What is the vertices y=3x^2-6x+4 ?

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

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vertices y=3x^2-6x+4

vertices

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

Minimum (1, 1)

Solution steps

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

The parabola parameters are:

a = 3, b = - 6, c = 4

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

x_{v} = - \frac{( - 6 )}{2 \cdot 3}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(1, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (1, 1)

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

x^{2} - x

y = - x^{2} + 6 x - 1

- (x - 30)^{2} + 40

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