To determine the vertical asymptote(s) of the function

$f(x) = \frac{x+3}{x^2 + 11x + 24},$

we first need to find the values of $x$ that make the denominator equal to zero, as vertical asymptotes occur when the denominator is zero and the numerator is not zero at those points.

Step 1: Factor the denominator

The denominator is $x^2 + 11x + 24$. We can factor this quadratic expression:

$x^2 + 11x + 24 = (x + 3)(x + 8).$

Step 2: Set the denominator equal to zero

To find the vertical asymptotes, set the denominator equal to zero:

$(x + 3)(x + 8) = 0.$

This gives us two potential vertical asymptotes:

$x + 3 = 0 \quad \text{or} \quad x + 8 = 0,$ which simplifies to:

$x = -3 \quad \text{or} \quad x = -8.$

Step 3: Check for cancellation

Now, check if the numerator $x + 3$ can cancel with the denominator. Since the numerator also contains $x + 3$, the factor $x + 3$ cancels out with the same factor in the denominator. Therefore, $x = -3$ is not a vertical asymptote; instead, it is a removable discontinuity.

Conclusion:

The only vertical asymptote is at $x = -8$, because the factor $x + 8$ does not cancel out.

Final Answer:

The function has a vertical asymptote at $x = -8$.

Would you like further explanation on this process, or do you have any other questions?

Related Questions:

1. What are horizontal asymptotes and how do we find them for this function?
2. How do removable discontinuities differ from vertical asymptotes?
3. Can a function have both vertical and horizontal asymptotes? Why or why not?
4. How do you determine horizontal asymptotes for rational functions?
5. What happens graphically at a removable discontinuity?

Tip:

Always check for common factors between the numerator and denominator to identify removable discontinuities before concluding the vertical asymptotes.