What is the vertices y=4x^2-16 ?

The vertices y=4x^2-16 is Minimum (0,-16)

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vertices y=4x^2-16

vertices

y = 4 x^{2} - 16

Minimum (0, - 16)

Solution steps

y = 4 x^{2} - 16

The parabola parameters are:

a = 4, b = 0, c = - 16

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

x_{v} = - \frac{0}{2 \cdot 4}

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 = 4

Minimum (0, - 16)

f (x) = x^{2} - 8 x + 17

y = 24000 x^{2} + 900 x - 150

16 - (12 x + 8)^{2}

y^{2} - 8 x - 6 y - 23 = 0

g (x) = (x + 3)^{2} + 5