What is the vertices y=3x^2+6x-1 ?

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

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vertices y=3x^2+6x-1

vertices

y = 3 x^{2} + 6 x - 1

Minimum (- 1, - 4)

Solution steps

y = 3 x^{2} + 6 x - 1

The parabola parameters are:

a = 3, b = 6, c = - 1

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} = - 4

Therefore the parabola vertex is

(- 1, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- 1, - 4)

(8 x - 23)^{2} - 10

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

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

f (x) = - 6 x^{2} - 6 x - 6

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