What is the vertices f(x)=-3x^2 ?

The vertices f(x)=-3x^2 is Maximum (0,0)

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vertices f(x)=-3x^2

vertices

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

Maximum (0, 0)

Solution steps

y = - 3 x^{2}

The parabola parameters are:

a = - 3, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( - 3 )}

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 = - 3

Maximum (0, 0)

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

f (x) = 108 - 5 x^{2} - 7 x

2 x^{2} - x - 1

2 x^{2} - x - 6

6 x^{2} - 13 x + 6