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

The vertices f(x)=-2x^2+2 is Maximum (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)=-2x^2+2

vertices

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

Maximum (0, 2)

Solution steps

y = - 2 x^{2} + 2

The parabola parameters are:

a = - 2, b = 0, c = 2

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

x_{v} = - \frac{0}{2 ( - 2 )}

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 = - 2

Maximum (0, 2)

y = - x^{2} + 4 x + 5

y = 2 x^{2} + 15 x

f (x) = - x^{2} + 10 x - 16

f (x) = - x^{2} + 5 x - 6

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