Find the Vertex Form y=4x^2+8x-4

$y = 4 x^{2} + 8 x - 4$

Complete the square for $4 x^{2} + 8 x - 4$ .

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

$a = 4$

$b = 8$

$c = - 4$

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

$d = \frac{8}{2 \cdot 4}$

Simplify the right side.

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

Factor $2$ out of $8$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $4$ .

$e = - 4 - \frac{64}{16}$

Divide $64$ by $16$ .

$e = - 4 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = - 4 - 4$

Subtract $4$ from $- 4$ .

$e = - 8$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(x + 1)}^{2} - 8$ .

$4 {(x + 1)}^{2} - 8$

Set $y$ equal to the new right side.

$y = 4 {(x + 1)}^{2} - 8$

Find the Vertex Form y=4x^2+8x-4

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