What is the vertices y=-x^2+3 ?

The vertices y=-x^2+3 is Maximum (0,3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=-x^2+3

vertices

y = - x^{2} + 3

Maximum (0, 3)

Solution steps

y = - x^{2} + 3

The parabola parameters are:

a = - 1, b = 0, c = 3

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

x_{v} = - \frac{0}{2 ( - 1 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(0, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (0, 3)

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

y = 5 x^{2} + 13 x - 6

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

x^{2} - 12 x + 44

2 y^{2} - 8 x - 24 y + 56 = 0