What is the vertices 100(1-x/(40))^2 ?

The vertices 100(1-x/(40))^2 is Minimum (40,0)

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vertices 100(1-x/(40))^2

vertices

100 (1 - \frac{x}{40})^{2}

Minimum (40, 0)

Solution steps

y = 100 (1 - \frac{x}{40})^{2}

y = 100 (1 - \frac{x}{40})^{2}

The parabola parameters are:

a = \frac{1}{16}, m = 40, n = 40

x_{v} = \frac{m + n}{2}

x_{v} = \frac{40 + 40}{2}

Simplify

x_{v} = 40

Plug in

x_{v} = 40

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(40, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{16}

Minimum (40, 0)

y = x^{2} + 12 x + 16

f (x) = - 5 x^{2} + 20 x + 60

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

f (x) = - x^{2} - 14 x - 42

y = 2 x^{2} + 4 x + 6

What is the vertices 100(1-x/(40))^2 ?
The vertices 100(1-x/(40))^2 is Minimum (40,0)