## Calculus 10th Edition

Vertical asymptote at $x=6$ and $x=-6.$
A function has vertical asymptotes at values that make the denominator only $0$. $h(x)=\dfrac{6x}{36-x^2}=\dfrac{6x}{(6-x)(6+x)}\to(6-x)(6+x)=0\to x=6$ or $x=-6.$