Both intercepts are #0#.
#sqrt(x^2+7) > 0# for all #x#, so there are no vertical asymptotes.
#lim_(xrarroo)f(x) = lim_(xrarroo)x/(xsqrt(1+7/x^2)) = 1#, so #y=1# is a horizontal asymptote on the right.
#lim_(xrarr-oo)f(x) = lim_(xrarr-oo)x/(-xsqrt(1+7/x^2)) = -1#, so #y = -1# is a horizontal asymptote on the left.
#f'(x) = 7/(x^2+7)^(3/2) > 0# for all #x#, so there are no local extrema.
#f''(x) = (-21x)/(x^2+7)^(5/2)# changes sign at #(0,0)#, so I have been taught that #(0,0)# is an inflection point. (Note that: #f'(0) != 0#, so some would say #(0,0)# is not an inflection point.)
graph{x/sqrt(x^2+7) [-16.01, 16.03, -7.95, 8.05]}
