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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=x^2+2

vertices

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

Minimum (0, 2)

Solution steps

y = x^{2} + 2

The parabola parameters are:

a = 1, b = 0, c = 2

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

x_{v} = - \frac{0}{2 \cdot 1}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(0, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 2)

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

x^{2} - 8 x + 12

X^{2} - 5 X - 24

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

4 t^{2} + 1