1. The problem asks to find the value of $f(-4)$ for the function $y = f(x)$, which is a parabola with vertex at approximately $(-2, -6)$ and passing through $(-4, -2)$. 2. The vertex form of a parabola is given by $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. Here, $h = -2$ and $k = -6$, so the function is $$f(x) = a(x + 2)^2 - 6.$$ 3. To find $a$, use the point $(-4, -2)$ which lies on the parabola: $$-2 = a(-4 + 2)^2 - 6$$ $$-2 = a(-2)^2 - 6$$ $$-2 = 4a - 6$$ 4. Solve for $a$: $$-2 + 6 = 4a$$ $$4 = 4a$$ $$\cancel{4} = 4a \Rightarrow 1 = a$$ 5. So, $a = 1$ and the function is $$f(x) = (x + 2)^2 - 6.$$ 6. Now find $f(-4)$: $$f(-4) = (-4 + 2)^2 - 6 = (-2)^2 - 6 = 4 - 6 = -2.$$ 7. Therefore, the value of $f(-4)$ is $-2$.