What is the vertices 3x^2-3x ?

The vertices 3x^2-3x is Minimum (1/2 ,-3/4)

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vertices 3x^2-3x

vertices

3 x^{2} - 3 x

Minimum (\frac{1}{2}, - \frac{3}{4})

Solution steps

y = 3 x^{2} - 3 x

The parabola parameters are:

a = 3, b = - 3, c = 0

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{3}{4}

Therefore the parabola vertex is

(\frac{1}{2}, - \frac{3}{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{3}{4})

y = 2 x^{2} + 12 x + 17

4 x^{2} + 8 x + 4

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

3 x^{2} + 6 x + 2

2 x^{2} - 3 x