Find the vertical asymptote(s) of the graph of the function. $$ f(x)=\frac{2-x}{(x-5)(x+2)} $$

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1. The given function is 
   $$
   f(x)=\frac{2-x}{(x-5)(x+2)}.
   $$
   
2. Vertical asymptotes occur where the denominator is zero provided that the numerator is not zero at those points.

3. Set the denominator equal to zero:
   $$
   (x-5)(x+2)=0.
   $$
   This gives:
   $$
   x-5=0 \quad \text{or} \quad x+2=0,
   $$
   resulting in:
   $$
   x=5 \quad \text{or} \quad x=-2.
   $$

4. Next, check that the numerator does not also equal zero at these $$x$$-values:
   - For $$ x=5 $$: 
     $$
     2-5=-3 \neq 0.
     $$
   - For $$ x=-2 $$:
     $$
     2-(-2)=4 \neq 0.
     $$

5. Since the denominator is zero and the numerator is non-zero at $$ x=5 $$ and $$ x=-2 $$, the function has vertical asymptotes at these points.

The vertical asymptotes are at $$ x=5 $$ and $$ x=-2 $$.
The vertical asymptotes are at $$ x=5 $$ and $$ x=-2 $$.

Find the vertical asymptote(s) of the graph of the function. \( f(x)=\frac{2-x}{(x-5)(x+2)} \)

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