The problem asks to analyze the function $y = 5 \cdot (\frac{1}{2})^x$.

1. Problem Description

The problem asks to analyze the function $y = 5 \cdot (\frac{1}{2})^x$.

2. Solution Steps

To understand the behavior of the function $y = 5 \cdot (\frac{1}{2})^x$, we can calculate some values for different values of $x$.

When $x = 0$, we have $y = 5 \cdot (\frac{1}{2})^0 = 5 \cdot 1 = 5$.

When $x = 1$, we have $y = 5 \cdot (\frac{1}{2})^1 = 5 \cdot \frac{1}{2} = 2.5$.

When $x = 2$, we have $y = 5 \cdot (\frac{1}{2})^2 = 5 \cdot \frac{1}{4} = 1.25$.

When $x = 3$, we have $y = 5 \cdot (\frac{1}{2})^3 = 5 \cdot \frac{1}{8} = 0.625$.

When $x = -1$, we have $y = 5 \cdot (\frac{1}{2})^{-1} = 5 \cdot 2 = 10$.

When $x = -2$, we have $y = 5 \cdot (\frac{1}{2})^{-2} = 5 \cdot 4 = 20$.

When $x = -3$, we have $y = 5 \cdot (\frac{1}{2})^{-3} = 5 \cdot 8 = 40$.

The function is an exponential decay. As $x$ increases, $y$ approaches

0. As $x$ decreases, $y$ increases without bound. The y-intercept is $(0, 5)$.

3. Final Answer

The function is $y = 5 \cdot (\frac{1}{2})^x$. It is an exponential decay function.