What is the vertices f(x)=2x^2-4x+1 ?

The vertices f(x)=2x^2-4x+1 is Minimum (1,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices f(x)=2x^2-4x+1

vertices

Minimum (1, - 1)

Solution steps

y = 2 x^{2} - 4 x + 1

The parabola parameters are:

a = 2, b = - 4, c = 1

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

x_{v} = - \frac{( - 4 )}{2 \cdot 2}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(1, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (1, - 1)

What is the vertices f(x)=2x^2-4x+1 ?
The vertices f(x)=2x^2-4x+1 is Minimum (1,-1)