Find the Vertex Form f(x)=3x^2-6x+2

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

Step 1

Write $f (x) = 3 x^{2} - 6 x + 2$ as an equation.

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

Step 2

Complete the square for $3 x^{2} - 6 x + 2$ .

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

$a = 3$

$b = - 6$

$c = 2$

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 $d = \frac{b}{2 a}$ .

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

Simplify the right side.

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $3$ out of $- 3$ .

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ by $1$ .

$d = - 1$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 2 - \frac{{(- 6)}^{2}}{4 \cdot 3}$

Simplify the right side.

Simplify each term.

Raise $- 6$ to the power of $2$ .

$e = 2 - \frac{36}{4 \cdot 3}$

Multiply $4$ by $3$ .

$e = 2 - \frac{36}{12}$

Divide $36$ by $12$ .

$e = 2 - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$e = 2 - 3$

Subtract $3$ from $2$ .

$e = - 1$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x - 1)}^{2} - 1$ .

$3 {(x - 1)}^{2} - 1$

Step 3

Set $y$ equal to the new right side.

$y = 3 {(x - 1)}^{2} - 1$

Find the Vertex Form f(x)=3x^2-6x+2

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