Answer
(a) $x=\frac{1}{3}$ (multiplicity 2) and $x=1$ (multiplicity 3).
(b) $x=1$ crosses the x-axis, $x=\frac{1}{3}$ touches the x-axis.
(c) $4$.
(d) $y=x^5$.
Work Step by Step
(a) For $f(x)=(x-\frac{1}{3})^2(x-1)^3$, we can list real zero as $x=\frac{1}{3}$ (multiplicity 2) and $x=1$ (multiplicity 3).
(b) At $x=1$ the graph crosses the x-axis, while at $x=\frac{1}{3}$ the graph touches the x-axis.
(c) The maximum number of turning points on the graph is given by $n-1=5-1=4$.
(d) As $n=5, a_5\gt0$, the end behaviors are rise to the right and fall to the left, similar to $y=x^5$.