Question

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

Answer 1

We are given initial value of the function =108.

Therefore, f(0) = 108.

Let us find some values of the function to graph it.

$\\f_1=f(1)=\dfrac{2}{3}f(0)= \dfrac{2}{3}\cdot 108=72,\\f_2=f(2)=\dfrac{2}{3}f(1)= \dfrac{2}{3}\cdot 72=48,\\ f_3=f(3)=\dfrac{2}{3}f(2)= \dfrac{2}{3}\cdot 48=32$

Therefore, some of the terms of the sequence would be

108, 72, 48, 32...

We can see that function is decreasing exponentially by factor of 2/3.

Therefore, function would be $y=108(2/3)^x$

Now, we can plot the graph for exponential function $y=108(2/3)^x$