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

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

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vertices f(x)=x^2-2

vertices

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

Minimum (0, - 2)

Solution steps

y = x^{2} - 2

The parabola parameters are:

a = 1, b = 0, c = - 2

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} = - 2

Therefore the parabola vertex is

(0, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 2)

x^{2} + y^{2} - 9 = 0

y^{2} = 20 x

\frac{x ^{2}}{49} + \frac{y ^{2}}{16} = 1

\frac{( y - 6 ) ^{2}}{64} - \frac{( x - 8 ) ^{2}}{36} = 1

9 x^{2} - 9 y^{2} + 18 x + 36 y = 63