Find the Holes in the Graph f(x)=(4x^2-4x)/(3x^2+4x-7)

$f (x) = \frac{4 x^{2} - 4 x}{3 x^{2} + 4 x - 7}$

Step 1

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

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

$f (x) = \frac{4 x (x) - 4 x}{3 x^{2} + 4 x - 7}$

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

$f (x) = \frac{4 x (x) + 4 x (- 1)}{3 x^{2} + 4 x - 7}$

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

$f (x) = \frac{4 x (x - 1)}{3 x^{2} + 4 x - 7}$

Step 2

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 7 = - 21$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

$f (x) = \frac{4 x (x - 1)}{3 x^{2} + 4 (x) - 7}$

Rewrite $4$ as $- 3$ plus $7$

$f (x) = \frac{4 x (x - 1)}{3 x^{2} + (- 3 + 7) x - 7}$

Apply the distributive property.

$f (x) = \frac{4 x (x - 1)}{3 x^{2} - 3 x + 7 x - 7}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{4 x (x - 1)}{(3 x^{2} - 3 x) + 7 x - 7}$

Factor out the greatest common factor (GCF) from each group.

$f (x) = \frac{4 x (x - 1)}{3 x (x - 1) + 7 (x - 1)}$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$f (x) = \frac{4 x (x - 1)}{(x - 1) (3 x + 7)}$

Step 3

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$f (x) = \frac{4 x (x - 1)}{(x - 1) (3 x + 7)}$

Rewrite the expression.

$f (x) = \frac{4 x}{3 x + 7}$

Step 4

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

$x - 1$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{4 x}{3 x + 7}$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $\frac{4 x}{3 x + 7}$ and simplify.

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

$\frac{4 \cdot 1}{3 \cdot 1 + 7}$

Simplify.

Multiply $4$ by $1$ .

$\frac{4}{3 \cdot 1 + 7}$

Simplify the denominator.

Multiply $3$ by $1$ .

$\frac{4}{3 + 7}$

Add $3$ and $7$ .

$\frac{4}{10}$

Cancel the common factor of $4$ and $10$ .

Factor $2$ out of $4$ .

$\frac{2 (2)}{10}$

Cancel the common factors.

Factor $2$ out of $10$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{2}{5}$

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

$(1, \frac{2}{5})$

Step 6

Find the Holes in the Graph f(x)=(4x^2-4x)/(3x^2+4x-7)

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