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

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

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

vertices

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

Minimum (0, - 3)

Solution steps

y = 2 x^{2} - 3

The parabola parameters are:

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

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} = - 3

Therefore the parabola vertex is

(0, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (0, - 3)

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

3 x^{2} - x + 5

y = x^{2} - 8 x + 17

y = 2 x^{2} - 4 x + 15

4 (x + 6) + 3 (2 + x + 3 x^{2}) - 2 x^{2}