Question

The table below represents the displacement of a horse from its barn as a function of time:

Answer 1

"Part A: What is the y-intercept of the function, and what does this tell you about the horse? (4 points)" The y-intercept is (0,8), which tells us that at the beginning the horse is 8 miles from the barn.

"Part B: Calculate the average rate of change of the function represented by the table between x = 1 to x = 3 hours, and tell what the average rate represents. (4 points)" The value of the function at x=1 is 58 and that at x=3 is 158. Thus, the change in the horse's distance from the barn is 158-58, or 100 feet. The time period involved here is 2 sec. Thus, the average rate of change of the horse's position with respect to time is

100 feet

average rate of change = ---------------- = 50 ft/sec

2 sec

If the horse were to move steadily at a fixed rate from 58 feet to 158 feet from the barn, its average rate would be 50 ft/sec.

"Part C: What would be the domain of the function if the horse continued to walk at this rate until it traveled 508 feet from the barn? (2 points)"

Here time begins at x=0 and ends at x=4 sec. Thus, the appropriate domain here is [0,4] sec.