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

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

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vertices y=2x^2-2

vertices

y = 2 x^{2} - 2

Minimum (0, - 2)

Solution steps

y = 2 x^{2} - 2

The parabola parameters are:

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

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

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

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 = 2

Minimum (0, - 2)

f (x) = \frac{2}{3} x - 5

\frac{2 x}{x ^{2} + 81}

\frac{3 x + 8}{2 x - 3}

lo g_{2} (x)

f (x) = \frac{1}{x + 2} - 1