What is the vertices f(x)=2x^2-4x+3 ?

The vertices f(x)=2x^2-4x+3 is Minimum (1,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=2x^2-4x+3

vertices

f (x) = 2 x^{2} - 4 x + 3

Minimum (1, 1)

Solution steps

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

The parabola parameters are:

a = 2, b = - 4, c = 3

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)

y = 18 x^{2} - 6 x + 4

f (x) = - x^{2} + 14 x - 48

f (x) = x^{2} + 11 x - 35

f (x) = 2 x^{2} + 4 x

y = 3 x^{2} - 6 x + 4

What is the vertices f(x)=2x^2-4x+3 ?
The vertices f(x)=2x^2-4x+3 is Minimum (1,1)