The given function is:

$r(x) = \frac{1}{x + 8}$

It is stated that this function has a vertical asymptote at $x = -8$, which occurs because the denominator $x + 8 = 0$ at $x = -8$.

Now, the problem asks for the equation of the vertical asymptote of the transformed function $r(-8x)$.

Step-by-Step Analysis

The new function $r(-8x)$ can be written as:

$r(-8x) = \frac{1}{-8x + 8}$

We need to find the value of $x$ that makes the denominator zero, which will give us the vertical asymptote. So, set the denominator equal to zero:

$-8x + 8 = 0$

Solve for $x$:

$-8x = -8$

$x = 1$

Conclusion

The equation of the vertical asymptote of the transformed function $r(-8x)$ is:

$x = 1$

Would you like further details or clarification on any step?

Related Questions:

1. How do you find horizontal asymptotes for rational functions?
2. What happens to the asymptotes when a function is reflected over the y-axis?
3. Can a function have more than one vertical asymptote?
4. What is the effect of multiplying the input of a function by a constant on its graph?
5. How do horizontal shifts affect the position of asymptotes?

Tip:

Always check how transformations such as reflections or stretches/compressions affect both the location of the asymptotes and the overall shape of the graph.