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

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

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

vertices

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

Maximum (0, - 1)

Solution steps

y = - x^{2} - 1

The parabola parameters are:

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

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

x_{v} = - \frac{0}{2 ( - 1 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(0, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (0, - 1)

- x^{2} - x + 2

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

y = + 16 - (20 x + 15)^{2}

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

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