What is the vertices x^2-8x+12 ?

The vertices x^2-8x+12 is Minimum (4,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices x^2-8x+12

vertices

x^{2} - 8 x + 12

Minimum (4, - 4)

Solution steps

y = x^{2} - 8 x + 12

The parabola parameters are:

a = 1, b = - 8, c = 12

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

x_{v} = - \frac{( - 8 )}{2 \cdot 1}

Simplify

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(4, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 4)

X^{2} - 5 X - 24

y = x^{2} + 8 x + 7

4 t^{2} + 1

f (x) = \frac{1}{2} (x - 1)^{2}

x^{2} + 1