Direct Variation

  1. Make sure that Inverse variation is off and Direct variation is on. Vary the k slider in the direct variation function and observe how the graph changes.
    1. Which values of k lead to steeper graphs?
    2. Which values of k lead to flatter graphs?
  2. Set k = 2. (To quickly set a slider to a specific number, type the number into the field to the right of the slider, and then press Enter.)
    1. What is the slope of the graph when k = 2?
    2. Try other values of k. How does k relate to the slope of the line?
    3. If k > 0, is it true that the y–values of a direct variation function will always increase when the x–values increase? Why or why not?
  3. Make sure k is set equal to 2. Click on the TABLE tab. Set the MIN to 0, the MAX to 10, the STEP to 1, and then click on the Compute button.
    1. The x column of the table shows values of x from 0 through 10. The ydirect column shows the value of y = 2x for each x value. What is the value of y when x = 3?
    2. How much does the value of y change from x = 3 to x = 4? How much does it change from x = 6 to x = 7?
    3. When x changes by one unit, how much do the y–values change? How does this relate to the constant of variation, k?
  4. Imagine that you were exchanging money from one form of currency into another, where $1.00 in Currency 1 equals $1.70 in Currency 2. To model this currency conversion, set k = 1.7 in the CONTROLS view, then click on the TABLE tab and find 1 in the x column. Then look in the ydirect column and notice that the output of the direct variation function is 1.7.
    1. How much is $8.00 in Currency 1 equal to in Currency 2?
    2. How much is $50.00 in Currency 1 worth in Currency 2? (You may need to adjust the MIN, MAX, and STEP in the table to see x = 50.)

Inverse Variation

  1. In the CONTROLS view, turn off Direct variation and turn on Inverse variation. Vary the value of k using the slider and observe how the graph changes. From left to right on the graph, does y increase or decrease?
  2. Set k = 2, and click on the TABLE tab.
    1. What is the difference between the y–values for x = 1 and x = 2?
    2. How does this compare to the difference between the y–values for x = 4 and x = 5?
    3. Explain how this relates to the shape of the graph.
  3. Return to the CONTROLS view to solve the following problem. If the force needed to lift an object is inversely proportional to the number of pulleys used, how much force is needed to lift an object weighing 5 Newtons? Set k = 5.0 to represent the weight of the object. In the TABLE view, the x column represents the number of pulleys used. The yinverse column shows how much force is needed to lift an object that weighs k Newtons using x pulleys.
    1. How many Newtons of force does it take to lift an object that weighs 5 Newtons using 2 pulleys?
    2. What if 4 pulleys are used?
    3. How much force is needed to lift an object that weighs 8 Newtons in a system with 6 pulleys?