What is the vertices f(x)=-2x^2-12x-16 ?

The vertices f(x)=-2x^2-12x-16 is Maximum (-3,2)

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vertices f(x)=-2x^2-12x-16

vertices

f (x) = - 2 x^{2} - 12 x - 16

Maximum (- 3, 2)

Solution steps

y = - 2 x^{2} - 12 x - 16

The parabola parameters are:

a = - 2, b = - 12, c = - 16

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

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

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 3, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- 3, 2)

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

f (x) = 2 x^{2} + 8 x + 9

y = - x^{2} + 8 x - 15

f (x) = - 2 x^{2} + 8 x - 5

f (x) = - 2 x^{2} + 8 x - 9

What is the vertices f(x)=-2x^2-12x-16 ?
The vertices f(x)=-2x^2-12x-16 is Maximum (-3,2)