What is the vertices x^2=4y ?

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

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vertices x^2=4y

vertices

x^{2} = 4 y

Minimum (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{1 \cdot 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}

Minimum (0, 0)

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

x^{2} - 6 x + y^{2} - 32 y = 0

(y - 2)^{2} = 8 (x - 1)

\frac{y ^{2}}{9} - \frac{x ^{2}}{4} = 1

y = 3 x^{2}