Find the slope of the graph of the function at the given point.
f(t) = t−4;    

1
2
, 16



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.

Find the slope of the graph of the function at the given point. f(t) = t - 4; (1/2, 16)

Learn how to find the slope of the linear function f(t) = t - 4 at the point (1/2, 16). This problem covers key calculus concepts, including derivatives of linear functions and understanding the constant slope. Suitable for students in Grades 11-12.

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