https://file.aiquickdraw.com/mathbot/file/1727067398629md9q4itw.jpg

Let's analyze the problem in the image step by step.

### Problem 16
The problem involves a curve \( c(t) \) representing some function and a tangent line at the point \( (1.95, 7.7) \). You are required to:
1. **Estimate a second point on the tangent line**.
2. **Calculate the rate of change of the function at the labeled point**.
3. **Calculate the percentage rate of change of the function at the labeled point**.

#### Part (a) - Estimate a second point on the tangent line
To find another point on the tangent line, use the slope of the line. The line is tangent at \( t = 1.95 \) with the point \( (1.95, 7.7) \) on the curve. If you can estimate the slope, you can find another point by using the point-slope form of the equation of the tangent line:

\[
c(t) = m(t - t_0) + c_0
\]

where:
- \( m \) is the slope (rate of change of the function at \( t = 1.95 \))
- \( (t_0, c_0) = (1.95, 7.7) \)

Let's estimate the slope by examining the graph. From the graph, estimate the rise over run (i.e., \( \frac{\Delta y}{\Delta x} \)).

#### Part (b) - Calculate the rate of change of the function at the labeled point
The rate of change at the point is the derivative of the function \( c'(t) \) at \( t = 1.95 \), which is also the slope of the tangent line. We would estimate this by checking the graph to see the change in \( c(t) \) (in thousand chips) as \( t \) (hours) changes.

#### Part (c) - Calculate the percentage rate of change
The percentage rate of change at the labeled point is given by:

\[
\text{Percentage Rate of Change} = \left(\frac{c'(t)}{c(t)}\right) \times 100
\]

This value tells you the relative change in \( c(t) \) per hour at the given time \( t = 1.95 \).

### Estimations Based on the Image
- **Slope (from the graph)**: 
    - Estimate the slope by looking at the tangent line and noting that the graph seems to have a rise of about 2 units for a run of about 1 unit. Thus, the slope is approximately \( m \approx 2 \).

- **Second Point**: 
    - You can use \( t = 2.95 \) (moving right by 1 unit horizontally) and calculate the corresponding \( c(t) \) using the slope.

Would you like me to provide the detailed calculation for each part? Also, let me know if you have any specific questions or if you need further clarification! 

### Additional Practice Questions
1. If the slope of the tangent line was \( m = 3 \), what would be the second point on the tangent line?
2. How does the rate of change at \( t = 1.95 \) compare to that at \( t = 2.5 \) if the function was \( c(t) = 0.5t^2 + 2 \)?
3. Calculate the average rate of change of \( c(t) \) between \( t = 1 \) and \( t = 3 \).
4. If \( c'(t) \) were negative at \( t = 2.5 \), what would that imply about the behavior of \( c(t) \)?
5. What is the significance of the second derivative \( c''(t) \) at \( t = 1.95 \) in the context of this problem?

### Tip:
When dealing with graphs, always try to estimate the slope first as it directly relates to the rate of change and provides insight into the behavior of the function around that point.

The figure has a tangent line drawn at a labeled point (1.95, 7.7). Estimate a second point on the tangent line, calculate the rate of change of the function at the labeled point, and calculate the percentage rate of change of the function at the labeled point.

This problem involves estimating a second point on the tangent line drawn at (1.95, 7.7) and calculating the rate of change of the function at that point. It also covers the percentage rate of change, using key calculus concepts such as derivatives and tangent lines.

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