What is the vertices y=6x^2-12x+1 ?

The vertices y=6x^2-12x+1 is Minimum (1,-5)

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vertices y=6x^2-12x+1

vertices

y = 6 x^{2} - 12 x + 1

Minimum (1, - 5)

Solution steps

y = 6 x^{2} - 12 x + 1

The parabola parameters are:

a = 6, b = - 12, c = 1

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

x_{v} = - \frac{( - 12 )}{2 \cdot 6}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 5

Therefore the parabola vertex is

(1, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 6

Minimum (1, - 5)