The polynomial given by 1.6t^ 2 - 28t + 200 calculates the heart rate, where t represents the elapsed time in minutes since exercise stopped.

1. What is the heart rate when the athlete first stops exercising?

2. What is the heart rate after 5 minutes?

3. Describe what happens to the heart rate after exercise stops.

I will assume that you have a typo and you meant

- 1.6t^2 - 28t + 200

since it would only decrease for the first 8 or 9 minutes, after that the heart rate would actually increase. This makes no sense.

which is it ?

To find the heart rate at a specific time using the given polynomial equation, we need to substitute the elapsed time in minutes since exercise stopped into the equation and solve for the heart rate.

1. To find the heart rate when the athlete first stops exercising (t = 0 minutes), substitute 0 into the equation:

Heart rate at t = 0 minutes = 1.6(0)^2 - 28(0) + 200 = 200

Therefore, the heart rate when the athlete first stops exercising is 200.

2. To find the heart rate after 5 minutes (t = 5 minutes), substitute 5 into the equation:

Heart rate at t = 5 minutes = 1.6(5)^2 - 28(5) + 200 = 1.6(25) - 140 + 200 = 40

Therefore, the heart rate after 5 minutes is 40.

3. To describe what happens to the heart rate after exercise stops, we need to analyze the equation. The given polynomial is a quadratic equation in the form of ax^2 + bx + c, where t represents the elapsed time in minutes since exercise stopped.

In this case, the coefficient of the t^2 term (a) is positive (1.6), which means the quadratic curve is concave upward. This suggests that initially, the heart rate will decrease until it reaches a minimum point, after which it will start increasing again.

Additionally, the coefficient of the t term (b) is negative (-28), indicating that the heart rate initially drops at a faster rate, but eventually levels off and begins to increase more gradually.

Therefore, after exercise stops, the heart rate decreases initially, reaches a minimum point, and then starts increasing over time.