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)

y^{2} - 6 y = 8 x - 1

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

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

f (x) = x^{2} + 10 x + 16

f (x) = x^{2} - 10 x + 25