ok, so in year 0, the number of rabbits is 6 which is 6*(4)^0
in year 1, the number of rabbits is 4*6 which is 6*(4)^1
in year 2, the number of rabbits is (4*6) from year one multiplyed by 4, so a total of rabbits is 4*6*4 which is 6*(4)^2
in year 3, the number of rabbits is 6*(4)^2 from year two multiplyed by 4, so a total of rabbits is 4*6*4*4 which is 6*(4)^3
The sequence will continue in the same way.
Therefore, for year x:
f(x) = 6 * (4)^x which is choice A