To find the vertical asymptotes of the given function

$f(x) = \frac{x^2 + 7x + 12}{2x + 6}$

we need to identify the values of $x$ that make the denominator zero, since vertical asymptotes occur where the denominator is zero and the numerator is non-zero.

Step 1: Solve for the denominator.

Set the denominator equal to zero and solve for $x$:

$2x + 6 = 0$

$2x = -6$

$x = -3$

Thus, $x = -3$ is a potential vertical asymptote.

Step 2: Check if the numerator is also zero at $x = -3$.

The numerator is:

$x^2 + 7x + 12$

Substitute $x = -3$ into the numerator:

$(-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0$

Since the numerator is also zero at $x = -3$, this means that there is a hole at $x = -3$, not a vertical asymptote.

Conclusion:

There are no vertical asymptotes for this function. Instead, there is a hole at $x = -3$.

Would you like a deeper explanation on any part of this, or have any further questions?

Here are some related questions you might consider:

1. How do horizontal asymptotes differ from vertical asymptotes?
2. What happens at a hole in the graph of a rational function?
3. Can a function have both vertical asymptotes and holes?
4. How do you find horizontal asymptotes of rational functions?
5. What happens if both the numerator and denominator are zero at the same point?

Tip: When both the numerator and denominator are zero at a given point, it results in a hole, not an asymptote!