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

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

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vertices y=x^2-1

vertices

y = x^{2} - 1

Minimum (0, - 1)

Solution steps

y = x^{2} - 1

The parabola parameters are:

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

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} = - 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 = 1

Minimum (0, - 1)

+ \frac{- 1}{2} x^{2} + 18

x^{2} - 4 x + 12

y = x^{2} - x

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

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