Find the Vertex Form f(x)=2x^2-8x+6

$f (x) = 2 x^{2} - 8 x + 6$

Step 1

Write $f (x) = 2 x^{2} - 8 x + 6$ as an equation.

$y = 2 x^{2} - 8 x + 6$

Step 2

Complete the square for $2 x^{2} - 8 x + 6$ .

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

$a = 2$

$b = - 8$

$c = 6$

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

$d = \frac{- 8}{2 \cdot 2}$

Simplify the right side.

Cancel the common factor of $- 8$ and $2$ .

Factor $2$ out of $- 8$ .

$d = \frac{2 \cdot - 4}{2 \cdot 2}$

Cancel the common factors.

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

$d = \frac{2 \cdot - 4}{2 (2)}$

Cancel the common factor.

$d = \frac{2 \cdot - 4}{2 \cdot 2}$

Rewrite the expression.

$d = \frac{- 4}{2}$

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

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

$e = 6 - \frac{{(- 8)}^{2}}{4 \cdot 2}$

Simplify the right side.

Simplify each term.

Raise $- 8$ to the power of $2$ .

$e = 6 - \frac{64}{4 \cdot 2}$

Multiply $4$ by $2$ .

$e = 6 - \frac{64}{8}$

Divide $64$ by $8$ .

$e = 6 - 1 \cdot 8$

Multiply $- 1$ by $8$ .

$e = 6 - 8$

Subtract $8$ from $6$ .

$e = - 2$

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

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

Step 3

Set $y$ equal to the new right side.

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

Find the Vertex Form f(x)=2x^2-8x+6

Related Questions

Support LCMGCF.com by disabling your adblocker.