What is the vertices f(x)=-x^2-2x-1 ?

The vertices f(x)=-x^2-2x-1 is Maximum (-1,0)

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vertices f(x)=-x^2-2x-1

vertices

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

Maximum (- 1, 0)

Solution steps

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

The parabola parameters are:

a = - 1, b = - 2, c = - 1

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 1, 0)

f (x) = - (x + 3) (x + 10)

y^{2} = - 32 x

y = 2 x^{2} + 16 x + 29

f (x) = - (x^{2} + 3 x - 4)

y = 4 x^{2} - 16

What is the vertices f(x)=-x^2-2x-1 ?
The vertices f(x)=-x^2-2x-1 is Maximum (-1,0)