Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=x-3x^{2/3}$, we have $y'=1-2x^{-1/3}$ and $y''=\frac{2}{3}x^{-4/3}$
Step 2. Letting $y'=0$, we have $1-2x^{-1/3}=0$, which gives $x=8$. Also $x=0$ is undefined in $y'$; thus $x=0,8$ are possible critical points. We have $y(0)=0, y(8)=-4$.
Step 3. Check signs of $y'$ to get $..(+)..(0)..(-)..(8)..(+)..$; thus the function increases on $(-\infty,0)$, $(8,\infty)$ and decreases on $(0,8)$ . This also means that $y(0)=0$ is a local maximum and $y(8)=-4$ is a local minimum.
Step 4. Since $y''\ne0$, there are no inflection points.
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 $x=0,27$. 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.
