What is the vertices x^2=-4y ?

The vertices x^2=-4y is Maximum (0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices x^2=-4y

vertices

x^{2} = - 4 y

Maximum (0, 0)

Solution steps

x^{2} = - 4 y

Rewrite

x^{2} = - 4 y

in the form

y = a x^{2} + b x + c

y = - \frac{x ^{2}}{4}

The parabola parameters are:

a = - \frac{1}{4}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( - \frac{1}{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 = - \frac{1}{4}

Maximum (0, 0)

x^{2} - y^{2} = 9

x^{2} + y^{2} + 4 x - 6 y + 12 = 0

x^{2} = 16 y

f (x) = - x^{2} + 6 x + 7

- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1