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Let's analyze the given problem to determine why \( x = -6 \) is not in the valid domain.

The function given is:
\[ f(x) = \sqrt{\frac{x^2 - 36}{x + 6}} \]

To find the intercept, we set \( f(x) = 0 \):
\[ \sqrt{\frac{x^2 - 36}{x + 6}} = 0 \]

This implies:
\[ \frac{x^2 - 36}{x + 6} = 0 \]

Simplifying the numerator:
\[ x^2 - 36 = 0 \]
\[ (x - 6)(x + 6) = 0 \]

So, the solutions are:
\[ x = 6 \quad \text{and} \quad x = -6 \]

However, we need to consider the domain of the original function \( f(x) \). The function \( f(x) \) includes a square root and a fraction. For the function to be defined:
1. The denominator \( x + 6 \neq 0 \)
2. The expression inside the square root must be non-negative.

The first condition \( x + 6 \neq 0 \) implies:
\[ x \neq -6 \]

Therefore, \( x = -6 \) makes the denominator zero, which makes the function undefined. Hence, \( x = -6 \) is not in the valid domain of the function.

The only valid solution within the domain is \( x = 6 \), which gives the intercept at \( (6, 0) \).

Do you want more details or have any questions?

Here are 8 related questions for further exploration:
1. What is the domain of the function \( f(x) \)?
2. How do you find the range of the function \( f(x) \)?
3. What happens to the function \( f(x) \) as \( x \) approaches -6?
4. How do you determine the vertical asymptotes of \( f(x) \)?
5. What are the intercepts of the function \( f(x) \)?
6. How do you graph the function \( f(x) \)?
7. What is the behavior of \( f(x) \) for very large positive and negative values of \( x \)?
8. How do you find the critical points of \( f(x) \)?

Tip: Always check the domain restrictions when dealing with rational functions and square roots.

Explore why x = -6 is not in the valid domain of a function involving square roots and rational expressions. This detailed explanation covers domain restrictions, intercepts, and the behavior of functions near critical points. Perfect for high school students studying functions and algebra.

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