Find the Vertex y=2x^2+4x-4

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

Rewrite the equation in vertex form.

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

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

$a = 2$

$b = 4$

$c = - 4$

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

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

Simplify the right side.

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

Factor $2$ out of $4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$d = 1$

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

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $4^{2}$ and $4$ .

Factor $4$ out of $4^{2}$ .

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

Cancel the common factors.

Factor $4$ out of $4 \cdot 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $2$ out of $4$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

$e = - 4 - \frac{2 \cdot 2}{2 (1)}$

Cancel the common factor.

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

Rewrite the expression.

$e = - 4 - \frac{2}{1}$

Divide $2$ by $1$ .

$e = - 4 - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$e = - 4 - 2$

Subtract $2$ from $- 4$ .

$e = - 6$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $2 {(x + 1)}^{2} - 6$ .

$2 {(x + 1)}^{2} - 6$

Set $y$ equal to the new right side.

$y = 2 {(x + 1)}^{2} - 6$

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 2$

$h = - 1$

$k = - 6$

Find the vertex $(h, k)$ .

$(- 1, - 6)$

Find the Vertex y=2x^2+4x-4

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