The problem asks us to analyze the function $f(x) = 5 \cdot (\frac{1}{2})^x$ and graph it.

1. Problem Description

The problem asks us to analyze the function $f(x) = 5 \cdot (\frac{1}{2})^x$ and graph it.

2. Solution Steps

First, let's find a few points on the graph by plugging in some values for $x$:

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

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

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

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

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

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

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

The points we have are (0, 5), (1, 2.5), (2, 1.25), (3, 0.625), (-1, 10), and (-2, 20). Since the graph is an exponential decay, the y-values approach 0 as x approaches infinity and the y-values approach infinity as x approaches negative infinity.

3. Final Answer

The solution involves understanding the function $f(x) = 5 \cdot (\frac{1}{2})^x$ and recognizing that it is an exponential decay function. We found several points on the graph: (0, 5), (1, 2.5), (2, 1.25), (3, 0.625), (-1, 10), and (-2, 20). By plotting these points and considering the behavior of exponential decay functions, the graph can be sketched. Because I am unable to draw on the graph I am unable to provide the finished answer in the context of the actual question. However, if you plot these points on the graph and follow the curve and behavior of an exponential decay function, that should answer the question.