What is the vertices (40x^3)/(3x) ?

The vertices (40x^3)/(3x) is Minimum (0,0)

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vertices (40x^3)/(3x)

vertices

\frac{40 x ^{3}}{3 x}

Minimum (0, 0)

Solution steps

y = \frac{40 x ^{3}}{3 x}

Rewrite

y = \frac{40 x ^{3}}{3 x}

in the form

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

y = \frac{40 x ^{2}}{3}

The parabola parameters are:

a = \frac{40}{3}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( \frac{40}{3} )}

Simplify

x_{v} = 0

Plug in

x_{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 = \frac{40}{3}

Minimum (0, 0)

f (x) = 3 x^{2} + 6 x + 1

y = x^{2} + 6 x + 5

f (x) = 3 x^{2} + 6 x - 1

t^{2} + 2

y = - x^{2} - 4 x + 1