# What are the steps to draw the graph of this function : f(x)=e^-(x^2-2*x+1/2) ?

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).`

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