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

The vertices f(x)=4x^2 is Minimum (0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=4x^2

vertices

f (x) = 4 x^{2}

Minimum (0, 0)

Solution steps

y = 4 x^{2}

The parabola parameters are:

a = 4, b = 0, c = 0

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} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (0, 0)

2 x^{2} + 2 x + 3 y = 0

y = 4 x - x^{2}

y = x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

x^{2} + 5 x + 6

f (x) = 2 x^{2} + 6 x - 1