Answer
See graph and explanations.
Work Step by Step
a. As $x\to 0^+, y\to0$ because $\lim_{x\to 0^+}\frac{3}{2}(\frac{x}{x-1})^{2/3}=\frac{3}{2}(\frac{0}{0-1})^{2/3}=0$;
b. As $x\to\pm\infty, y\to3/2$ (asymptote $y=3/2$) because $\lim_{x\to\pm\infty}\frac{3}{2}(\frac{x}{x-1})^{2/3}=\lim_{x\to\pm\infty}\frac{3}{2}(\frac{1}{1-1/x})^{2/3}=\frac{3}{2}(\frac{1}{1-0})^{2/3}=\frac{3}{2}$;
c. As $x\to 1, y\to\infty$ (asymptote $x=1$) because $\lim_{x\to 1}\frac{3}{2}(\frac{x}{x-1})^{2/3}=\frac{3}{2}(\frac{1}{1-1})^{2/3}=\infty$;
As $x\to -1, y\to\frac{3}{2\sqrt[3] 4}$ because $\lim_{x\to -1}\frac{3}{2}(\frac{x}{x-1})^{2/3}=\frac{3}{2}(\frac{-1}{-1-1})^{2/3}=\frac{3}{2\sqrt[3] 4}$;
