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The problem is asking to find the vertical asymptotes of the given function:

\[
f(x) = \frac{x^2 - x - 2}{x^2 + x - 6}
\]

### Step 1: Factor both the numerator and the denominator.

- The numerator \(x^2 - x - 2\) can be factored as:
  \[
  x^2 - x - 2 = (x - 2)(x + 1)
  \]

- The denominator \(x^2 + x - 6\) can be factored as:
  \[
  x^2 + x - 6 = (x - 2)(x + 3)
  \]

Thus, the function becomes:
\[
f(x) = \frac{(x - 2)(x + 1)}{(x - 2)(x + 3)}
\]

### Step 2: Cancel out common factors (if any).

There is a common factor of \(x - 2\) in both the numerator and the denominator, so we cancel it out, leaving:
\[
f(x) = \frac{x + 1}{x + 3} \quad \text{for} \quad x \neq 2
\]

### Step 3: Determine the vertical asymptotes.

Vertical asymptotes occur when the denominator is zero, provided the factor does not cancel out with the numerator. For this function, the denominator is \(x + 3\). Setting it equal to zero:
\[
x + 3 = 0
\]
\[
x = -3
\]

Thus, there is a vertical asymptote at \(x = -3\).

### Conclusion:
The correct answer is:

**a. \(x = -3\)**

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Would you like more details or have any questions?

Here are 5 related questions for practice:
1. What is the horizontal asymptote of the function \(f(x) = \frac{x+1}{x+3}\)?
2. How do you find horizontal asymptotes of a rational function?
3. Can a rational function have more than one vertical asymptote?
4. What is the domain of the function \(f(x) = \frac{x+1}{x+3}\)?
5. What happens to the function as \(x\) approaches the vertical asymptote?

**Tip:** Vertical asymptotes occur when the denominator is zero and the factor doesn't cancel with the numerator!

Tentukan asimtot tegak grafik fungsi f(x) = (x^2 - x - 2) / (x^2 + x - 6)

Learn how to find the vertical asymptote of the function f(x) = (x^2 - x - 2) / (x^2 + x - 6). This problem covers important algebraic concepts like factoring quadratics and determining vertical asymptotes of rational functions. Suitable for students in Grades 9-12.

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