The images you uploaded are of the same type, showing the rational function $f(x) = \frac{2x^2 + 7x + 3}{x^2 - x - 12}$ and part (b) asking about vertical asymptotes.

To summarize part (b):

Vertical Asymptotes

A rational function has vertical asymptotes where the denominator equals zero and the numerator does not.

1. Factor the denominator: $x^2 - x - 12 = (x - 4)(x + 3)$ This shows the function has potential vertical asymptotes at $x = 4$ and $x = -3$.

2. Check the numerator at these values: $2x^2 + 7x + 3$ At $x = 4$, the numerator is $2(4)^2 + 7(4) + 3 = 63$, which is non-zero. Hence, there is a vertical asymptote at $x = 4$.

   At $x = -3$, the numerator is $2(-3)^2 + 7(-3) + 3 = 0$, so there's no vertical asymptote here, but a hole at $x = -3$.

Thus, the vertical asymptote is at $x = 4$, and the function has a hole at $x = -3$.

Would you like a deeper explanation of the limits or the behavior of the function around these points? Here are five follow-up questions:

1. What are the horizontal asymptotes for this function?
2. Can you explain how limits help determine the type of asymptote?
3. What does it mean graphically when there's a hole in the function?
4. How do you find the range of this function?
5. How can the function's end behavior be analyzed using limits?

Tip: Always check for both vertical asymptotes and holes in rational functions by testing where both the denominator and numerator are zero.