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. $x = -3$

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!