Let's do a little detective work.....
x^2 - 16 factors as ( x + 4) ( x - 4) so we have vertical asypmtotes at x = - 4 and x = 4
This rules out the first graph
Note that when x = 0 the y intercept = -6/16 = 3/8 .....this rules out the third graph which has a negative y intercept
Now.....let's find the zeroes of the function (where the grpah intercepts the x axis ) by solving this
x^ 2 + 5x - 6 = 0
(x - 1) ( x + 6) = 0
Setting both of these to 0 and solving produces x = 1 and x = -6
Note that the second graph meets all the criteria
1. Vertical asymptotes at x = -4 and x = 4
2. Positive y intercept
3. Zeroes (roots) at x = -6 and x = 1