`X^2-4x+5`

We know that;

`(x-2)^2 = x^2-4x+4`

`(x-2)^2+1 = x^2-4x+4+1`

`(x-2)^2+1 = x^2-4x+5`

`y=(x^2-4x+5)`

`y=(x-2)^2+1`

We know that `(x-2)>= 0` always .So the minimum of y will be obtained when `(x-2) =0`

`(x-2) = 0`

`X=2`

Minimum `y=(x-2)^2+1=(2-2)^2+1=1`

*So the required answers are;*

`x^2-4x+5 = (x-2)^2+1` **minimum point on curve is 1.**

`x^2 - 4x + 5`

To express in (x + a)^2 + b form we need to complete the square.

`(x^2 - 4x +c)^2 + 5` To find c, divide -4 by 2 and square it.

`(x^2 - 4x +4) + 5-4`

`(x - 2)^2 + 1`

Hence in the form (a+x)^2 + b **we have**: `(x-2)^2 + 1`

This makes **(2, 1) the minimum** point.

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