What is the vertices y=-3x^2-12x-9 ?

The vertices y=-3x^2-12x-9 is Maximum (-2,3)

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vertices y=-3x^2-12x-9

vertices

y = - 3 x^{2} - 12 x - 9

Maximum (- 2, 3)

Solution steps

y = - 3 x^{2} - 12 x - 9

The parabola parameters are:

a = - 3, b = - 12, c = - 9

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

x_{v} = - \frac{( - 12 )}{2 ( - 3 )}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(- 2, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (- 2, 3)

x^{2} = 8 (y + 4)

y = 3 x^{2} - 12 x + 8

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

y = x^{2} - 4 x + 7

x = (y - 3)^{2}

What is the vertices y=-3x^2-12x-9 ?
The vertices y=-3x^2-12x-9 is Maximum (-2,3)