Answer
(a) $y = 0$ is a horizontal asymptote.
(b) $f$ is decreasing on the interval $(0, \infty)$
$f$ is increasing on the interval $(-\infty,0)$
(c) The local maximum is $(0,1)$
(d) The graph is concave down on this interval: $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$
The graph is concave up on this interval: $(-\infty,-\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2}, \infty)$
The points of inflection are $(-\frac{\sqrt{2}}{2}, \frac{1}{\sqrt{e}})$ and $(\frac{\sqrt{2}}{2}, \frac{1}{\sqrt{e}})$
(e) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = e^{-x^2}$
$f(x)$ is defined for all values of $x$
There are no vertical asymptotes.
$\lim\limits_{x \to -\infty}f(x) = 0$
$\lim\limits_{x \to \infty}f(x) = 0$
$y = 0$ is a horizontal asymptote.
(b) We can find the points where $f'(x) = 0$:
$f'(x) = -2x~e^{-x^2} = 0$
$x = 0$
When $x \gt 0$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(0, \infty)$
When $x \lt 0$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(-\infty,0)$
(c) $f(0) = e^{-(0)^2} = 1$
The local maximum is $(0,1)$
There is no local minimum.
(d) We can find the points where $f''(x) = 0$:
$f''(x) = -2~e^{-x^2}+(-2x)(-2x)e^{-x^2}$
$f''(x) = -2~e^{-x^2}+4x^2e^{-x^2}$
$f''(x) = (4x^2-2)e^{-x^2} = 0$
$4x^2-2 = 0$
$x^2 = \frac{1}{2}$
$x = \pm \frac{\sqrt{2}}{2}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(-\infty,-\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2}, \infty)$
$f(-\frac{\sqrt{2}}{2}) = e^{-(-\frac{\sqrt{2}}{2})^2} = \frac{1}{\sqrt{e}}$
$f(\frac{\sqrt{2}}{2}) = e^{-(\frac{\sqrt{2}}{2})^2} = \frac{1}{\sqrt{e}}$
The points of inflection are $(-\frac{\sqrt{2}}{2}, \frac{1}{\sqrt{e}})$ and $(\frac{\sqrt{2}}{2}, \frac{1}{\sqrt{e}})$
(e) We can see a sketch of the graph below.