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}

Rewrite

y = (2 x)^{2}

in the form

y = a x^{2} + b x + c

y = 4 x^{2}

The parabola parameters are:

a = 4, b = 0, c = 0

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

x_{v} = - \frac{0}{2 \cdot 4}

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 = 4

Minimum (0, 0)

y = 2 x^{2} - 19 x + 9

4 x^{2} - 20 x - 24 y + 97 = 0

y - 2 = (x - 1)^{2}

y = (x - 5)^{2}

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