What are the horizontal and vertical asymptotes of r(x)=(6x+1)/(16x^2)+1?

Horiz. asy.:  experiment with letting x grow larger and larger (without bound).  Eventually the given expression will approach some limiting value.

If we continue to increase x in (6x+1)/(16x^2), the fraction will approach 0.  Try it!  

If we continue to increase x in (6x+1)/(16x^2) + 1, the expression will approach 0 + 1, or 1.

Which one did you mean?

Thus, the horiz. asy. would be y=0 or y=1. 

Vertical asy.:  Identify the x-value, if any, at which the denominator of this fraction = 0.

If you meant (6x+1)/(16x^2)                 (without the +1), the fraction is undefined at x=0.  Thus, the vert. asy. is the vertical line x=0.

The answer would be 13.5 minutes.