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

just took it it's D

Answer 2

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)=(2)/(3)f(0)= (2)/(3)\cdot 108=72,\\f_2=f(2)=(2)/(3)f(1)= (2)/(3)\cdot 72=48,\\ f_3=f(3)=(2)/(3)f(2)= (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`