What is the vertices x^2=16y ?

The vertices x^2=16y is Minimum (0,0)

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vertices x^2=16y

vertices

x^{2} = 16 y

Minimum (0, 0)

Solution steps

x^{2} = 16 y

Rewrite

x^{2} = 16 y

in the form

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

y = \frac{1 \cdot x ^{2}}{16}

The parabola parameters are:

a = \frac{1}{16}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( \frac{1}{16} )}

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}{16}

Minimum (0, 0)

y^{2} = - 10 x

\frac{x ^{2}}{25} + \frac{y ^{2}}{25} = 1

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

- x^{2} + 9 y^{2} + 4 x + 54 y + 68 = 0

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