Transcript PowerPoint 2

Graphing Exponential Growth Functions
An exponential function involves the expression b x where the base b is a
positive number other than 1. In this lesson you will study exponential
functions for which b > 1.
To see the basic shape of the graph of an exponential function such as f(x) = 2^x,
you can make a table of values and plot points, as shown below.
x behavior
f(x) = of
2 xthe graph.
Notice the end
2 –3which
= 18 means
As x +, –3
f(x) +,
that the graph
2 –2up= to1 the right.
–2 moves
4
–1 = 1means
0, 2which
2
As x –, –1
f(x)
that the
graph has the line y = 00 as an asymptote.
0
2 = 1
1 is a line
2 1 =that
2 a graph
An asymptote
approaches as
2 you move
2 2 = 4away from the
origin.
3
23 = 8
Graphing Exponential Growth Functions
ACTIVITY
Developing
Concepts
INVESTIGATING GRAPHS OF EXPONENTIAL FUNCTIONS
1
1
Graph y =
• 2 x and y = 3 • 2 x. Compare
3
the graphs with the graph of y = 2 x.
2
Graph y = – 1 • 2 x and y = –5 • 2 x.
5
Compare the graphs with the graph of
y = 2 x.
3
Describe the effect of a on the graph of
y = a • 2 x when a is positive and when
a is negative.
Graphing Exponential Growth Functions
In the previous slide you may have observed the following about the graph
y = a • 2 x:
The graph passes through the point (0, a).
That is, the y-intercept is a.
The x-axis is an asymptote of the graph.
The domain is all real numbers.
The range is y > 0 if a > 0 and y < 0 if a < 0.
The characteristics of the graph of y = a • 2 x listed above are true of
the graph of y = a b x. If a > 0 and b > 1, the function y = a b x is an
exponential growth function.
Graphing Exponential Functions of the Form y = ab x
Graph the function.
y=
1
• 3x
2
SOLUTION
( 12) and (1, 32). Then, from left
Plot 0,
to right, draw a curve that begins just
above the x-axis, passes through the
two points, and moves up to the right.
Graphing Exponential Functions of the Form y = ab x
Graph the function.
1
y=
• 3x
2
SOLUTION
( 12) and (1, 32). Then, from left
Plot 0,
to right, draw a curve that begins just
above the x-axis, passes through the
two points, and moves up to the right.
()
3
y= –
2
x
SOLUTION
( 32). Then,
Plot (0, –1) and 1, –
from left to right, draw a curve that
begins just below the x-axis,
passes through the two points,
and moves down to the right.
Graphing a General Exponential Function
To graph a general exponential function,
y = ab x – h + k
begin by sketching the graph of y = ab x. Then translate the graph horizontally
by h units and vertically by k units.
Graph y = 3 • 2 x – 1 – 4. State the domain and range.
SOLUTION
Begin by lightly sketching the graph of y = 3 • 2 x, which
passes through (0, 3) and (1, 6). Then translate
the graph 1 unit to the right and 4 units down.
Notice that the graph passes through
(1, –1) and (2, 2). The graph’s asymptote is the line y
= – 4. The domain is all real numbers, and the
range is y > – 4.
Using Exponential Growth Models
When a real-life quantity increases by a fixed percent each year (or other
time period), the amount y of quantity after t years can be modeled by this
equation:
y = a(1 + r) t
In this model, a is the initial amount and r is the percent increase expressed
as a decimal.
The quantity 1 + r is called the growth factor.
Modeling Exponential Growth
INTERNET HOSTS In January, 1993, there were about 1,313,000 Internet
hosts. During the next five years, the number of hosts increased by about
100% per year.
Write a model giving the number h (in millions) of hosts t years after 1993.
About how many hosts were there in 1996?
SOLUTION
The initial amount is a = 1.313 and the percent increase is r = 1. So, the
exponential growth model is:
h = a(1 + r) t
Write exponential growth model.
= 1.313(1 + 1) t
Substitute for a and r.
= 1.313 • 2 t
Simplify.
Using this model, you can estimate the number of hosts in 1996 (t = 3) to
be h = 1.313 • 2 3  10.5 million.
Modeling Exponential Growth
INTERNET HOSTS In January, 1993, there were about 1,313,000 Internet
hosts. During the next five years, the number of hosts increased by about
100% per year.
Graph the model.
SOLUTION
The graph passes through the points (0, 1.313)
and (1, 2.626). It has the t-axis as an asymptote.
To make an accurate graph, plot a few other
points. Then draw a smooth curve through
the points.
Modeling Exponential Growth
INTERNET HOSTS In January, 1993, there were about 1,313,000 Internet
hosts. During the next five years, the number of hosts increased by about
100% per year.
Use the graph to estimate the year when there
were 30 million hosts.
SOLUTION
Using the graph, you can estimate that the
number of hosts was 30 million sometime
during 1997 (t  4.5).
Modeling Exponential Growth
In the previous example the annual percent increase was 100%. This
translated into a growth factor of 2, which means that the number of
Internet hosts doubled each year.
People often confuse percent increase and growth factor, especially when
a percent increase is 100% or more.
For example, a percent increase of 200% means that a quantity tripled,
because the growth factor is 1 + 2 = 3.
When you hear or read reports of how a quantity has changed, be sure to pay
attention to whether a percent increase or a growth factor is being discussed.
Modeling Exponential Growth
In January, 1980, there were about 2,180,000 workers worked at home.
During the next ten years, the number of workers working at home
increased 5% per year.
a) Write a model giving the number w (in millions) of workers working at
home t years after 1980. About how many workers worked at home in
1983?
b) Graph the model.
c) Use the graph to estimate the year when there were 3.22 million
workers who worked at home.
Using Exponential Growth Models
COMPOUND INTEREST Exponential growth functions are used in real-life
situations involving compound interest. Compound interest is interest paid
on the initial investment, called the principal, and on previously earned
interest. (Interest paid only on the principal is called simple interest.)
Although interest earned is expressed as an annual percent, the interest is
usually compounded more frequently than once per year. Therefore, the
formula y = a(1 + r) t must be modified for compound interest problems.
COMPOUND INTEREST
Consider an initial principal P deposited in an account that pays interest at
an annual rate r (expressed as a decimal), compounded n times per year.
The amount A in the account after t years can be modeled by this equation:
( )
A=P 1 +
r
n
nt
Finding the Balance in an Account
FINANCE You deposit $1000 in an account that pays 8% annual interest.
Find the balance after 1 year if the interest is compounded with the given
frequency.
annually
SOLUTION
With interest compounded annually, the balance at the end of 1 year is:
nt
( )
0.08
A = 1000 (1 +
1 )
r
A=P 1 +
n
Write compound interest model.
1•1
P = 1000, r = 0.08, n = 1, t = 1
= 1000(1.08)1
Simplify.
= 1080
Use a calculator.
The balance at the end of 1 year is $1080.
Finding the Balance in an Account
FINANCE You deposit $1000 in an account that pays 8% annual interest.
Find the balance after 1 year if the interest is compounded with the given
frequency.
quarterly
SOLUTION
With interest compounded quarterly, the balance at the end of 1 year is:
nt
( )
0.08
A = 1000 (1 +
4 )
A=P 1 +
r
n
Write compound interest model.
4•1
P = 1000, r = 0.08, n = 4, t = 1
= 1000(1.02)4
Simplify.
 1082.43
Use a calculator.
The balance at the end of 1 year is $1082.43.
Finding the Balance in an Account
FINANCE You deposit $1000 in an account that pays 8% annual interest.
Find the balance after 1 year if the interest is compounded with the given
frequency.
daily
SOLUTION
With interest compounded daily, the balance at the end of 1 year is:
nt
( )
0.08
A = 1000 (1 +
365 )
r
A=P 1 +
n
1000(1.000219) 365
 1083.28
Write compound interest model.
365 • 1
P = 1000, r = 0.08, n = 365, t = 1
Simplify.
Use a calculator.
The balance at the end of 1 year is $1083.28.
Finding the Balance in an Account
FINANCE You deposit $1500 in an account that pays 6% annual interest.
Find the balance after 1 year if the interest is compounded with the given
frequency.
a) annually
b) semiannually
c) quarterly