Find the vertical asymptote(s) of the graph of the function. \( f(x)=\frac{2-x}{(x-3)(x+2)} \)

To find the vertical asymptotes of the function \( f(x) = \frac{2-x}{(x-3)(x+2)} \), you need to determine where the denominator is equal to zero since vertical asymptotes occur at these points (provided the numerator isn't also zero at those points). Set the denominator equal to zero: \[ (x-3)(x+2) = 0. \] This gives the equations \( x - 3 = 0 \) and \( x + 2 = 0 \), leading to solutions: \[ x = 3 \quad \text{and} \quad x = -2. \] Thus, the vertical asymptotes of the function are at \( x = 3 \) and \( x = -2 \). In summary, the graph of \( f(x) = \frac{2-x}{(x-3)(x+2)} \) has vertical asymptotes at these two x-values: \( x = 3 \) and \( x = -2 \).