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

The vertices y=-x^2+4x-4 is Maximum (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

Maximum (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 ( - 1 )}

Simplify

- \frac{4}{2 ( - 1 )} : 2

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

Maximum (2, 0)

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

y = x^{2} + 16 x + 71

f (x) = 2 x^{2} - 12 x + 1

f (x) = x^{2} + 9 x + 8

8 x^{2} + 12 x + 3