Question

Which is the graph of the sequence defined by the function f(x + 1) = Two-thirdsf(x) if the initial value of the sequence is 108?

One a coordinate plane, 4 points are plotted. The points are (0, 108), (1, 36), (2, 12), (3, 4).

One a coordinate plane, 4 points are plotted. The points are (1, 108), (2, 36), (3, 12), (4, 4).

One a coordinate plane, 4 points are plotted. The points are (0, 108), (1, 72), (2, 48), (3, 32).

One a coordinate plane, 4 points are plotted. The points are (1, 108), (2, 72), (3, 48), (4, 32).

Answer 1

We have f(n+1) = (2/3) f(n) and initial value f(1) = 108. So

f(2) = (2/3)·108 = 72
f(3) = (2/3)·72 = 48
f(4) = (2/3)·48 = 32

Thus the points are (1,108), (2,72), (3,48), (4,32) — the fourth graph.