we have the function
[tex]f(x)=-2x^2-8x-3[/tex]
This is a vertical parabola open downward (the leading coefficient is negative)
The vertex is a maximum
The axis of symmetry is the x-coordinate of the vertex
so
Convert the given equation into vertex form
y=a(x-h)^2+k
where
(h,k) is the vertex
x=h ----> axis of symmetry
step 1
Factor -2
[tex]f(x)=-2(x^2+4x)-3[/tex]
step 2
Complete the square
[tex]\begin{gathered} f(x)=-2(x^2+4x+4-4)-3 \\ f(x)=-2(x^2+4x+4)-3+8 \\ f(x)=-2(x^2+4x+4)+5 \end{gathered}[/tex]
step 3
Rewrite as perfect squares
[tex]f(x)=-2(x+2)^2+5[/tex]
The vertex is the point (-2,5)
The axis of symmetry is x=-2
see the attached figure below
