What is the vertices y=3x^2-12x+8 ?

The vertices y=3x^2-12x+8 is Minimum (2,-4)

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vertices y=3x^2-12x+8

vertices

y = 3 x^{2} - 12 x + 8

Minimum (2, - 4)

Solution steps

y = 3 x^{2} - 12 x + 8

The parabola parameters are:

a = 3, b = - 12, c = 8

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

x_{v} = - \frac{( - 12 )}{2 \cdot 3}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(2, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (2, - 4)

f (x) = - x^{2} - 4 x + 1

y = x^{2} - 4 x + 7

x = (y - 3)^{2}

y^{2} = x + 1

f (x) = - x^{2} + 8 x - 12