Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=x^{1/3}(x-4)=x^{4/3}-4x^{1/3}$, we have $y'=\frac{4}{3}x^{1/3}-\frac{4}{3}x^{-2/3}$ and $y''=\frac{4}{9}x^{-2/3}+\frac{8}{9}x^{-5/3}$
Step 2. Letting $y'=0$, we have $\frac{4}{3}x^{1/3}-\frac{4}{3}x^{-2/3}=0$, which gives $x=1$. Also, $x=0$ is undefined in $y'$; thus $x=0,1$ are possible critical points. We have $y(0)=0, y(1)=-3$.
Step 3. Check signs of $y'$ to get $..(-)..(0)..(-)..(1)..(+)..$; thus the function decreases on $(-\infty,0)$, $(0,1)$ and increases on $(1,\infty)$ . This also means that $y(1)=-3$ is the only local minimum.
Step 4. Letting $y''=0$, we have $x=-2$ and we also need to consider $x=0$, where $y''$ is undefined. Check the signs of $y''$ to get $..(+)..(-2)..(-)..(0)..(+)..$, indicating $x=-2, 0$ are inflection points with the function is concave up for $x\lt-2$ or $x\gt0$ and concave down for $-2\lt x\lt0$.
Step 5. The end behaviors of the functions are $x\to-\infty, y\to\infty$ and $x\to\infty, y\to\infty$
Step 6. The x-intercept can be found by letting $y=0$, which gives $x=0,4$. The y-intercept can be found by letting $x=0$ which gives $y=0$.
Step 7. Graph the function based on the above information as shown in the figure.
