Answer:
a) Exponential decay
b,) The parent function is shifted up 1 unit.
Explanation:
Assuming your function is
![y = ( (1)/(3) )^(x) + 1](https://img.qammunity.org/2021/formulas/mathematics/high-school/qmhh6n4e60cuz351zo7kv220m39ci5glbr.png)
Then we can compare this to:
![y = {b}^(x) + c](https://img.qammunity.org/2021/formulas/mathematics/high-school/dty1n8zskassgxd6jpyt6qzzcu4l0cdf31.png)
If
![0 \: < \: b \: < 1](https://img.qammunity.org/2021/formulas/mathematics/high-school/iar77grrk9xqme4ct2l9hed2kqbdwig13m.png)
then y is an exponential decay function.
If |b|>1, then y is an exponential growth.
We can see that:
![0 \: < \: (1)/(3) \: < \: 1](https://img.qammunity.org/2021/formulas/mathematics/high-school/ksgrtl8s6gzxnjvsekv8y1rquu1ntfrlhu.png)
Therefore the given function is an exponential decay.
The parent function is shifted up 1 unit.