What is the vertices x^2-16 ?

The vertices x^2-16 is Minimum (0,-16)

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vertices x^2-16

vertices

x^{2} - 16

Minimum (0, - 16)

Solution steps

y = x^{2} - 16

The parabola parameters are:

a = 1, b = 0, c = - 16

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} = - 16

Therefore the parabola vertex is

(0, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 16)

2 x^{2} + 2 x - 4

x^{2} + 2 x + 3

y = - x^{2}

x^{2} - 4 x + 4

t^{2}