What is the vertices y=500(0.04-x^2) ?

The vertices y=500(0.04-x^2) is Maximum (0,20)

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vertices y=500(0.04-x^2)

vertices

y = 500 (0.04 - x^{2})

Maximum (0, 20)

Solution steps

y = 500 (0.04 - x^{2})

Rewrite

y = 500 (0.04 - x^{2})

in the form

y = a x^{2} + b x + c

y = - 500 x^{2} + 20

The parabola parameters are:

a = - 500, b = 0, c = 20

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

x_{v} = - \frac{0}{2 ( - 500 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 20

Therefore the parabola vertex is

(0, 20)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 500

Maximum (0, 20)

- 5 x^{2} - 40 x - 77

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

y = 5 x^{2} - 2 x - 1

f (x) = x^{2} - 10 x + 21

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

What is the vertices y=500(0.04-x^2) ?
The vertices y=500(0.04-x^2) is Maximum (0,20)