What is the vertices-4x^2 ?

The vertices-4x^2 is Maximum (0,0)

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vertices-4x^2

vertices

- 4 x^{2}

Maximum (0, 0)

Solution steps

y = - 4 x^{2}

The parabola parameters are:

a = - 4, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( - 4 )}

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 = - 4

Maximum (0, 0)

x^{2} - 4 x - 7

f (x) = - 3 x^{2} + 4 x - 1

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

f (x) = - 9.5 x^{2} - 47.5 x + 63

f (x) = 3 (x - 1) (x + 2)