ out of $4 x^{2}$ .

$f (x) = \frac{3 x (x - 5)}{4 x (x) + 4 x}$

Factor $4 x$ out of $4 x$ .

$f (x) = \frac{3 x (x - 5)}{4 x (x) + 4 x (1)}$

Factor $4 x$ out of $4 x (x) + 4 x (1)$ .

$f (x) = \frac{3 x (x - 5)}{4 x (x + 1)}$

Step 3

Cancel the common factor of $x$ .

Cancel the common factor.

$f (x) = \frac{3 x (x - 5)}{4 x (x + 1)}$

Rewrite the expression.

$f (x) = \frac{3 (x - 5)}{4 (x + 1)}$

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 $\frac{3 (x - 5)}{4 (x + 1)}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{3 (x - 5)}{4 (x + 1)}$ and simplify.

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

$\frac{3 (0 - 5)}{4 (0 + 1)}$

Simplify.

Subtract $5$ from $0$ .

$\frac{3 \cdot - 5}{4 (0 + 1)}$

Add $0$ and $1$ .

$\frac{3 \cdot - 5}{4 \cdot 1}$

Multiply $3$ by $- 5$ .

$\frac{- 15}{4 \cdot 1}$

Multiply $4$ by $1$ .

$\frac{- 15}{4}$

Move the negative in front of the fraction.

$- \frac{15}{4}$

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

$(0, - \frac{15}{4})$

Step 6

Find the Holes in the Graph f(x)=(3x^2-15x)/(4x^2+4x)

Related Questions

Support LCMGCF.com by disabling your adblocker.