The table below represents the displacement of a horse from its barn as a function of time:
time(hours) x.
0
1
2
3
4
displacement from barn (feet) y
8
58
108
158
208
part a: what is the y-intercept of the function, and what does this tell you about the horse? 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. part c: what would be the domain of the function of the horse continued to walk at this rate until it traveled 508 feet from the barn?
will rate

A. The y-intercept (b) of a linear equation is obtained when x = 0. Therefore from the given table,

y - intercept = 8

Since at time zero the displacement is 8 ft, this means that the horse was already outside the barn initially.

B. The average rate of change of the function represents the slope of the linear equation (m). This can be calculated using the formula:

average rate of change = m = (y2 – y1) / (x2 – x1)

m = (158 – 58) / (3 – 1)

m = 50

C.  Since we have determine the y-intercept and the slope, we can formulate the linear equation:

y = m x + b

y = 50 x + 8

The domain is the value of x. When y = 508, x is equivalent to

508 = 50 x + 8

x = 10 hrs

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .