Answer
a. Zeros:$\quad -\displaystyle \frac{1}{2}, \ \ 2$(double)
b. see image
Work Step by Step
$P(x)= 2x^{3}-7x^{2}+4x+4,$
$a_{0}=4 \quad $p: $\pm 1,\pm 2,\pm 4$
$a_{n}=2,\qquad $q: $\pm 1,\pm 2$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try $\pm 1,\pm 2,$...
$\begin{array}{r|rrrrrrrrrrr}\hline
2 & 2 & -7 & 4 & 4 & & & \\
& & 4 & -6 & -4 & & & \\
\hline & 2 & -3 & -2 & 0 & & & \\
& & \swarrow & & & & & \\
2 & 2 & -3 & -2 & & & & \\
& & 4 & 2 & & & & \\
\hline & 2 & 1 & 0 & & & & \\
\end{array}$
$P(x)=(x-2)^{2}(2x+1)$
Zeros:$ -\displaystyle \frac{1}{2}, \ \ 2$(double)
$P(0)=4\qquad $(y intercept)
Calculate $P(0.5)=4.5$
to figure whether the graph turns before or after the y-intercept.
(Here, the graph rises through the y-intercept.)
The leading coefficient is positive, so the left far end is pointing down.
From left to right, the graph comes from far below x,
rising crosses x at (-1/2,0), and continues rising,
rises through the y-intercept at (0,6), turns,
falls to the next (double) zero at (2,0), which it touches and turns back up
and continues rising
