What is the vertices x^2=-12y ?

The vertices x^2=-12y is Maximum (0,0)

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vertices x^2=-12y

vertices

x^{2} = - 12 y

Maximum (0, 0)

Solution steps

x^{2} = - 12 y

Rewrite

x^{2} = - 12 y

in the form

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

y = - \frac{x ^{2}}{12}

The parabola parameters are:

a = - \frac{1}{12}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( - \frac{1}{12} )}

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}{12}

Maximum (0, 0)

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

x^{2} + 8 y + 2 x - 23 = 0

y = x^{2} - 10 x + 24

2 x^{2} - 5 x + 5

f (x) = 2 x^{2} + 8 x - 3