Answer
The graph, observing from left to right,
- falls on $(-\infty,-3)$, is concave up,
- at $(-3,1)$ reaches a local minimum, turns,
- starts increasing on the interval $(-3,-1),$ still concave up,
- at $(-2,5)$ changes concavity to "down" (inflection point), and continues rising to
- the point $(-1,5)$, a local maximum, from where
- it falls on the interval $(-1,\infty)$, concave down.
Work Step by Step
$y=f(x)=1-9x-6x^{2}-x^{3}$
$f'(x)=-9-12x-3x^{2}=-3(x^{2}+4x+3)=-3(x+3)(x+1)$
$f''(x)=-12-6x=-6(x+2)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=-3,-1\qquad$... (critical points)
Applying the 2nd derivative test,
$\left\{\begin{array}{lllll}
x=-3, & f''(0)=6\gt 0 & \Rightarrow(\min), & f(0)=1 & (-3,1)\\
x=-1, & f''(0)=-24 \lt 0 & \Rightarrow(\max), & f(-1)=5 & (-1,5)
\end{array}\right.$
$f''(x)=0$ for $x=-2,\qquad f(-2)=3,\quad(-2,3)$ is a point of inflection.
For graphing purposes:
The graph of $f'(x)$ is a parabola that opens down. The graph is above the x-axis between the zeros, and below for the other two intervals.
$\left[\begin{array}{cccccccccc}
f'(x): & -\infty & -- & -3 & ++ & -1 & -- & \infty\\
f(x): & & \searrow & | & \nearrow & | & \searrow &
\end{array}\right]$
Concavity and points of inflection:
$\left[\begin{array}{ccccccc}
f''(x): & -\infty & ++ & -2 & -- & \infty & \\
f(x): & & \cup & inf. & \cap & &
\end{array}\right]$
