\( f(x)=\frac{x+3}{(x-1)(x+3)} \quad \) Find the verical asymptotes and holes of the func

To find the vertical asymptotes and holes of the function \( f(x)=\frac{x+3}{(x-1)(x+3)} \), we start by examining the denominator. A vertical asymptote occurs where the denominator equals zero, but the numerator is not zero at the same point. For this function, set the denominator \( (x-1)(x+3) = 0 \). This gives us \( x = 1 \) and \( x = -3 \). However, at \( x = -3 \), both the numerator and denominator are zero, indicating a hole rather than an asymptote. Thus, there's a hole at \( x = -3 \) and a vertical asymptote at \( x = 1 \). To remember this concept, visualize holes as "gaps" where the function isn't defined due to cancellation between numerator and denominator. The vertical asymptotes are where the function shoots off to infinity!