What is the vertices y=-x^2 ?

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

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vertices y=-x^2

vertices

y = - x^{2}

Maximum (0, 0)

Solution steps

y = - x^{2}

The parabola parameters are:

a = - 1, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( - 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

Maximum (0, 0)

x^{2} - 4 x + 4

t^{2}

f (x) = - x^{2} - 6 x + 3

x^{2} - 2 x - 3

y = \frac{- 3 x ^{2} - 2 x + 6}{5}