Answer
See graph
Work Step by Step
$y$ = $f(x)$ = $x^{4}-4x$ = $x(x^{3}-4)$
$A.$ $f$ is a polynomial so $D$ = $R$
$B.$ $y$-intercept = $f(0)$ = $0$, $x$-intercepts are $0$ and $\sqrt[3] 4$
$C.$ No symmetry
$D.$ No asympote
$E.$ $f'(x)$ = $4x^{3}-4$ = $4(x^{3}-1)$ = $4(x-1)(x^{2}+x+1)$ $\gt$ => $x$ $\gt$ $1$ so $f$ is increasing on $(1,\infty)$ and decreasing on $(-\infty,1)$
$F.$ No local maximum value , local minimum value $f(1)$ = $-3$
$G.$ $f''(x)$ = $12x^{2}$ $\gt$ $0$ =for all $x$, so $f$ is concave upward on $(-\infty,\infty)$
No inflection point
