The graph below shows the graph of $y=f(x)$.

1. Determine the axial intercepts and asymptote for $y = f^{-1}(x)$.

The points $(-4, 0)$ and $(0, 3)$ on the graph of $y=f(x)$ correspond to the points $(0, -4)$ and $(3, 0)$ on the graph of $y=f^{-1}(x)$.
The horizontal asymptote $y=4$ for $f(x)$ corresponds to the vertical asymptote $x=4$ for $f^{-1}(x)$.

2. Sketch the graph of $y=f^{-1}(x)$.

Draw a smooth curve that passes through the y-intercept at $(0, -4)$ and the x-intercept at $(3, 0)$. As $x$ approaches $4$ from the left, the curve should rise steeply, approaching the vertical line $x=4$ asymptotically.

3. State the geometrical relationship between the two graphs.

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y=f(x)$ in the line $y=x$.

1. Find the range of the inner function $g(x)$.

By completing the square for $g(x)$:
$g(x) = x^2 + 6x + 5$
$g(x) = (x + 3)^2 - 9 + 5$
$g(x) = (x + 3)^2 - 4$
Since $(x + 3)^2 \ge 0$ for all real $x$, the range of $g$ is $g(x) \ge -4$.

2. Determine the range of the composite function by observing the graph of $f(x)$ over the restricted domain.

For the composite function $fg(x)$, the range of $g$ acts as the domain for $f$. Thus, we need to find the range of $f(x)$ for $x \ge -4$.
From the given graph, $f(-4) = 0$.
As $x$ increases from $-4$, the curve strictly rises and approaches the horizontal asymptote $y = 4$ from below.
Therefore, the values of $f(x)$ start from $0$ and go up to, but do not include, $4$.

1. Identify the root of $f(x) = 0$ from the given graph.

From the graph of $y = f(x)$, the curve crosses the $x$-axis at the point $(-4, 0)$.
Therefore, the only solution to $f(x) = 0$ is $x = -4$.

2. Use the root to find the value of $g(a)$.

We are given that $fg(a) = 0$, which means $f(g(a)) = 0$.
Comparing this with $f(-4) = 0$, the input to $f$ must be $-4$.
Thus, $g(a) = -4$.

(Note: Checking for consistency, $g(a) = a^2 + 6a + 5 = -4 \implies (a+3)^2 = 0 \implies a = -3$, which aligns with the condition that $a$ is a negative constant.)

1. Express $x$ in terms of $y$ by completing the square for the quadratic expression.

Let $y = h(x) = x^2 + 6x + 5$.
Completing the square:
$y = (x+3)^2 - 3^2 + 5$
$y = (x+3)^2 - 4$
$(x+3)^2 = y + 4$
$x+3 = \pm\sqrt{y+4}$
$x = -3 \pm\sqrt{y+4}$

2. Determine the correct sign for the inverse function using the given domain of $h$.

The domain of $h$ is $x < -4$.
This implies $x + 3 < -1$, which is negative.
Therefore, we must choose the negative square root to satisfy this condition:
$x = -3 - \sqrt{y+4}$
So, $h^{-1}(x) = -3 - \sqrt{x+4}$.

3. Find the domain of $h^{-1}$ by determining the range of $h$.

The domain of $h^{-1}$ is equal to the range of $h$.
For $x < -4$:
$x+3 < -1$
$(x+3)^2 > 1$
$(x+3)^2 - 4 > 1 - 4$
$h(x) > -3$
Since the range of $h$ is $h(x) > -3$, the domain of $h^{-1}$ is $x > -3$.