To find the vertical asymptotes of the rational function:

$f(x) = \frac{x^2 - 9x + 14}{x^2 - 5x + 6}$

Steps:

1. Vertical asymptotes occur where the denominator is zero, but the numerator is not zero at the same points. So, we first solve for when the denominator equals zero.

The denominator is:

$x^2 - 5x + 6 = 0$

2. Solve the quadratic equation:

$x^2 - 5x + 6 = 0$

We can factor it:

$(x - 2)(x - 3) = 0$

So, the denominator is zero when $x = 2$ and $x = 3$.

3. Check if the numerator is zero at these values: The numerator is $x^2 - 9x + 14$. Substituting $x = 2$ and $x = 3$ into the numerator:

   • For $x = 2$: $2^2 - 9(2) + 14 = 4 - 18 + 14 = 0$
   • For $x = 3$: $3^2 - 9(3) + 14 = 9 - 27 + 14 = -4$

   So, at $x = 2$, the numerator is also zero, meaning there is no vertical asymptote at $x = 2$ (it is a removable discontinuity).

   At $x = 3$, the numerator is not zero, so there is a vertical asymptote at $x = 3$.

Conclusion:

The vertical asymptote is at $x = 3$, which corresponds to the second option: $x = 3$.

Would you like further clarification on any part of this process?

Here are some related questions to explore further:

1. How do horizontal asymptotes differ from vertical asymptotes?
2. What is a removable discontinuity in a function?
3. How do you find the horizontal asymptote for this function?
4. Can there be more than one vertical asymptote for a rational function?
5. What are the steps to factor a quadratic expression efficiently?

Tip: Always check if the numerator cancels out any of the roots of the denominator to identify removable discontinuities!