What is the vertices f(x)=3x^2-2 ?

The vertices f(x)=3x^2-2 is Minimum (0,-2)

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vertices f(x)=3x^2-2

vertices

f (x) = 3 x^{2} - 2

Minimum (0, - 2)

Solution steps

y = 3 x^{2} - 2

The parabola parameters are:

a = 3, b = 0, c = - 2

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

x_{v} = - \frac{0}{2 \cdot 3}

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 = 3

Minimum (0, - 2)

- x^{2} - 6 x - 9

f (x) = x^{2} - 5

y = x^{2} + 3 x - 5

y = x^{2} + 3 x + 2

y = 4 x^{2} - 16 x + 22