What is the vertices y=4x^2-8 ?

The vertices y=4x^2-8 is Minimum (0,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=4x^2-8

vertices

y = 4 x^{2} - 8

Minimum (0, - 8)

Solution steps

y = 4 x^{2} - 8

The parabola parameters are:

a = 4, b = 0, c = - 8

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

x_{v} = - \frac{0}{2 \cdot 4}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 8

Therefore the parabola vertex is

(0, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (0, - 8)

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

y = (x + 2)^{2} - 3

f (x) = 2 x^{2} - 8 x - 5

f (x) = 2 x^{2} - 8 x + 6

y = + 0.5 x^{2} - 6 x + 21