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

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

Rewrite the equation in vertex form.

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

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

$a = - 4$

$b = 8$

$c = 2$

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

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

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 $4$ and $- 4$ .

Factor $4$ out of $4$ .

$d = \frac{4 (1)}{- 4}$

Move the negative one from the denominator of $\frac{1}{- 1}$ .

$d = - 1 \cdot 1$

Multiply $- 1$ 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 $e = c - \frac{b^{2}}{4 a}$ .

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $- 4$ .

$e = 2 - \frac{64}{- 16}$

Divide $64$ by $- 16$ .

$e = 2 - - 4$

Multiply $- 1$ by $- 4$ .

$e = 2 + 4$

Add $2$ and $4$ .

$e = 6$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 4 {(x - 1)}^{2} + 6$ .

$- 4 {(x - 1)}^{2} + 6$

Set $y$ equal to the new right side.

$y = - 4 {(x - 1)}^{2} + 6$

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

$a = - 4$

$h = 1$

$k = 6$

Find the vertex $(h, k)$ .

$(1, 6)$

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

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