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

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

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vertices f(x)=x^2-6

vertices

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

Minimum (0, - 6)

Solution steps

y = x^{2} - 6

The parabola parameters are:

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

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

x_{v} = - \frac{0}{2 \cdot 1}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 6

Therefore the parabola vertex is

(0, - 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 6)

x^{2} + 5 x - 1

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

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

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

2 x^{2} + 5 x - 3