What is the vertices y=3x^2-3x-6 ?

The vertices y=3x^2-3x-6 is Minimum (1/2 ,-27/4)

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vertices y=3x^2-3x-6

vertices

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

Minimum (\frac{1}{2}, - \frac{27}{4})

Solution steps

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

The parabola parameters are:

a = 3, b = - 3, c = - 6

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

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

Simplify

- \frac{- 3}{2 \cdot 3} : \frac{1}{2}

x_{v} = \frac{1}{2}

Plug in

x_{v} = \frac{1}{2}

to find the

y_{v}

value

y_{v} = - \frac{27}{4}

Therefore the parabola vertex is

(\frac{1}{2}, - \frac{27}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (\frac{1}{2}, - \frac{27}{4})

f (x) = x^{2} - 3 x + 10

- 2 x^{2} + 12 x - 20

y^{2} + 2 x + 6 y + 17 = 0

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

- (x + 2)^{2} + 6

What is the vertices y=3x^2-3x-6 ?
The vertices y=3x^2-3x-6 is Minimum (1/2 ,-27/4)