Find the Vertex -2x^2+160x

$- 2 x^{2} + 160 x$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- 2 x^{2} + 160 x$ .

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

$a = - 2$

$b = 160$

$c = 0$

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

$d = \frac{160}{2 \cdot - 2}$

Simplify the right side.

Cancel the common factor of $160$ and $2$ .

Factor $2$ out of $160$ .

$d = \frac{2 \cdot 80}{2 \cdot - 2}$

Cancel the common factors.

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

$d = \frac{2 \cdot 80}{2 (- 2)}$

Cancel the common factor.

$d = \frac{2 \cdot 80}{2 \cdot - 2}$

Rewrite the expression.

$d = \frac{80}{- 2}$

Cancel the common factor of $80$ and $- 2$ .

Factor $2$ out of $80$ .

$d = \frac{2 (40)}{- 2}$

Move the negative one from the denominator of $\frac{40}{- 1}$ .

$d = - 1 \cdot 40$

Multiply $- 1$ by $40$ .

$d = - 40$

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

$e = 0 - \frac{160^{2}}{4 \cdot - 2}$

Simplify the right side.

Simplify each term.

Raise $160$ to the power of $2$ .

$e = 0 - \frac{25600}{4 \cdot - 2}$

Multiply $4$ by $- 2$ .

$e = 0 - \frac{25600}{- 8}$

Divide $25600$ by $- 8$ .

$e = 0 - - 3200$

Multiply $- 1$ by $- 3200$ .

$e = 0 + 3200$

Add $0$ and $3200$ .

$e = 3200$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 2 {(x - 40)}^{2} + 3200$ .

$- 2 {(x - 40)}^{2} + 3200$

Set $y$ equal to the new right side.

$y = - 2 {(x - 40)}^{2} + 3200$

Step 2

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

$a = - 2$

$h = 40$

$k = 3200$

Step 3

Find the vertex $(h, k)$ .

$(40, 3200)$

Step 4

Find the Vertex -2x^2+160x

Related Questions

Support LCMGCF.com by disabling your adblocker.