What is the vertices f(x)=x^2-8 ?

The vertices f(x)=x^2-8 is Minimum (0,-8)

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vertices f(x)=x^2-8

vertices

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

Minimum (0, - 8)

Solution steps

y = x^{2} - 8

The parabola parameters are:

a = 1, b = 0, c = - 8

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} = - 8

Therefore the parabola vertex is

(0, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 8)

x^{2} - 24 y + 48 = 0

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

3 x^{2} + 8 y = 0

4 x^{2} - 9 y^{2} = 36