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Learning Objectives

At the end of this lesson, students will be able to:

  • Distinguish direct and inverse variation.
  • Graph inverse variation equations.
  • Write inverse variation equations.
  • Solve real-world problems using inverse variation equations.

Vocabulary

Terms introduced in this lesson:

variation
direct variation
inverse variation
joint variation
constant of proportionality
increase, decrease

Teaching Strategies and Tips

This lesson focuses on inverse variation models and graphing inverse variation equations. Use it to motivate rational functions, which are covered in the next six lessons.

Remind students about having learned direct variation in chapter Graphs of Equations and Functions.

  • Point out that direct variation is a linear relationship.
  • The x and y-intercepts are 0.
  • The slope of the line is the only parameter, denoted by k, and called the constant of proportionality.
  • It takes only one more point to determine the direct variation.

Some examples of direct variation relationships are:

  • Height of a person and the length of their shadow on flat ground.
  • Circumference and radius of the circle.
  • Weight of an object on a spring and the amount the spring has stretched.

In the examples and Review Questions, have students decide on a variation model first and then solve for the constant of proportionality using the given information. This determines the equation of the variation which is necessary for answering the rest of the problem.

Use Example 1 to illustrate the graph of an inverse variation.

  • Construct a similar table of values. Allow students to observe the function’s behavior numerically.

Remind students of scientific notation in Example 6.

In applied problems such as Examples 5 and 6, emphasize that direct variations are ubiquitous and significant in the real-world.

In Review Questions 1-4, encourage students to apply stretches to the basic graph y = \frac{1}{x}.

Example:

Graph the following inverse variation relationship.

y = \frac{10}{x}.

Hint: Since y = \frac{10}{x} = 10 \cdot \frac{1}{x}, the graph can be obtained from that of y = \frac{1}{x} by stretching by a factor of 10.

Error Troubleshooting

Remind students in Example 1 that dividing by zero is undefined.

Remind students in Example 6 to square the 5.3 which is in parentheses:

K = 740(5.3 \times 10^{-11})^2 = 740 \cdot 5.3^2 \cdot 10^{-22}

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Feb 22, 2012

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Apr 29, 2014
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