What is the vertices f(x)=-2x^2-16x-31 ?

The vertices f(x)=-2x^2-16x-31 is Maximum (-4,1)

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vertices f(x)=-2x^2-16x-31

vertices

f (x) = - 2 x^{2} - 16 x - 31

Maximum (- 4, 1)

Solution steps

y = - 2 x^{2} - 16 x - 31

The parabola parameters are:

a = - 2, b = - 16, c = - 31

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

x_{v} = - \frac{( - 16 )}{2 ( - 2 )}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 4, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- 4, 1)

- x^{2} + 2 x + 2

f (x) = - 2 x^{2} - 16 x - 37

- x^{2} + 2 x + 4

4 x^{2} - 8 x - 2

2 x (x - 9) - (x - 9)

What is the vertices f(x)=-2x^2-16x-31 ?
The vertices f(x)=-2x^2-16x-31 is Maximum (-4,1)