To find the slope of the graph of the function $f(t) = t - 4$ at the given point $\left(\frac{1}{2}, 16\right)$, we need to compute the derivative of the function, which represents the slope of the tangent line to the graph at any point.

The function $f(t) = t - 4$ is a linear function. Its derivative with respect to $t$ is:

$f'(t) = \frac{d}{dt}(t - 4) = 1.$

Since the derivative of a linear function is constant, the slope of the graph of $f(t)$ at any point is always 1.

Therefore, the slope of the graph of the function at the point $\left(\frac{1}{2}, 16\right)$ is:

$\boxed{1}.$

Would you like more details on this solution, or do you have any other questions?

Here are some related questions for further exploration:

1. What is the equation of the tangent line to the graph at $\left(\frac{1}{2}, 16\right)$?
2. How does the slope of a linear function compare to that of a quadratic function at any given point?
3. What are the critical points of a linear function, if any?
4. How do you find the slope of a function at a point if the function is not linear?
5. How does the slope of a function relate to the concept of instantaneous rate of change?

Tip: The slope of a linear function is the same everywhere on the graph because it is constant. For non-linear functions, you need to calculate the derivative at a specific point to find the slope.