What is the vertices f(x)=-2x^2-16x-30 ?

The vertices f(x)=-2x^2-16x-30 is Maximum (-4,2)

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vertices f(x)=-2x^2-16x-30

vertices

f (x) = - 2 x^{2} - 16 x - 30

Maximum (- 4, 2)

Solution steps

y = - 2 x^{2} - 16 x - 30

The parabola parameters are:

a = - 2, b = - 16, c = - 30

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} = 2

Therefore the parabola vertex is

(- 4, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- 4, 2)

x = - 5 y^{2}

y = 6 x^{2} - 15 x

f (x) = - (x - 5)^{2} - 4

y = \frac{x ^{2} 5 x 8}{x 3}

x^{2} - 4 x

What is the vertices f(x)=-2x^2-16x-30 ?
The vertices f(x)=-2x^2-16x-30 is Maximum (-4,2)