What is the vertices y=(2x-1)^2 ?

The vertices y=(2x-1)^2 is Minimum (1/2 ,0)

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vertices y=(2x-1)^2

vertices

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

Minimum (\frac{1}{2}, 0)

Solution steps

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

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

The parabola parameters are:

a = 4, m = \frac{1}{2}, n = \frac{1}{2}

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( \frac{1}{2} ) + ( \frac{1}{2} )}{2}

Simplify

\frac{( \frac{1}{2} ) + ( \frac{1}{2} )}{2} : \frac{1}{2}

x_{v} = \frac{1}{2}

Plug in

x_{v} = \frac{1}{2}

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(\frac{1}{2}, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (\frac{1}{2}, 0)

- 3 x^{2} - 6 x + 2

f (x) = 5 x^{2} - 10 x + 4

(y - 9)^{2} = - 4 (x - 2)

(x + 2)^{2} = - 8 (y + 3)

y = x^{2} + 3 x + 4