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The problem is asking for the domain of the function:

\[
m(x) = \frac{x - 8}{x^2 + 6x - 27}
\]

### Steps to find the domain:

1. **Denominator Restriction**: The denominator \(x^2 + 6x - 27\) must not be equal to zero, as division by zero is undefined. So, we need to solve the equation:

   \[
   x^2 + 6x - 27 = 0
   \]

2. **Solve for \(x\)** using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   For \(a = 1\), \(b = 6\), and \(c = -27\):
   \[
   x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-27)}}{2(1)}
   \]
   Simplifying:
   \[
   x = \frac{-6 \pm \sqrt{36 + 108}}{2}
   \]
   \[
   x = \frac{-6 \pm \sqrt{144}}{2}
   \]
   \[
   x = \frac{-6 \pm 12}{2}
   \]
   So, \(x = \frac{-6 + 12}{2} = 3\) and \(x = \frac{-6 - 12}{2} = -9\).

3. **Domain Exclusion**: The values \(x = 3\) and \(x = -9\) make the denominator zero, so these values are excluded from the domain.

4. **Conclusion**: The domain of the function in interval notation is all real numbers except \(x = 3\) and \(x = -9\). Therefore, the domain is:

   \[
   (-\infty, -9) \cup (-9, 3) \cup (3, \infty)
   \]

Let me know if you'd like further details or clarifications!

Here are some related questions for you:
1. What would the domain be if the denominator were \(x^2 + 6x - 25\)?
2. How do you determine if a function is continuous on its domain?
3. What happens to the graph of \(m(x)\) at \(x = 3\) and \(x = -9\)?
4. How do you find the range of this rational function?
5. Can there be other restrictions on the domain besides division by zero?

**Tip**: Always check for both division by zero and square roots when determining the domain of a function!

Give the domain for m(x) = (x - 8) / (x^2 + 6x - 27) in interval notation.

This problem demonstrates how to find the domain of the rational function m(x) = (x - 8) / (x^2 + 6x - 27). By solving the quadratic equation in the denominator, students will learn about domain restrictions and interval notation. Suitable for students in Grades 10-12.

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