To find the relative extreme points of the temperature function $T(t) = -0.3t^2 + 2.4t + 98.5$ over the interval $[0, 8]$, we first need to determine the critical points by taking the derivative and setting it to zero.

1. Calculate the derivative: $T'(t) = -0.6t + 2.4$

2. Set the derivative equal to zero to find critical points: $-0.6t + 2.4 = 0$ $0.6t = 2.4$ $t = 4$

3. Evaluate $T(t)$ at the critical point and endpoints:

   • At $t = 0$: $T(0) = -0.3(0)^2 + 2.4(0) + 98.5 = 98.5$
   • At $t = 4$: $T(4) = -0.3(4)^2 + 2.4(4) + 98.5 = -0.3(16) + 9.6 + 98.5 = -4.8 + 9.6 + 98.5 = 103.3$
   • At $t = 8$: $T(8) = -0.3(8)^2 + 2.4(8) + 98.5 = -0.3(64) + 19.2 + 98.5 = -19.2 + 19.2 + 98.5 = 98.5$

4. Summarize the values:

   • $T(0) = 98.5$
   • $T(4) = 103.3$ (relative maximum)
   • $T(8) = 98.5$

The relative extreme point is at $t = 4$ with a maximum temperature of $103.3$ °F. The function decreases before and after this point within the interval.

Graph Sketch

To sketch the graph, plot the points:

• (0, 98.5)
• (4, 103.3)
• (8, 98.5)

The parabola opens downwards, with a vertex at $(4, 103.3)$.

Summary

• Relative maximum at $t = 4$ with $T(4) = 103.3$ °F.
• The temperature decreases on either side of this point within the interval $[0, 8]$.

Would you like more details or have any questions? Here are some related questions you might consider:

1. What is the significance of finding relative extreme points in this context?
2. How would the graph change if the coefficient of $t^2$ were positive?
3. What does the negative coefficient of $t^2$ indicate about the temperature trend?
4. Can you explain the role of derivatives in finding extreme points?
5. How would you interpret the results if $T(t)$ represented a different scenario, like a medication effect over time?

Tip: Always check the endpoints of your interval when looking for relative extrema!