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

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1. First, set the denominator equal to zero:
   $$
   (x-3)(x+2) = 0
   $$
2. Solve each factor for $$ x $$:
   $$
   x-3=0 \quad \Rightarrow \quad x=3
   $$
   $$
   x+2=0 \quad \Rightarrow \quad x=-2
   $$
3. Confirm that the numerator is not equal to zero at these $$ x $$-values:
   - For $$ x=3 $$:
     $$
     2-3=-1 \quad (\neq 0)
     $$
   - For $$ x=-2 $$:
     $$
     2-(-2)=4 \quad (\neq 0)
     $$
4. Since the numerator is nonzero at $$ x=3 $$ and $$ x=-2 $$, the graph of $$ f(x) $$ has vertical asymptotes at these values.

The vertical asymptotes are at $$ x=3 $$ and $$ x=-2 $$.
The vertical asymptotes are at $$ x=3 $$ and $$ x=-2 $$.

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

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