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

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

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vertices f(x)= 1/2 (x-1)^2

vertices

f (x) = \frac{1}{2} (x - 1)^{2}

Minimum (1, 0)

Solution steps

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

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

The parabola parameters are:

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

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{2}

Minimum (1, 0)

x^{2} + 1

2 x^{2} - 5 x + 3

2 X^{2} + 3 X - 20

2 X^{2}

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

What is the vertices f(x)= 1/2 (x-1)^2 ?
The vertices f(x)= 1/2 (x-1)^2 is Minimum (1,0)