Compare a linear function to an exponential function.
Complete the table below, then graph each function on the same chart.
x y = 2x y = 2x
0 0 1
1 2 2
2 4 4
3 6 8
4 8 16
5 10 32
In changing from x = 1 to x = 2 in the equation y = 2x, y increases by what amount?
2
In changing from x = 1 to x = 2 in the equation y = 2x, y increases by what amount?
2
In changing from x = 2 to x = 3 in the equation y = 2x, y increases by what amount?
6 5
4 3
2 1
0 0 5 10 15 20 25 30 35
y = 2x
6 5
4 3
2 1
0 0 5 10 15 20 25 30 35
y = 2x
2
In changing from x = 2 to x = 3 in the equation y = 2x, y increases by what amount?
Properties/Characteristics of Exponential Functions
The basic form of an exponential equation is:
The variable x is the exponent in the statement above. It can be any real number. The base is b.
Can the base number b, be any value or are there restrictions on it? Do the following four problems to find out.
1.) For this problem, the base value is 1. What happens when this value is chosen? Complete the chart to see what happens:
x
y
= 1
x-2 1 Conclusion based upon the results to the left:
-1 1 no matter the value of x, y is equal to one.
0 1
1 1
2 1
3 1
2.) For this problem, the base value is a negative number. What happens when this value is chosen? Complete the chart to see what happens:
x
y
= (-2)
x-3 -0.125
-2.5 #NUM!
-2 0.25 Conclusion based upon the results to the left:
x
-1.5 #NUM! whole number values work for x, but other values such as 3/2 do not…
-1 -0.5
-0.5 #NUM!
0 1
0.5 #NUM!
1 -2
1.5 #NUM!
2 4
2.5 #NUM!
3 -8
3.5 #NUM!
4 16
4.5 #NUM!
5 -32
3.) For this problem, the base value is a number less than 1 but greater than 0. Graph the results using the XY Scatter (smoothed lines) option.
x
y
= (2/3)
x-4 5.0625
-3.5 4.133513941
-2 2.25
-0.5 1.224744871
0 1
12 10
8 6
4 2
0 -2
-4 0
1 2 3 4 5 6
As x increases, y approaches ___.
1 0.666666667 2 0.444444444 3 0.296296296 4 0.197530864 5 0.131687243 6 0.087791495 7 0.058527663 8 0.039018442 9 0.026012295 10 0.01734153 11 0.01156102 12 0.007707347
4.) For this problem, the base value is a number greater than 1. Graph the results using the XY Scatter (smoothed lines) option.
x
y
= 1.5
x-6 0.087791495 -5 0.131687243 -4 0.197530864 -3 0.296296296 -2 0.444444444 -1 0.666666667 0 1 10 8 6 4 2 0 -2 -4 -6 0 10 20 30 40 50 60
Note the y-intercept. Does it change if the b value
changes?
As x decreases, y approaches ___.
This is an example of exponential growth. Such as money compounded in a savings account.
12 10 8 6 4 2 0 -2 -4 0 1 2 3 4 5 6
1 1.5
2 2.25
3 3.375
4 5.0625
5 7.59375
6 11.390625
7 17.0859375
8 25.62890625
9 38.44335938
10 57.66503906
Note the y-intercept. Does it change if the b value
changes?
10 8
6 4
2 0
-2 -4
-6 0
Using Microsoft Excel:
–Graph the function y = 3x for x = -3 to 3 (in 0.5 increments)
–Graph the function y = (1/3)x for x = -3 to 3 (in 0.5 increments)
Graphing Exponential Functions
0 5 10 15 20 25 30
Application Problems Involving Exponential Functions
1.) The amount (in milligrams) of a drug in the body t hours after taking a pill is given by the function :
a) Graph this function for t = 0 to 12 hours, in 1-hr increments. b) Determine the amount of drug left in the body after 10 hours.
t A
0 25
1 21.25
2 18.062 3 15.353
4 13.05
5 11.093 6 9.4287 7 8.0144 8 6.8123 9 5.7904 10 4.9219 11 4.1836
12 3.556
14 12
10 8
6 4
2 0
0 5 10 15 20 25 30
Series1
t
t
2.) In 1980 there were about 203,000 cell phone subscribers in the U.S. Since then, there has been an exponential growth in the number of cell phone subscribers.
This growth can be modeled using the function:
where C represents the number of cell subscribers (in millions) and t represents the year.
a) Graph this function for t = 1981 to 2007 in 1 year increments. b) Determine the number of cell phone subscribers in 2005.
1981 0.0407 1982 0.0608 1983 0.0909 1984 0.1359 1985 0.2031 1986 0.3037 1987 0.454 1988 0.6787 1989 1.0147 1990 1.517 1991 2.2679 1992 3.3905 2010 2005 2000 1995 1990 1985 1980 0 200 400 600 800 1000 1200 1400 1600
19801993 5.0688 1994 7.5778 1995 11.329 1996 16.937
1997 25.32
1998 37.854 1999 56.592 2000 84.604 2001 126.48 2002 189.09 2003 282.69 2004 422.63 2005 631.83 2006 944.58 2007 1412.2
2010 2005
2000 1995
1990 1985
Application: Compound Interest
The compound interest formula is an example of an exponential function because the exponent is a variable.
A is the accumulated amount.
P is the principle (original amt. invested).
t is the time in years.
r is the interest rate per year.
n is the number of compounding periods per year.
Make a graph that demonstrates how an initial invest would grow over a time frame of 25 years. Principle = $500
The interest is 4.25% compounded daily.
Principle 500
Interest Rate 4.25%
Number of Compounding Periods 365 per year
Time Amount
1 $ 521.71
2 $ 544.36
3 $ 567.99
P is the principle (original amt. invested).
n is the number of compounding periods per year.
Amount
30 25
20 15
10 5
4 $ 592.65
5 $ 618.38
6 $ 645.22
7 $ 673.23
8 $ 702.46
9 $ 732.96
10 $ 764.78
11 $ 797.98
12 $ 832.62
13 $ 868.77
14 $ 906.48
15 $ 945.84
16 $ 986.90
17 $ 1,029.74
18 $ 1,074.45
19 $ 1,121.09
20 $ 1,169.77
21 $ 1,220.55
22 $ 1,273.54
23 $ 1,328.83
24 $ 1,386.52
25 $ 1,446.71
Amount
30 25
20 15
10 5
Amount
30 25
20 15
10 5
The natural exponential e.
We use the irrational number p when working with applications involving circles.
Another commonly used irrational number, e, is used when working with certain natural applications.
Complete the table below to determine the approximate decimal represenation of e.
n
1 2
2 2.25
3 2.37037037
10 2.59374246
1,000 2.716923932
10,000 2.718145927
Based upon the pattern established in the completed table above,
it would appear that the value of e as n grows larger and larger, approaches what approximate value?
2.71
n n
1
Application: Atmospheric Pressure
The formula for the atmospheric pressure (in millimeters of mercury) is:
h is the height in meters.
Complete the table below using the pressure formula.
Using the computed information make a graph to show how pressure changes as height increases from 0 m to 2000 m.
h P
0 760
100 750.184
200 740.495
300 730.931
400 721.49
500 712.171
600 702.973
700 693.893
800 684.931
900 676.085
1000 667.353
1100 658.733
1200 650.225
Atmospheric Pressue in mm of mercury for heights from 0 to 2000 meters.
2500 2000
1500 1000
500 0
500 550 600 650 700 750 800
Height in meters.
Pressure in mm of mercury.
h
e
Using the computed information make a graph to show how pressure changes as height increases from 0 m to 2000 m.
Atmospheric Pressue in mm of mercury for heights from 0 to 2000 meters.
2500 2000
1500 1000
500 0
500 550 600 650 700 750 800
Height in meters.
1300 641.827
1400 633.537
1500 625.354
1600 617.277
1700 609.305
1800 601.435
1900 593.667
2000 585.999
Atmospheric Pressue in mm of mercury for heights from 0 to 2000 meters.
2500 2000
1500 1000
500 0
500 550 600 650 700 750 800
Height in meters.
Atmospheric Pressue in mm of mercury for heights from 0 to 2000 meters.
2500 2000
1500 1000
500 0
500 550 600 650 700 750 800
Height in meters.
Application: Determine how well a certain chemical dissolves at various temperature levels.
The number of grams of a chemical that will dissolve in a solution is given by the formula:
where t is the temperature in degrees Celsius.
Create a graph to show how the quantity (in grams) of this chemical that will dissolve in a solution over a temperature range of 0 to 200 degrees Celsius.
t C
0 100
10 122.1
20 149.2
50 271.8
100 738.9
200 5460
Amount of chemical (in grams) that will dissolve in a solution.
250 200
150 100
50 0
0 1000 2000 3000 4000 5000 6000
Temperature in degrees Celsius
Grams of chemical that will dissolve.
t
e
Amount of chemical (in grams) that will dissolve in a solution.
250 200
150 100
50 0
0 1000 2000 3000 4000 5000 6000
Temperature in degrees Celsius
Amount of chemical (in grams) that will dissolve in a solution.
250 200
150 100
50 0
0 1000 2000 3000 4000 5000 6000
Temperature in degrees Celsius
Amount of chemical (in grams) that will dissolve in a solution.
250 200
150 100
50 0
0 1000 2000 3000 4000 5000 6000
Temperature in degrees Celsius