To solve this problem, let’s analyze the given functions and determine the $y$-intercepts of each function.

Step 1: Find $a$ (the $y$-intercept of $f(x)$)

The function $f(x) = 3(2)^x$. The $y$-intercept occurs when $x = 0$.

$f(0) = 3(2)^0 = 3 \cdot 1 = 3$

Thus, $a = 3$.

Step 2: Find $b$ (the $y$-intercept of $g(x)$)

The function $g(x) = 3(2)^x - 6$. Similarly, the $y$-intercept occurs when $x = 0$.

$g(0) = 3(2)^0 - 6 = 3 \cdot 1 - 6 = 3 - 6 = -3$

Thus, $b = -3$.

Step 3: Calculate $a - b$

Now that we know $a = 3$ and $b = -3$, we can find $a - b$:

$a - b = 3 - (-3) = 3 + 3 = 6$

Final Answer

The value of $a - b$ is 6.

Would you like further details on any part of this solution? Here are some additional related questions to consider:

1. How would the result change if $f(x)$ or $g(x)$ had different constants?
2. What is the significance of the base of the exponential function in $f(x)$ and $g(x)$?
3. How would we find the $y$-intercepts if $f(x)$ and $g(x)$ were linear functions?
4. What does shifting the function $f(x)$ by a constant do to its $y$-intercept?
5. How could we calculate the difference between the intercepts if the functions were trigonometric?

Tip: When finding the $y$-intercept of a function, set $x = 0$ and evaluate the function at that point.