What is the vertices-x^2-4 ?

The vertices-x^2-4 is Maximum (0,-4)

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vertices-x^2-4

vertices

- x^{2} - 4

Maximum (0, - 4)

Solution steps

y = - x^{2} - 4

The parabola parameters are:

a = - 1, b = 0, c = - 4

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} = - 4

Therefore the parabola vertex is

(0, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (0, - 4)

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

y = 2 x^{2} - 8 x + 13

f (x) = - 2 x^{2} - 12 x - 16

f (x) = (x + 4)^{2} + 1

f (x) = 2 x^{2} + 8 x + 9