What is the vertices x^2=8y ?

The vertices x^2=8y is Minimum (0,0)

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vertices x^2=8y

vertices

x^{2} = 8 y

Minimum (0, 0)

Solution steps

x^{2} = 8 y

Rewrite

x^{2} = 8 y

in the form

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

y = \frac{1 \cdot x ^{2}}{8}

The parabola parameters are:

a = \frac{1}{8}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( \frac{1}{8} )}

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}{8}

Minimum (0, 0)

y = x^{2} - x - 2

- 2 x^{2} - 20 x - 39

y = 2 x^{2} + 5 x - 3

x^{2} - 5 x + 6

2 x^{2} + 3