What is the vertices-2x^2+12x ?

The vertices-2x^2+12x is Maximum (3,18)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices-2x^2+12x

vertices

- 2 x^{2} + 12 x

Maximum (3, 18)

Solution steps

y = - 2 x^{2} + 12 x

The parabola parameters are:

a = - 2, b = 12, c = 0

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{12}{2 ( - 2 )}

Simplify

- \frac{12}{2 ( - 2 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 18

Therefore the parabola vertex is

(3, 18)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (3, 18)

f (x) = 4 x^{2} - 7 x - 2

2 y^{2} + 8 y + x + 1 = 0

y^{2} - 12 y - 4 x + 28 = 0

f (x) = (x - 2)^{2} - 9

(3 x - 4) (4 x - 3) - 2 x