## Calculus 10th Edition

Vertical asymptotes at $x=8$ and $x=-8.$
A function has vertical asymptotes at values that make only the denominator $0$. $g(x)=\dfrac{2x+1}{x^2-64}=\dfrac{2x+1}{(x-8)(x+8)}\to(x-8)(x+8)=0\to x=8$ or $x=-8.$