Which is the graph of the sequence defined by the function f(x+1)=2/3f(x) if the initial value of the sequence is 108

First four terms of this sequence are:[tex] f_0=f(0)=108,\\f_1=f(1)=\dfrac{2}{3}f(0)= \dfrac{2}{3}\cdot 108=72,\\f_2=f(2)=\dfrac{2}{3}f(1)= \dfrac{2}{3}\cdot 72=48,\\ f_3=f(3)=\dfrac{2}{3}f(2)= \dfrac{2}{3}\cdot 48=32 [/tex].This sequence is decreasing and seems to be exponential. See whether you can determine such a and b, that [tex] f(x)=a\cdot b^x [/tex]:1. [tex] f(0)=a\cdot b^0=a,\\ f(0)=108 [/tex], then a=108,2. [tex] f(1)=108\cdot b^1=108 b,\\ f(1)=72 [/tex], then [tex] b=\dfrac{72}{108} =\dfrac{2}{3} [/tex].3. Check for x=2 and x=3:a) x=2:[tex] f(2)=108\cdot \left(\dfrac{2}{3}\right)^2=108\cdot \dfrac{4}{9} =48 [/tex] (true),b) x=3: [tex] f(3)=108\cdot \left(\dfrac{2}{3}\right)^3=108\cdot \dfrac{8}{27} =32 [/tex] (true).4. Check for all x:[tex] f(x+1)=108\cdot \left(\dfrac{2}{3}\right)^{x+1},\\ f(x)=108\cdot \left(\dfrac{2}{3}\right)^x [/tex].Equate f(x+1) and 2/3f(x):[tex] 108\cdot \left(\dfrac{2}{3}\right)^{x+1}=\dfrac{2}{3}\cdot 108\cdot \left(\dfrac{2}{3}\right)^x ,\\\left(\dfrac{2}{3}\right)^{x+1}=\left(\dfrac{2}{3}\right)^{x+1} [/tex] is true equality for all x.Answer: the graph is the graph of the function [tex]f(x)=108\cdot \left(\dfrac{2}{3}\right)^x [/tex].