We have the exponential function:

$y = 4(0.97)^t$

Now, we will calculate $y$ for $t = 1, 2, 3$.

1. For $t = 1$:
   $y = 4(0.97)^1 = 4 \times 0.97 = 3.88$

2. For $t = 2$:
   $y = 4(0.97)^2 = 4 \times (0.97 \times 0.97) = 4 \times 0.9409 = 3.7636$

3. For $t = 3$:
   $y = 4(0.97)^3 = 4 \times (0.97 \times 0.97 \times 0.97) = 4 \times 0.912273 = 3.6491$

Final values:

• $y(1) = 3.88$
• $y(2) = 3.7636$
• $y(3) = 3.6491$

Would you like a graph or further explanations?

Related Questions:

1. What is the general behavior of an exponential decay function like this?
2. How does the base $0.97$ affect the rate of decay?
3. What is the half-life of this function?
4. What happens to $y$ as $t \to \infty$?
5. Can this function be rewritten in logarithmic form?

Tip:

In an exponential function $y = a(b)^t$, if $0 < b < 1$, the function represents exponential decay, meaning it decreases over time.