The problem asks us to graph the function $f(x) = 2 \cdot (\frac{1}{2})^x$ on the provided coordinate plane.

1. Problem Description

The problem asks us to graph the function $f(x) = 2 \cdot (\frac{1}{2})^x$ on the provided coordinate plane.

2. Solution Steps

To graph the function, we will calculate the value of $f(x)$ for different values of $x$ and then plot the points on the graph.

Let's choose some values for $x$: $x = -2, -1, 0, 1, 2, 3$.

For $x = -2$:

$f(-2) = 2 \cdot (\frac{1}{2})^{-2} = 2 \cdot (2)^2 = 2 \cdot 4 = 8$

So, the point is $(-2, 8)$.

For $x = -1$:

$f(-1) = 2 \cdot (\frac{1}{2})^{-1} = 2 \cdot (2)^1 = 2 \cdot 2 = 4$

So, the point is $(-1, 4)$.

For $x = 0$:

$f(0) = 2 \cdot (\frac{1}{2})^0 = 2 \cdot 1 = 2$

So, the point is $(0, 2)$.

For $x = 1$:

$f(1) = 2 \cdot (\frac{1}{2})^1 = 2 \cdot \frac{1}{2} = 1$

So, the point is $(1, 1)$.

For $x = 2$:

$f(2) = 2 \cdot (\frac{1}{2})^2 = 2 \cdot \frac{1}{4} = \frac{1}{2} = 0.5$

So, the point is $(2, 0.5)$.

For $x = 3$:

$f(3) = 2 \cdot (\frac{1}{2})^3 = 2 \cdot \frac{1}{8} = \frac{1}{4} = 0.25$

So, the point is $(3, 0.25)$.

Now we plot the points $(-2, 8), (-1, 4), (0, 2), (1, 1), (2, 0.5), (3, 0.25)$ on the coordinate plane and draw the curve.

3. Final Answer

The graph of $f(x) = 2 \cdot (\frac{1}{2})^x$ is a decreasing exponential function that passes through the points $(-2, 8), (-1, 4), (0, 2), (1, 1), (2, 0.5), (3, 0.25)$. The graph will approach the x-axis (y=0) as x goes to infinity. The y-intercept is at $(0,2)$.

Note that without the ability to draw the graph here, the answer consists of calculating the points and a textual description of the graph.