To plot the graph of the sequence defined by the function [tex]\( f(x+1) = \frac{2}{3} f(x) \)[/tex] with an initial value of [tex]\( f(0) = 108 \)[/tex], let's first generate the sequence by calculating its terms step-by-step.
1. Initial Value:
[tex]\[
f(0) = 108
\][/tex]
2. First Term:
[tex]\[
f(1) = \frac{2}{3} \cdot f(0) = \frac{2}{3} \cdot 108 = 72
\][/tex]
3. Second Term:
[tex]\[
f(2) = \frac{2}{3} \cdot f(1) = \frac{2}{3} \cdot 72 = 48
\][/tex]
4. Third Term:
[tex]\[
f(3) = \frac{2}{3} \cdot f(2) = \frac{2}{3} \cdot 48 = 32
\][/tex]
5. Fourth Term:
[tex]\[
f(4) = \frac{2}{3} \cdot f(3) = \frac{2}{3} \cdot 32 = 21.3333
\][/tex]
6. Fifth Term:
[tex]\[
f(5) = \frac{2}{3} \cdot f(4) = \frac{2}{3} \cdot 21.3333 = 14.2222
\][/tex]
7. Sixth Term:
[tex]\[
f(6) = \frac{2}{3} \cdot f(5) = \frac{2}{3} \cdot 14.2222 = 9.4815
\][/tex]
8. Seventh Term:
[tex]\[
f(7) = \frac{2}{3} \cdot f(6) = \frac{2}{3} \cdot 9.4815 = 6.3210
\][/tex]
9. Eighth Term:
[tex]\[
f(8) = \frac{2}{3} \cdot f(7) = \frac{2}{3} \cdot 6.3210 = 4.2140
\][/tex]
10. Ninth Term:
[tex]\[
f(9) = \frac{2}{3} \cdot f(8) = \frac{2}{3} \cdot 4.2140 = 2.8093
\][/tex]
11. Tenth Term:
[tex]\[
f(10) = \frac{2}{3} \cdot f(9) = \frac{2}{3} \cdot 2.8093 = 1.8729
\][/tex]
Thus, the sequence is:
[tex]\[
\{108, 72, 48, 32, 21.3333, 14.2222, 9.4815, 6.3210, 4.2140, 2.8093, 1.8729\}
\][/tex]
To graph this sequence, you would plot the points [tex]\( (0, 108) \)[/tex], [tex]\( (1, 72) \)[/tex], [tex]\( (2, 48) \)[/tex], [tex]\( (3, 32) \)[/tex], [tex]\( (4, 21.3333) \)[/tex], [tex]\( (5, 14.2222) \)[/tex], [tex]\( (6, 9.4815) \)[/tex], [tex]\( (7, 6.3210) \)[/tex], [tex]\( (8, 4.2140) \)[/tex], [tex]\( (9, 2.8093) \)[/tex], and [tex]\( (10, 1.8729) \)[/tex] on a graph with the x-axis representing the term number and the y-axis representing the value of the term.
The resulting graph will show a rapidly decreasing sequence as each subsequent value is [tex]\(\frac{2}{3}\)[/tex] of the previous one. This exponential decay creates a curve that descends sharply at first and then more gradually as it approaches zero.
