What is the vertices x^2-y=0 ?

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

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vertices x^2-y=0

vertices

x^{2} - y = 0

Minimum (0, 0)

Solution steps

x^{2} - y = 0

Rewrite

x^{2} - y = 0

in the form

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

y = x^{2}

The parabola parameters are:

a = 1, b = 0, c = 0

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

x_{v} = - \frac{0}{2 \cdot 1}

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 = 1

Minimum (0, 0)

y = x^{2} + 8 x + 12

f (x) = - x^{2} - 2 x - 1

f (x) = - (x + 3) (x + 10)

y^{2} = - 32 x

y = 2 x^{2} + 16 x + 29