Find the Vertex Form 6x-4+x^2

$6 x - 4 + x^{2}$

Step 1

Write $6 x - 4 + x^{2}$ as an equation.

$y = 6 x - 4 + x^{2}$

Step 2

Simplify $6 x - 4 + x^{2}$ .

Move $- 4$ .

$y = 6 x + x^{2} - 4$

Reorder $6 x$ and $x^{2}$ .

$y = x^{2} + 6 x - 4$

Step 3

Complete the square for $x^{2} + 6 x - 4$ .

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

$a = 1$

$b = 6$

$c = - 4$

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

$d = \frac{6}{2 \cdot 1}$

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

Factor $2$ out of $6$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $3$ by $1$ .

$d = 3$

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

$e = - 4 - \frac{6^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = - 4 - \frac{36}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = - 4 - \frac{36}{4}$

Divide $36$ by $4$ .

$e = - 4 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = - 4 - 9$

Subtract $9$ from $- 4$ .

$e = - 13$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 3)}^{2} - 13$ .

${(x + 3)}^{2} - 13$

Step 4

Set $y$ equal to the new right side.

$y = {(x + 3)}^{2} - 13$

Find the Vertex Form 6x-4+x^2

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