Answer
a. Zeros:$ -3, -1,2$
b. see image
Work Step by Step
$P(x)= -x^{3}-2x^{2}+5x+6,$
$a_{0}=6 \quad $p: $\pm 1,\pm 2,\pm 3,\pm 6$
$a_{n}=-1,\qquad $q: $\pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try $\pm 1,\pm 2,\pm 3$...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1 & -1 & -2 & 5 & 6 & & & \\
& & 1 & 1 & -6 & & & \\
\hline & -1 & -1 & 6 & 0 & & & \\
& & \swarrow & & & & & \\
2 & -1 & -1 & 6 & & & & \\
& & -2 & -6 & & & & \\
\hline & -1 & -3 & 0 & & & & \\
\end{array}$
$P(x)=(x+1)(x-2)(-x-3)=-(x+1)(x-2)(x+3)$
Zeros:$ -3, -1,2$
$P(0)=6\qquad $(y intercept)
Calculate $P(-0.5)=3.125$ and $P(0.5)=7.875$
to figure whether the graph turns before or after the y-intercept.
(Here, the graph rises through the y-intercept.)
The leading coefficient is negative, so the left far end is pointing up.
From left to right, the graph comes from far above x,
falling, crosses x at (-3,0), dives down, turns,
crosses x again at (-1,0),
rises through the y-intercept at (0,6), turns,
falls to the next zero at (2,0),
from where it continues falling.