Question

Sketch the graph of y=e^(2x-x^2). Include an analysis going through various aspects of the function such as domain, symmetry, asymptotes, etc.

I thought I was to take the first derivative and set it to zero to find my critical numbers first. So I got f ' (x) = (2-2x)e^(2x-x^2)
When I set it to zero is where I'm lost.
If I divide by (2-2x) I have 0=e^(2x-x^2) and then if I take the ln of each side to bring down the exponent I have an undefined on the left side, right?

Answer 1

Set y=0 for roots --> there exists no real value of x for which the equation may be solved (i.e. there are no roots for this question). --> y=0 is a horizontal asymptote for the graph of this function.
Set x=0 for y-intercept --> y=1
Since e^a > 0 (for all a in the real number domain), y>0.
When x=2, y=1.
f'(x)=(2-2x)e^(2x-(x^2)) --> x=1 is a turning point (max) [verify that f''(x)<0].
max turning point occurs at (1, e).
For x>2, y descends exponentially (since x^2 rises quicker than 2x).
For 1<x<2, y rises exponentially as x --> 1^+.
For 0<x<1, y dips exponentially as x --> 0^+.
For -1<x<0, y rises exponentially as x --> (-1)^+.
For -2<x<-1, y dips exponentially as x --> (-2)^+.
For x<-2, y dips exponentially as x --> -infinity.

So the graph resembles the following:



It's an asymmetric curve, which is positively-skewed.