What is the vertices y=2x^2 ?

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

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vertices y=2x^2

vertices

y = 2 x^{2}

Minimum (0, 0)

Solution steps

y = 2 x^{2}

The parabola parameters are:

a = 2, b = 0, c = 0

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} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (0, 0)

x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

3 x^{2} - x - 2

f (x) = 4 x^{2} + 2

x^{2} + 3

(2 x - 1) (5 x - 6)