Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 49

(a) there is no vertical asymptotes, f is defined and differentiable everywhere.
When x->+-oo, -x^2->-oo and f(x)->+0, so there is one horizontal asymptote y=0.

(b) f'(x) = -2x*e^(-x^2). It is >0 for x<0 and <0 for x>0,
so f(x) is increases on (-oo, 0) and decreases on (0, +oo).

(c) therefore there is one local maximum x=0. f(0)=1.

(d) f''(x) = e^(-x^2)*(-2 + 4x^2).
This is <0 for x on (-1/sqrt(2), 1/sqrt(2)) , f is concave downward there. And f''>0 on (-oo, -1/sqrt(2)) and on (1/sqrt(2), +oo), f is concave upward there.

(e)