What is the vertices y=x^2-4x+4 ?

The vertices y=x^2-4x+4 is Minimum (2,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=x^2-4x+4

vertices

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

Minimum (2, 0)

Solution steps

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

The parabola parameters are:

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

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(2, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 0)

f (x) = - 2 x^{2} - 3 x + 1

f (x) = x^{2} - 4 x + 6

4 + 2 x - x^{2}

y^{2} = 40 x

y^{2} = 16 x - 8