What is the vertices x^2+16 ?

The vertices x^2+16 is Minimum (0,16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices x^2+16

vertices

x^{2} + 16

Minimum (0, 16)

Solution steps

y = x^{2} + 16

The parabola parameters are:

a = 1, b = 0, c = 16

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} = 16

Therefore the parabola vertex is

(0, 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 16)

y = x^{2} + 4 x

f (x) = (x - 4)^{2} - 3

f (x) = 2 x^{2} - 3 x - 5

y = - 2 x^{2} - 16 x - 37

x = 3 y^{2}