What is the vertices y=(6x)/7*(4x)/3 ?

The vertices y=(6x)/7*(4x)/3 is Minimum (0,0)

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vertices y=(6x)/7*(4x)/3

vertices

y = \frac{6 x}{7} \cdot \frac{4 x}{3}

Minimum (0, 0)

Solution steps

y = \frac{6 x}{7} \cdot \frac{4 x}{3}

Rewrite

y = \frac{6 x}{7} \cdot \frac{4 x}{3}

in the form

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

y = \frac{8 x ^{2}}{7}

The parabola parameters are:

a = \frac{8}{7}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( \frac{8}{7} )}

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{8}{7}

Minimum (0, 0)

y^{2} - 6 y - 4 x + 5 = 0

5 x (2 x + 4) - 3 x

f (x) = - 4 x^{2} - 12 x + 5

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

y = - 2 x^{2} + 5

What is the vertices y=(6x)/7*(4x)/3 ?
The vertices y=(6x)/7*(4x)/3 is Minimum (0,0)