What is the vertices-3x^2-6x+2 ?

The vertices-3x^2-6x+2 is Maximum (-1,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices-3x^2-6x+2

vertices

- 3 x^{2} - 6 x + 2

Maximum (- 1, 5)

Solution steps

y = - 3 x^{2} - 6 x + 2

The parabola parameters are:

a = - 3, b = - 6, c = 2

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

x_{v} = - \frac{( - 6 )}{2 ( - 3 )}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(- 1, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (- 1, 5)

f (x) = 5 x^{2} - 10 x + 4

(y - 9)^{2} = - 4 (x - 2)

(x + 2)^{2} = - 8 (y + 3)

y = x^{2} + 3 x + 4

y = 24 x - 3 x^{2}