Lesson 3-3
Real-World Example 1 Find Rate of Change
MOVIES Use the table to find the rate of
change. Then explain its meaning. Number of Total Cost
Movie Tickets ($)
change in y dollars x y
rate change = change in x
tickets 3 22.50
change in cost 6 45.00
= 9 67.50
change in number of tickets
45 – 22.5
= 6–3
22.5 7.5
= or
3 1
7.5
The rate of change is 1 . This means that each ticket costs $7.50.
Real-World Example 2 Variable Rate of Change
POPULATION The graph shows the density of
population for the state of Idaho in various years.
a. Find the rates of change for 1930-1960
and 1990-2000.
1930 – 1960:
change in density 8.1 5.4 people
= Substitute.
change in time 1960 1930 years
2 .7
= or 0.09 Simplify.
30
Over this 30-year period, density increased by 2.7 people per square mile, for a rate
of change of 0.09 people per square mile per year.
1990-2000:
change in density 15.6 12.2
= Substitute.
change in time 2000 1990
3 .4
= or 0.34 Simplify.
10
Over this 10-year period, the density increased by 3.4 people per square mile, for a
rate of change of 0.34 people per square mile per year.
b. Explain the meaning of the rate of change in each case.
For 1930-1960, on average, there were 0.09 more people per square mile each year than the last.
For 1990-2000, on average, there were 0.34 more people per square mile each year than the last.
c. How are the different rates of change shown on the graph?
There is a greater vertical change for 1990-2000 than for 1930-1960. Therefore, the section of the
graph for 1990-2000 is steeper.
Example 3 Constant Rates of Change
Determine whether each function is linear. Explain.
a. b.
x y x y
–2 2 0 1
–1 7 4 3
0 12 8 6
1 17 12 10
2 22 16 15
x y x y
+1 –2 2 +5 +4 0 1 +2
–1 7 +5 4 3 +3
+1 +4
0 12 8 6
+1 +5 +4 +4
1 17 12 10
+1 2 22 +5 +4 16 15 +5
Since the x-values increase by the same Since the y-values do not increase
amount and the y-values increase by the at the same rate, this rate of change
same amount, the rate of change is is not constant. Thus, the function
constant. Thus, the function is linear. is not linear.
Example 4 Positive, Negative, and Zero Slope
Find the slope of a line that passes through each pair of points.
a. (–1, 1) and (2, 2)
Let (–1, 1) = (x1, y1) and (2, 2) = (x2, y2).
y 2 y1 rise y
m=
x 2 x1 run
2–1 (–1, 1)
= 2 – (–1) Substitute. (2, 2) x
O
1
= Simplify.
3
1
The slope is 3.
b. (1, 3) and (4, 1)
Let (1, 3) = (x1, y1) and (4, 1) = (x2, y2). y
y y rise
m= 2 1
x 2 x1 run
1 3
(1, 3)
= Substitute.
4 1
–2 (4, 1)
= 3 Simplify. O x
2
The slope is –3.
c. (–3, 6) and (4, 6)
Let (–3, 6) = (x1, y1) and (4, 6) = (x2, y2). y
y y rise
m= 2 1
x 2 x1 run (–3, 6) (4, 6)
6–6
= 4 – (–3) Substitute.
0
= –1 or 0 Simplify.
O x
Example 5 Undefined Slope
Find the slope of the line that passes through (-3, 5.5) and (-3, 2.5).
y 2 y1 rise
m=
x 2 x1 run
2 .5 5 .5
= Substitute.
3 3
–3
= 0 or undefined Simplify.
Example 6 Find Coordinates Given the Slope
Find the value of r so that the line through (-6, -3) and (-1, r) has a slope of 2.
y 2 y1
m= Slope Formula
x 2 x1
r 3
2= Let (-6, -3) = (x1, y1) and (-1, r) = (x2, y2).
1 6
2 r 3 2
= Write 2 as 1 ; subtract.
1 5
2(5) = 1(r + 3) Find the cross products.
10 = r + 3 Simplify.
10 – 3 = r + 3 – 3 Subtract 3 from each side.
7=r Simplify.
So, the line goes through (-1, 7)