The following is a good example of the sort of communication requested on submitted assignments. The idea is to communicate your thinking and describe your work.
Class: Precalculus 1
Subject: f(x) Notation; The Generalized Modeling Process, Exercises 3-4
Date: **/**/****
Exercise 3.
The graph I sketched of y = f(x) has 5 index marks in increments of 20 on the y vertical dependent axis, labeled f(x), with 0 at the bottom and 80 at the top of the axis. The increments are numbered 0, 20, 40, 60, and 80. The horizontal independent x axis, labeled x value, has 6 index marks in increments of 2. The increments are 0, 2, 4, 6, 8 and 10, with 0 at the left end of the axis and 10 on the right end.
The top of the curve starts at data points ( 2,80 ) at the upper left corner of the chart. It continues downwards in a smooth, fairly steep curve to the lower right of the chart and stops at data point ( 10,25 ).
The value of x for which f(x) = 60. The value of x is 3. I calculated this by finding 60 on the y axis and then drawing a line parallel with the x axis until the line touched the curve. From where the line touched the curve I drew a vertical line down to the x axis. The point where the vertical line crossed the x axis is the value of x.
The value f(7). The value of f(7) is 30. To determine this, I located 7 on the x axis and drew a vertical line upwards until it touched the curve. Then I drew a line parallel with the x axis until it crossed the y axis. The point where the horizontal line crossed the y axis is the value of f(7).
The difference between f(7) and f(9). The difference between f(7) and f(9) is –5. I located 7 and 9 on the x axis and drew a vertical line from each upwards until it touched the curve. I then drew a horizontal line from where the points touched the curve to the y axis. The line for f(7) crossed the y axis at 30, and the line for f(9) crossed the y axis at 25. I then subtracted 30 from 25 to get –5.
The difference in x values between the points where f(x) = 70 and where f(x) = 30 is -4.5. To determine this I drew a horizontal line from the 70 and 30 indexes on the y axis until they touched the curve. I then drew vertical lines downwards from where they touched the curve until they crossed the x axis. The f(x) = 70 vertical line crossed the x axis at index 2.5. The f(x) = 30 vertical line crossed the x axis at index 7. I then subtracted 7 from 2.5.
Exercise 4.
y = T(3)
y = T(5)
y = (( T(5)) – (( T(3))
y = [(( T(5)) – (( T(3))] / 2
150 = T(t)
To find the length of time required for the temperature to fall from 80 to 30 we could draw a graph with the temperature on the y axis and time on the x axis. After the time vs. temperature coordinates were plotted and the curve drawn, we can find the length of time by locating 80 on the y axis and drawing a horizontal line until it touches the curve, at which point we draw a vertical line to the x axis. We do the same procedures again, only this time using 30 on the y axis. After the points are established on the x axis, the time it took the temperature to drop can be determined by subtracting the time from where the 80 vertical line crossed the x axis from the time where the 30 vertical line crossed the x axis.