What is the vertices y=18x^2-6x+4 ?

The vertices y=18x^2-6x+4 is Minimum (1/6 , 7/2)

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vertices y=18x^2-6x+4

vertices

y = 18 x^{2} - 6 x + 4

Minimum (\frac{1}{6}, \frac{7}{2})

Solution steps

y = 18 x^{2} - 6 x + 4

The parabola parameters are:

a = 18, b = - 6, c = 4

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

x_{v} = - \frac{( - 6 )}{2 \cdot 18}

Simplify

- \frac{- 6}{2 \cdot 18} : \frac{1}{6}

x_{v} = \frac{1}{6}

Plug in

x_{v} = \frac{1}{6}

to find the

y_{v}

value

y_{v} = \frac{7}{2}

Therefore the parabola vertex is

(\frac{1}{6}, \frac{7}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 18

Minimum (\frac{1}{6}, \frac{7}{2})

f (x) = - x^{2} + 14 x - 48

f (x) = x^{2} + 11 x - 35

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

y = 3 x^{2} - 6 x + 4

y = 3 x^{2} + 2 x - 1

What is the vertices y=18x^2-6x+4 ?
The vertices y=18x^2-6x+4 is Minimum (1/6 , 7/2)