What is the vertices 2pir^2 ?

The vertices 2pir^2 is Minimum (0,0)

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vertices 2pir^2

vertices

2 π r^{2}

Minimum (0, 0)

Solution steps

y = 2 π r^{2}

The parabola parameters are:

a = 2 π, b = 0, c = 0

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

r_{v} = - \frac{0}{2 \cdot 2 π}

Simplify

r_{v} = 0

Plug in

r_{v} = 0

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2 π

Minimum (0, 0)

f (x) = - 3 x^{2} + 6 x

y = 14 (x - 4) (x + 10)

x^{2} + 8 x - 80

f (x) = x^{2} - 6 x - 16

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