The function you provided is:

$f(x) = \frac{x - 1}{x^2 + 4x + 3}$

To find the vertical asymptotes:

Vertical asymptotes occur when the denominator equals zero, as long as the numerator does not also equal zero at the same points. We need to solve the equation:

$x^2 + 4x + 3 = 0$

Factor the quadratic:

$x^2 + 4x + 3 = (x + 3)(x + 1)$

Now, set each factor equal to zero:

$x + 3 = 0 \quad \text{or} \quad x + 1 = 0$

Solving these:

$x = -3 \quad \text{or} \quad x = -1$

Therefore, the vertical asymptotes are $x = -3$ and $x = -1$.

Answer:

The correct option is:

a. $x = -3$, $x = -1$

Do you want further details or have any questions?

Related Questions:

1. How do we determine the horizontal asymptotes of this function?
2. What happens to the function as $x \to \infty$?
3. Can this function have any oblique asymptotes?
4. How does the behavior of the function near $x = -3$ and $x = -1$ compare?
5. What is the domain of the function?

Tip:

Always check both the numerator and the denominator when finding asymptotes to ensure no cancellations occur.