Compare the graphs of y=k/x for k=1 and k=5. Sketch all three graphs: y=1/x, y=5/x, and y=-5/x.

Given the function $y = -\frac{4}{x} - 1$.
The general form of a hyperbola is $y = \frac{k}{x-p} + q$.
The vertical asymptote is $x=p$.
The horizontal asymptote is $y=q$.
In this function, $p=0$ and $q=-1$.
The equation of the vertical asymptote is $\boxed{x=0}$.
The equation of the horizontal asymptote is $\boxed{y=-1}$.

To determine the x-intercept, set $y=0$ and solve for $x$.

Multiply both sides by $x$:

The coordinates of the x-intercept are $\boxed{(-4, 0)}$.

Sketch the graph of $y = -\frac{4}{x} - 1$ on a set of axes. Indicate the coordinates of the x-intercept as well as the asymptotes.

Description of the sketch:

• Draw the vertical asymptote at $x=0$ (the y-axis).
• Draw the horizontal asymptote at $y=-1$.
• Plot the x-intercept at $(-4, 0)$.
• Since the constant $k=-4$ is negative, the branches of the hyperbola will be in the second and fourth quadrants relative to the asymptotes.
• Plot additional points to guide the sketch:
  • If $x=-1$, $y = -\frac{4}{-1} - 1 = 4 - 1 = 3$. Point: $(-1,3)$.
  • If $x=1$, $y = -\frac{4}{1} - 1 = -4 - 1 = -5$. Point: $(1,-5)$. The graph will approach the asymptotes but never touch them.

For the function $y = -\frac{4}{x} - 1$:
The domain is all real numbers except where the denominator is zero. So, $x \neq 0$.
The range is all real numbers except the horizontal asymptote. So, $y \neq -1$.
The domain is $\boxed{x \in \mathbb{R}, x \neq 0}$.
The range is $\boxed{y \in \mathbb{R}, y \neq -1}$.

Describe the different transformations of $y = \frac{4}{x}$ to $y = -\frac{4}{x} - 1$.

1. From $y = \frac{4}{x}$ to $y = -\frac{4}{x}$: This is a reflection across the x-axis (or y-axis).
2. From $y = -\frac{4}{x}$ to $y = -\frac{4}{x} - 1$: This is a vertical translation downwards by 1 unit.

Determine the line of symmetry with a negative gradient.
For a hyperbola of the form $y = \frac{k}{x-p} + q$, the lines of symmetry pass through the intersection of the asymptotes $(p,q)$ and have gradients 1 and $-1$.
From 2(a), the asymptotes are $x=0$ and $y=-1$. Their intersection is $(0, -1)$.
We need the line of symmetry with a negative gradient, so the slope $m = -1$.
Using the point-slope form $y - y_1 = m(x - x_1)$:

The line of symmetry with a negative gradient is $\boxed{y = -x - 1}$.