Answer
General shape:
Work Step by Step
Step 2
$y'=-(x^{2}-x+2)=-(x-2)(x+1)$
$y''=-(2x-1)$
Step 3
$y'=0$ when $ x=-1,2\qquad$... critical points.
Step 4
The graph of $y'$ is a parabola that opens down, so it is positive between the zeros, and negative on the other two intervals.
$ \left[\begin{array}{llllllll}
y': & & -- & | & ++ & | & -- & \\
& (-\infty & & -1 & & 2 & & \infty)\\
y: & & \searrow & & \nearrow & & \searrow &
\end{array}\right]$
Step 5
For concavity,
$y''=-(2x-1)$,
$y''=0$ for $x=\displaystyle \frac{1}{2}$
$ \left[\begin{array}{ccccc}
y': & & ++ & | & -- & \\
& (-\infty & & 1/2 & & \infty)\\
y: & & \cup & inf l.& \cap &
\end{array}\right]$
