we have
[tex]f(x)=-2(x+3)^{2}-1[/tex]
we know that
the equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
[tex](h,k)[/tex] is the vertex
If [tex]a > 0[/tex] ------> then the parabola open upward (vertex is a minimum)
If [tex]a < 0[/tex] ------> then the parabola open downward (vertex is a maximum)
In this problem
the vertex is the point [tex](-3,-1)[/tex]
[tex]a=-2[/tex]
so
[tex]-2 < 0[/tex] ------> then the parabola open downward (vertex is a maximum)
The domain is the interval-------> (-∞,∞)
that means------> all real numbers
The range is the interval--------> (-∞, -1]
[tex]y\leq-1[/tex]
that means
all real numbers less than or equal to [tex]-1[/tex]
therefore
the answer is
a) the vertex is the point [tex](-3,-1)[/tex]
b) the domain is all real numbers
c) the range is [tex]y\leq-1[/tex]
see the attached figure to better understand the problem
