What is the vertices f(x)=6x^2-1 ?

The vertices f(x)=6x^2-1 is Minimum (0,-1)

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vertices f(x)=6x^2-1

vertices

f (x) = 6 x^{2} - 1

Minimum (0, - 1)

Solution steps

y = 6 x^{2} - 1

The parabola parameters are:

a = 6, b = 0, c = - 1

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

x_{v} = - \frac{0}{2 \cdot 6}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(0, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 6

Minimum (0, - 1)

f (x) = - 2 x^{2} + 16 x - 31

f (x) = - x^{2} - 6 x - 10

y = 2 x (x - 4) + 7

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

f (x) = x^{2} + 6 x - 8