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

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

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

vertices

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

Minimum (2, 0)

Solution steps

y = (x - 2)^{2}

y = (x - 2)^{2}

The parabola parameters are:

a = 1, m = 2, n = 2

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

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

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(2, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 0)

y = 4 x^{2} - 12 x + 10

x^{2} - 12 x + 16 y + 68 = 0

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

f (x) = x^{2} - 4 x + 7

y = x^{2} - 6 x + 10