\( f(x)=\frac{x+3}{(x-1)(x+3)} \quad \) Find the vertical asymptotes and holes of the function. \( \begin{array}{l}\text { Hole: }(-3,-1 / 4) \\ \text { VA: } x=1,-3\end{array} \) \( \begin{array}{l}\text { Hole: None; } \\ \text { VA: } x=1,-3\end{array} \) Hole: \( (-3,-1 / 4) \) VA: \( x=1 \)

The given function is   f(x) = (x + 3) ⁄ [(x – 1)(x + 3)]. Step 1. Factor and cancel common factors.   Notice that (x + 3) appears in both the numerator and the denominator.   However, we can only cancel (x + 3) if we remember that x = –3 is not allowed in the original function. After canceling, we have   f(x) = 1 ⁄ (x – 1)  for x ≠ –3. Step 2. Identify the hole.   The cancellation indicates there is a removable discontinuity (a hole) at x = –3.   To find the y-coordinate of the hole, plug x = –3 into the simplified function:     f(–3) = 1 ⁄ (–3 – 1) = 1 ⁄ (–4) = –1/4.   Thus, there is a hole at (–3, –1/4). Step 3. Find the vertical asymptote.   The vertical asymptote(s) come from the factors in the denominator that were not canceled.   The denominator of the simplified function is x – 1.   Set x – 1 = 0, which gives x = 1.   Thus, there is a vertical asymptote at x = 1. Final Answer:   Hole: (–3, –1/4)   Vertical Asymptote: x = 1.