What is the vertices y=-3x^2-36x-104 ?

The vertices y=-3x^2-36x-104 is Maximum (-6,4)

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vertices y=-3x^2-36x-104

vertices

y = - 3 x^{2} - 36 x - 104

Maximum (- 6, 4)

Solution steps

y = - 3 x^{2} - 36 x - 104

The parabola parameters are:

a = - 3, b = - 36, c = - 104

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

x_{v} = - \frac{( - 36 )}{2 ( - 3 )}

Simplify

x_{v} = - 6

Plug in

x_{v} = - 6

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(- 6, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (- 6, 4)

f (x) = x^{2} - 6 x + 11

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

f (x) = x^{2} + 10 x + 21

y = (9 x - 2) x 3

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

What is the vertices y=-3x^2-36x-104 ?
The vertices y=-3x^2-36x-104 is Maximum (-6,4)