Answer
a. Zeros:$ \displaystyle \frac{1}{3}, \ \ 3$(double)
b. see image
Work Step by Step
$P(x)= 3x^{3}+17x^{2}+21x-9,$
$a_{0}=9 \quad $p: $\pm 1,\pm 3,\pm 9,\qquad $
$a_{n}=3,\qquad $q: $\pm 1,\pm 3$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$with synthetic division, we try $\pm 1,\pm 3,$...
$\begin{array}{r|rrrrrrrrrrr}\hline
-3 & 3 & 17 & 21 & -9 & & & \\
& & -9 & -24 & 9 & & & \\
\hline & 3 & 8 & -3 & 0 & & & \\
& & \swarrow & & & & & \\
-3 & 3 & 8 & -3 & & & & \\
& & -9 & 3 & & & & \\
\hline & 3 & -1 & 0 & & & & \\
\end{array}$
$P(x)=(x-3)^{2}(3x-1)$
Zeros:$ \displaystyle \frac{1}{3}, \ \ 3$(double)
$P(0)=-9\qquad $(y intercept)
Here, the graph rises through the y-intercept, as it has not yet reached any zero.
The leading coefficient is positive, so the left far end is pointing down.
From left to right, the graph comes from far below x,
rises through the y-intercept at (0,6),
crosses x at (1/3,0), rising, turns and ,
falls to the next (double) zero at (3,0), which it touches and turns back up
from where it continues rising
