Find the Vertex Form y=3x^2-12x+18

$y = 3 x^{2} - 12 x + 18$

Step 1

Complete the square for $3 x^{2} - 12 x + 18$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 3$

$b = - 12$

$c = 18$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 12}{2 \cdot 3}$

Simplify the right side.

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

$d = \frac{2 \cdot - 6}{2 \cdot 3}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 3$ .

$d = \frac{2 \cdot - 6}{2 (3)}$

Cancel the common factor.

$d = \frac{2 \cdot - 6}{2 \cdot 3}$

Rewrite the expression.

$d = \frac{- 6}{3}$

Cancel the common factor of $- 6$ and $3$ .

Factor $3$ out of $- 6$ .

$d = \frac{3 \cdot - 2}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

$d = \frac{3 \cdot - 2}{3 \cdot 1}$

Rewrite the expression.

$d = \frac{- 2}{1}$

Divide $- 2$ by $1$ .

$d = - 2$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 18 - \frac{{(- 12)}^{2}}{4 \cdot 3}$

Simplify the right side.

Simplify each term.

Raise $- 12$ to the power of $2$ .

$e = 18 - \frac{144}{4 \cdot 3}$

Multiply $4$ by $3$ .

$e = 18 - \frac{144}{12}$

Divide $144$ by $12$ .

$e = 18 - 1 \cdot 12$

Multiply $- 1$ by $12$ .

$e = 18 - 12$

Subtract $12$ from $18$ .

$e = 6$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x - 2)}^{2} + 6$ .

$3 {(x - 2)}^{2} + 6$

Step 2

Set $y$ equal to the new right side.

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

Find the Vertex Form y=3x^2-12x+18

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