Find Where Undefined/Discontinuous (4y^2-18y+18)/(y^3-6y^2+9y)

$\frac{4 y^{2} - 18 y + 18}{y^{3} - 6 y^{2} + 9 y}$

Step 1

Set the denominator in $\frac{4 y^{2} - 18 y + 18}{y^{3} - 6 y^{2} + 9 y}$ equal to $0$ to find where the expression is undefined.

$y^{3} - 6 y^{2} + 9 y = 0$

Step 2

Solve for $y$ .

Factor the left side of the equation.

Factor $y$ out of $y^{3} - 6 y^{2} + 9 y$ .

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

$y \cdot y^{2} - 6 y^{2} + 9 y = 0$

Factor $y$ out of $- 6 y^{2}$ .

$y \cdot y^{2} + y (- 6 y) + 9 y = 0$

Factor $y$ out of $9 y$ .

$y \cdot y^{2} + y (- 6 y) + y \cdot 9 = 0$

Factor $y$ out of $y \cdot y^{2} + y (- 6 y)$ .

$y (y^{2} - 6 y) + y \cdot 9 = 0$

Factor $y$ out of $y (y^{2} - 6 y) + y \cdot 9$ .

$y (y^{2} - 6 y + 9) = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$y (y^{2} - 6 y + 3^{2}) = 0$

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

$6 y = 2 \cdot y \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 3$ .

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y = 0$

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

Set $y$ equal to $0$ .

$y = 0$

Set ${(y - 3)}^{2}$ equal to $0$ and solve for $y$ .

Set ${(y - 3)}^{2}$ equal to $0$ .

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

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

Set the $y - 3$ equal to $0$ .

$y - 3 = 0$

Add $3$ to both sides of the equation.

$y = 3$

The final solution is all the values that make $y {(y - 3)}^{2} = 0$ true.

$y = 0, 3$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$y = 0, y = 3$

Step 4

Find Where Undefined/Discontinuous (4y^2-18y+18)/(y^3-6y^2+9y)

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