What is the vertices y=(9x-2)x3 ?

The vertices y=(9x-2)x3 is Minimum (1/9 ,-1/3)

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vertices y=(9x-2)x3

vertices

y = (9 x - 2) x 3

Minimum (\frac{1}{9}, - \frac{1}{3})

Solution steps

y = (9 x - 2) x \cdot 3

y = (9 x - 2) x \cdot 3

The parabola parameters are:

a = 27, m = \frac{2}{9}, n = 0

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

x_{v} = \frac{( \frac{2}{9} ) + 0}{2}

Simplify

\frac{( \frac{2}{9} ) + 0}{2} : \frac{1}{9}

x_{v} = \frac{1}{9}

Plug in

x_{v} = \frac{1}{9}

to find the

y_{v}

value

y_{v} = - \frac{1}{3}

Therefore the parabola vertex is

(\frac{1}{9}, - \frac{1}{3})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 27

Minimum (\frac{1}{9}, - \frac{1}{3})

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

f (x) = x^{2} - 4 x + 1

y = 3 x^{2} + 5 x - 4

y = x^{2} + 4

x^{2} - 6 x + 5

What is the vertices y=(9x-2)x3 ?
The vertices y=(9x-2)x3 is Minimum (1/9 ,-1/3)