So let's assume that n years pass. Then the function would be:
![y_n=18\cdot(0.65)^n](https://img.qammunity.org/2023/formulas/mathematics/college/1a01rrtcpzq9bd776syli0x4hcrg999aee.png)
Now let's assume that another year passes. Then the exponent of the function will be n+1 and we'll have:
![y_(n+1)=18\cdot(0.65)^(n+1)](https://img.qammunity.org/2023/formulas/mathematics/college/2p71n06fb1pmzke75lvizpdyp60ibpprm2.png)
If we take the value of the function at x=n+1 and we divide it by the value of the function one year before we can find the percentage of increase or decrease:
![(y_(n+1))/(y_n)=(18\cdot(0.65)^(n+1))/(18\cdot(0.65)^n)=0.65](https://img.qammunity.org/2023/formulas/mathematics/college/iofmx0iyzxkn0jtsf0vc8lnuz9hbpuodqs.png)
This means that every year the value of the function is the 65% of the value of the prrevious year. This means that every year there is aa decrease of 100%-65%=35%.
This means that the answer for the first blank space is 35% and the answer for the second is decrease.