Answer
$(a)$
Local maximum: $(-3,0)$
Local minimum: $(-4,-1)$
$(b)$
Absolute maximum: $(-3,0)$
Absolute minimum: none.
$(c)$
See graph.
Work Step by Step
$g(x)=-x^{2}-6x-9,\quad x\in[-4,\infty)$
$g'(x)=-2x-6=-2(x+3)$
$g$ and $g'$ are defined on $[-4,\infty)$
$g'(x)=0$ for $ x=-3\qquad$ ... critical point.
$g'(-3.5)=1\gt 0$
$g'(0)=-6\lt 0$
$g(1)=1,\quad g(2)=0$
$g':\quad \begin{array}{llllll}
-4 & & -3 & & \infty & \\
[ & ++ & | & -- & ) & \\
\hline & \nearrow & 0 & \searrow & & \\
-1 & & & & & \\
& & & & &
\end{array}$
