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The problem is asking how to define \( f(6) \) in order to make the function \( f(x) = \frac{x^2 - 3x - 18}{x - 6} \) continuous at \( x = 6 \).

Let's first analyze this function.

### Step 1: Factor the numerator
The numerator is \( x^2 - 3x - 18 \), and we need to factor it to simplify the function. We are looking for two numbers that multiply to \( -18 \) and add to \( -3 \).

\[
x^2 - 3x - 18 = (x - 6)(x + 3)
\]

### Step 2: Simplify the function
Now that we have factored the numerator, the function becomes:

\[
f(x) = \frac{(x - 6)(x + 3)}{x - 6}
\]

For \( x \neq 6 \), we can cancel the \( (x - 6) \) terms:

\[
f(x) = x + 3 \quad \text{for} \quad x \neq 6
\]

### Step 3: Find the limit as \( x \) approaches 6
Since we simplified \( f(x) \) to \( x + 3 \), we can find the limit of \( f(x) \) as \( x \) approaches 6:

\[
\lim_{x \to 6} f(x) = \lim_{x \to 6} (x + 3) = 6 + 3 = 9
\]

### Step 4: Define \( f(6) \)
In order for \( f(x) \) to be continuous at \( x = 6 \), the value of \( f(6) \) must equal the limit as \( x \) approaches 6. Therefore, we define \( f(6) = 9 \).

Thus, the correct answer is:

\[
f(6) = 9
\]

Would you like more details or have any questions?

Here are five related questions:
1. What is the definition of continuity at a point?
2. Why can't we directly substitute \( x = 6 \) into the original function?
3. How do you factor quadratic expressions?
4. What does it mean for a function to be discontinuous at a point?
5. Can you explain the difference between a removable and non-removable discontinuity?

**Tip:** Always check for potential simplifications in rational functions before calculating limits!

How would you define f(6) in order to make f(x) = (x^2 - 3x - 18) / (x - 6) continuous at 6?

Explore how to define f(6) to ensure the function f(x) = (x^2 - 3x - 18) / (x - 6) is continuous at x = 6. This guide details factoring techniques, simplification of rational functions, and applying continuity principles, suitable for high school students learning algebra.

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