What is the vertices y=18x^2-14x+9 ?

The vertices y=18x^2-14x+9 is Minimum (7/18 , 113/18)

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vertices y=18x^2-14x+9

vertices

y = 18 x^{2} - 14 x + 9

Minimum (\frac{7}{18}, \frac{113}{18})

Solution steps

y = 18 x^{2} - 14 x + 9

The parabola parameters are:

a = 18, b = - 14, c = 9

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

x_{v} = - \frac{( - 14 )}{2 \cdot 18}

Simplify

- \frac{- 14}{2 \cdot 18} : \frac{7}{18}

x_{v} = \frac{7}{18}

Plug in

x_{v} = \frac{7}{18}

to find the

y_{v}

value

y_{v} = \frac{113}{18}

Therefore the parabola vertex is

(\frac{7}{18}, \frac{113}{18})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 18

Minimum (\frac{7}{18}, \frac{113}{18})

f (x) = - 3 x^{2} + 18 x - 8

y = x^{2} - 6 x + 8

y = (x - 1)^{2} - 2

y = - 4 x^{2} - 12 x - 8

y = + x^{2} + 8 x + 9

What is the vertices y=18x^2-14x+9 ?
The vertices y=18x^2-14x+9 is Minimum (7/18 , 113/18)