Find Where Undefined/Discontinuous (16-4y^2)/(3y^2-12y+12)

$\frac{16 - 4 y^{2}}{3 y^{2} - 12 y + 12}$

Step 1

Set the denominator in $\frac{16 - 4 y^{2}}{3 y^{2} - 12 y + 12}$ equal to $0$ to find where the expression is undefined.

$3 y^{2} - 12 y + 12 = 0$

Step 2

Solve for $y$ .

Factor the left side of the equation.

Factor $3$ out of $3 y^{2} - 12 y + 12$ .

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

$3 (y^{2}) - 12 y + 12 = 0$

Factor $3$ out of $- 12 y$ .

$3 (y^{2}) + 3 (- 4 y) + 12 = 0$

Factor $3$ out of $12$ .

$3 y^{2} + 3 (- 4 y) + 3 \cdot 4 = 0$

Factor $3$ out of $3 y^{2} + 3 (- 4 y)$ .

$3 (y^{2} - 4 y) + 3 \cdot 4 = 0$

Factor $3$ out of $3 (y^{2} - 4 y) + 3 \cdot 4$ .

$3 (y^{2} - 4 y + 4) = 0$

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

$3 (y^{2} - 4 y + 2^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 y = 2 \cdot y \cdot 2$

Rewrite the polynomial.

$3 (y^{2} - 2 \cdot y \cdot 2 + 2^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 2$ .

$3 {(y - 2)}^{2} = 0$

Divide each term in $3 {(y - 2)}^{2} = 0$ by $3$ and simplify.

Divide each term in $3 {(y - 2)}^{2} = 0$ by $3$ .

$\frac{3 {(y - 2)}^{2}}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 {(y - 2)}^{2}}{3} = \frac{0}{3}$

Divide ${(y - 2)}^{2}$ by $1$ .

${(y - 2)}^{2} = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

${(y - 2)}^{2} = 0$

Set the $y - 2$ equal to $0$ .

$y - 2 = 0$

Add $2$ to both sides of the equation.

$y = 2$

Step 3