Find the Vertex -2x^2+16x-28

$- 2 x^{2} + 16 x - 28$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- 2 x^{2} + 16 x - 28$ .

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

$a = - 2$

$b = 16$

$c = - 28$

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

$d = \frac{16}{2 \cdot - 2}$

Simplify the right side.

Cancel the common factor of $16$ and $2$ .

Factor $2$ out of $16$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $2$ out of $8$ .

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

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$d = - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$d = - 4$

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

$e = - 28 - \frac{16^{2}}{4 \cdot - 2}$

Simplify the right side.

Simplify each term.

Raise $16$ to the power of $2$ .

$e = - 28 - \frac{256}{4 \cdot - 2}$

Multiply $4$ by $- 2$ .

$e = - 28 - \frac{256}{- 8}$

Divide $256$ by $- 8$ .

$e = - 28 - - 32$

Multiply $- 1$ by $- 32$ .

$e = - 28 + 32$

Add $- 28$ and $32$ .

$e = 4$

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

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

Set $y$ equal to the new right side.

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

Step 2

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

$a = - 2$

$h = 4$

$k = 4$

Step 3

Find the vertex $(h, k)$ .

$(4, 4)$

Step 4

Find the Vertex -2x^2+16x-28

Related Questions

Support LCMGCF.com by disabling your adblocker.