What is the vertices f(x)=-x^2-8x ?

The vertices f(x)=-x^2-8x is Maximum (-4,16)

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vertices f(x)=-x^2-8x

vertices

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

Maximum (- 4, 16)

Solution steps

y = - x^{2} - 8 x

The parabola parameters are:

a = - 1, b = - 8, c = 0

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

x_{v} = - \frac{( - 8 )}{2 ( - 1 )}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = 16

Therefore the parabola vertex is

(- 4, 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 4, 16)

y = x^{2} - 2 x + 5

y = x^{2} - 2 x + 1

x^{2} - y = 0

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

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

What is the vertices f(x)=-x^2-8x ?
The vertices f(x)=-x^2-8x is Maximum (-4,16)