## Calculus 10th Edition

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