#### Answer

$\dfrac{(x-3)}{(x-6)(x+2)}.$

#### Work Step by Step

Zero at $3$ implies numerator is equal to $0$ when $x=3\to(x-3)=0.$
Vertical asymptote at $-2$ and $6$ implies a denominator of equation $(x+2)(x-6).$
Putting everything together gives us $\dfrac{(x-3)}{(x+2)(x-6)}.$