What is the vertices-3x^2-6x+2 ?

The vertices-3x^2-6x+2 is Maximum (-1,5)

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vertices-3x^2-6x+2

vertices

Maximum (- 1, 5)

Solution steps

y = - 3 x^{2} - 6 x + 2

The parabola parameters are:

a = - 3, b = - 6, c = 2

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

x_{v} = - \frac{( - 6 )}{2 ( - 3 )}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(- 1, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (- 1, 5)