What is the vertices x^2+12y=0 ?

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

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vertices x^2+12y=0

vertices

x^{2} + 12 y = 0

Maximum (0, 0)

Solution steps

x^{2} + 12 y = 0

Rewrite

x^{2} + 12 y = 0

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)

- 4.9 t^{2} + 9.8 t

- x^{2} + 4 x + 3

y = x^{2} - 4 x - 12

(x + 4)^{2}

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