You need to evaluate the asymptotes of the given function, such that:

`lim_(x->+-oo) e^(-(x^2 - 2x + 1/2)) = lim_(x->oo) 1/(e^(x^2 - 2x + 1/2))`

`lim_(x->+-oo) 1/(e^(x^2 - 2x + 1/2)) = 1/oo = 0`

**Since evaluating the limit of the function to `+-oo` yields 0, hence, there exists horizontal asymptotes, `y = 0` , of the graph of function to `+-oo.` **

**Since there exists horizontal asymptotes, then there are no oblique asymptotes.**

**Since there are no restrictions to domain of the function, there are no vertical asymptotes.**

You need to evaluate the first derivative of the function to test if there are extreme points, such that:

`f'(x) = e^( x^2 - 2x + 0.5 )*((x^2 - 2x + 1/2)')/e^(2x^2 - 4x + 1)`

`f'(x) = (e^(x^2 - 2x + 1/2)*(2x - 2))/e^(2x^2 - 4x + 1)`

You need to solve for x the equation `f'(x) = 0` , such that:

`(e^(x^2 - 2x + 1/2)*(2x - 2))/e^(2x^2 - 4x + 1) = 0 => 2x - 2 = 0 => x = 1`

Since `f'(x) < 0` for `x in (-oo,1)` and `f'(x) > 0` for `x in (1,oo),` hence, the function reaches its minimum at `(1,sqrt e).`