Constant Rate of Change
and direct variation
Essential Question:
What does a linear graph that is a straight line tell you about the rate of change?
The graph show the relationship
between time and distance of a car
traveling 55 miles per hour.
Travel TIme
y
distance (mi)
Unit 8 Day 4
A. Choose any two points on the graph and find the rate of change.
400
300
200
100
(4, 220)
(3, 165)
(2, 110)
(1, 55)
1
2
3
x
4 5 6
time (hr)
B. Repeat part a with different points.
What is the rate of change
C. What do you notice about the rate of change between any
two points?
The graph for the Travel Time is a straight line. Relationships that have straight­line graphs are called linear relationships. Notice that as the time in hours increase by 1, the distance increases by 55.
Time (hr)
0
Distance (mi)
0 55 110 165 220
1
2
3
4
change in distance = 55 or 55 mph
change in time 1 Constant Rate of Change
Constant Rate of Change
Not a Constant Rate of Change
The rate of change between any two points in a linear relationship is the same or constant. We describe this as a constant rate of change.
Some linear relationships are proportional. If the ratios are equal, the linear relationship is proportional. If the ratios are not equal,
the linear relationship is nonproportional. number of people x
1 2 3 4
4 8 12 16
Cost of parking y 4 = 4 16 = 4 8 = 4 12 = 4
Number of people x 1 2 3 4 Number of people x
1
Cost of tickets y 13 = 11 31 =10
= 13 22 =11.3 40 Number of people x 1 2 3 4 Cost of tickets y
13 22 31
cost of parking y
2
3
4
The ratios are equal, so this is proportional. 40
The ratios are not equal, so this is nonproportional. A special type of linear equation that describes constant rate of change is called a direct variation. This graph always passes through the origin (0, 0) and is a proportional relationship.
The constant of variation is k in the equation y = kx. You can y You can use any point on the line, or on a table, find k by k = . x
or from a word problem to find the value of k. Let's try find the direct variation in each situation:
information time (mb)
(s)
x
y
2.5
3.75
10
15
10
40
y = kx k = y k = x The cost of cheese varies directly with the number of pounds bought. Suppose 2 pounds of cheese cost $8.40. Write an equation that could be used to find the unit cost of cheese.
y = kx
replace k with . y = x
k = y
x
Find the cost of 3.5 pounds of cheese.