#### Answer

A. The domain is $(-\infty, \infty)$
B. The y-intercept is $0$
The x-intercepts are $0$ and $\sqrt[3] 4$
C. The function is not odd or even.
D. $\lim\limits_{x \to -\infty} (x^4-4x) = \infty$
$\lim\limits_{x \to \infty} (x^4-4x) = \infty$
There are no asymptotes.
E. The function is increasing on the intervals $(1, \infty)$
The function is decreasing on the interval $(-\infty,1)$
F. $(1,-3)$ is a local minimum.
G. The graph is concave up on the intervals $(-\infty, 0) \cup (0, \infty)$
There is no point of inflection.
H. We can see a sketch of the curve below.

#### Work Step by Step

$y = x^4-4x$
A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = (0)^4-4(0) = 0$
The y-intercept is $0$
When $y = 0$:
$x^4+4x = 0$
$x(x^3-4) = 0$
$x = 0$ or $x = \sqrt[3] 4$
The x-intercepts are $0$ and $\sqrt[3] 4$
C. The function is not odd or even.
D. $\lim\limits_{x \to -\infty} (x^4-4x) = \infty$
$\lim\limits_{x \to \infty} (x^4-4x) = \lim\limits_{x \to \infty} (x)(x^3-4)=\infty$
There are no asymptotes.
E. We can find the values of $x$ such that $y' = 0$:
$y' = 4x^3-4 = 0$
$x^3-1 = 0$
$x = 1$
The function is increasing on the intervals $(1, \infty)$
The function is decreasing on the interval $(-\infty,1)$
F. When $x = 1$, then $y = (1)^4-4(1)= -3$
$(1,-3)$ is a local minimum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = 12x^2 = 0$
$x = 0$
When $x \lt 0$ or $x \gt 0$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty, 0) \cup (0, \infty)$
There is no point of inflection since the concavity does not change at $x=0$
H. We can see a sketch of the curve below.