Answer
(a) $x=\frac{1}{3}$ (multiplicity 2), $x=1$ (multiplicity 3).
(b) crosses the x-axis at $x=1$, touches at $x=\frac{1}{3}$.
(c) $4$.
(d) $y=x^5$.
Work Step by Step
(a) Given $f(x)=(x-\frac{1}{3})^2(x-1)^3$, we can find zero(s) $x=\frac{1}{3}$ (multiplicity 2), $x=1$ (multiplicity 3).
(b) The graph crosses the x-axis at $x=1$, touches at $x=\frac{1}{3}$.
(c) The maximum number of turning points on the graph is $n-1=5-1=4$.
(d) The end behavior is similar to those of $y=x^5$.
