What is the vertices f(x)=2x^2-8x ?

The vertices f(x)=2x^2-8x is Minimum (2,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=2x^2-8x

vertices

f (x) = 2 x^{2} - 8 x

Minimum (2, - 8)

Solution steps

y = 2 x^{2} - 8 x

The parabola parameters are:

a = 2, b = - 8, c = 0

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

x_{v} = - \frac{( - 8 )}{2 \cdot 2}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 8

Therefore the parabola vertex is

(2, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (2, - 8)

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

y^{2} = - 24 x

f (x) = 2 x^{2} - 12 x + 23

x^{2} + 6 x + 7

f (x) = 6 x^{2} - x + 1