What is the vertices f(x)=x^2-4 ?

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

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vertices f(x)=x^2-4

vertices

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

Minimum (0, - 4)

Solution steps

y = x^{2} - 4

The parabola parameters are:

a = 1, b = 0, c = - 4

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} = - 4

Therefore the parabola vertex is

(0, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 4)

- x^{2} + 9 y^{2} + 4 x + 54 y + 68 = 0

x = - \frac{1}{8} y^{2}

4 x^{2} - y^{2} = 16

x^{2} + y^{2} - 4 x - 4 y = 0

x^{2} = - 4 y