Date: 5-8-12
Topic: 8-10 Inverse
Variation
Objective
Essential Question: What is an inverse variation? Give an
example.
To use inverse variation to solve problems.
The table shows the time, t, that it takes a car to travel a
distance of 40 mi at the speed of r mi/h.
Rate in mi/h: r
Time in hours: t
20
2
30
43
40
1
50
45
You can see that
Notice that if the speed is increased, the time is decreased, so
that the product is always 40. You can say that the time varies
inversely as the rate. This example illustrates an inverse
variation.
Inverse Variation
An inverse variation is a function defined by an equation
of the form
𝑥𝑦 𝑘, where 𝑘 is a nonzero constant,
or
𝑘
𝑦
, 𝑤ℎ𝑒𝑟𝑒 𝑥 ≠
𝑥
You say that y varies inversely as x or that y is inversely
proportional to x. The constant k is the constant of
variation.
Summary
1
The graph of an inverse function is not a straight line, since the
equation
not linear. The term xy is of degree 2.
Example 1
Solution
Graph the equation
Calculate a table of values. Then graph the points.
The graph of
shown in Example 1 is called a
hyperbola. Since neither x nor y can have the value 0. the
graph does not intersect either the x-axis or the y-axis.
For every nonzero value of k, the graph of 𝑥𝑦
hyperbola.
𝑘 is a
When k is positive, the branches of the graph are in
Quadrants I and III.
When k is negative, the branches of the graph are in
Quadrants II and IV.
2
Exercise 1
Make a table of values and graph the equation,
Be sure to draw a smooth curve through the points.
x
y
x
y
6
3
2
1
1
2
3
6
3
Let ( , ) and ( , ) be two ordered pairs of the same
inverse variation. Since the coordinates must satisfy the
equation
, you know that
,
and
or
You can compare the equations for direct variation and inverse
variation.
Direct Variation
Inverse Variation
, or
The equations above show that for direct variation the
quotients of the coordinates are constant and for inverse
variation the products of the coordinates are constant.
One example of an inverse variation is the law of the lever. A
lever is a bar pivoted at a point called the fulcrum. If masses
m1 and m2 are placed at distances d1 and d2 from the fulcrum,
and the bar is balanced, then
4
Example 2
Solution
If a 24 g mass is 30 cm from the fulcrum of a lever, how far
from the fulcrum is a 45 g mass that balances the 24 g mass?
Let
,
, and
Use
,
.
 the distance of the 45 g mass from the fulcrum is 16 cm.
Exercise 1
Let
Let
Homework
,
,
, and
, and
, find
, find
.
.
P 399 Written Exercises: 1, 5, 11’ 15, 19
P 400 Problems: 1-17, every 4th.
5