* Evaluate exponential functions.
* Find the equation of an exponential function.
* Use compound interest formulas.
* Evaluate exponential functions with base e.
2. Concepts & Objectives
⚫ Exponential Functions
⚫ Evaluate exponential functions.
⚫ Find the equation of an exponential function.
⚫ Use compound interest formulas.
⚫ Evaluate exponential functions with base e.
3. Properties of Exponents
⚫ Recall that for a variable x and integers a and b:
+
=
a b a b
x x x
−
=
a
a b
b
x
x
x
( ) =
b
a ab
x x
= b
a b a
x x
= =
a b
x x a b
5. Simplifying Exponents
⚫ Example: Simplify
⚫ 1.
⚫ 2.
⚫ 3.
−
3 2 2
2 5
25
5
x y z
xy z
( )
−
4
2 3
2r s t
−
2 3 1 2
5 6
y y
− − − −
= 3 1 2 2 2 5
5x y z
( ) ( )
−
=
4 2 4 3
4 4
2 r s t
( )( )
−
=
2 1
3 2
5 6 y
− −
= 2 4 3
5x y z
−
= 8 12 4
16r s t
=
1
6
30y
6. Exponential Functions
⚫ If a > 0 and a 1, then
defines the exponential function with base a.
⚫ Example: Graph
⚫ Domain: (–∞, ∞)
⚫ Range: (0, ∞)
⚫ y-intercept: (0, 1)
( )= x
f x a
( )=2x
f x
7. Exponential Functions (cont.)
Characteristics of the graph of :
1. The points are on the graph.
2. If a > 1, then f is an increasing function; if 0 < a < 1, then
f is a decreasing function.
3. The x-axis is a horizontal asymptote.
4. The domain is (–∞, ∞), and the range is (0, ∞).
( )= x
f x a
( ) ( )
−
1
1, , 0,1 , 1,a
a
8. Exponential Growth/Decay
⚫ A function that models exponential growth grows by a
rate proportional to the amount present. For any real
number x and any positive real numbers a and b such
that b ≠ 1, an exponential growth function has the form
where
⚫ a is the initial or starting value of the function
⚫ b is the growth factor or growth multiplier per unit x
( ) x
f x ab
=
9. Exponential vs. Linear
⚫ Consider companies A and B. Company A has 100 stores
and expands by opening 50 new stores a year, so its
growth can be represented by the function
Company B has 100 stores and expands by increasing
the number of stores by 50% each year, so its growth
can be represented by the function
( ) 100 50
A x x
= +
( ) ( )
100 1 0.5
x
B x = +
10. Exponential vs. Linear (cont.)
⚫ A few years of growth for these companies are
illustrated in the table:
Year Company A Company B
0 100 + 50(0) = 100 100(1+0.5)0 = 100
1 100 + 50(1) = 150 100(1+0.5)1 = 150
2 100 + 50(2) = 200 100(1+0.5)2 = 225
3 100 + 50(3) = 250 100(1+0.5)3 = 337.5
x A(x) = 100 + 50x B(x) = 100(1+0.5)x
11. Exponential vs. Linear (cont.)
⚫ Graphing the two functions over 5 years shows the
difference even more clearly.
B(x)
A(x)
12. Writing Exponential Functions
⚫ Given two data points, how do we write an exponential
model?
1. If one of the data points has the form (0, a) (the y-
intercept), then a is the initial value. Substitute a
into the equation y = a(b)x, and solve for b with the
second set of values.
2. Otherwise, substitute both points into two
equations with the form and solve the system.
3. Using a and b found in the steps 1 or 2, write the
exponential function in the form f(x) = a(b)x.
13. Writing Exponential Functions
Example: In 2006, 80 deer were introduced into a wildlife
refuge. By 2012, the population had grown to 180 deer.
The population was growing exponentially. Write an
exponential function N(t) representing the population (N)
of deer over time t.
14. Writing Exponential Functions
Example: In 2006, 80 deer were introduced into a wildlife
refuge. By 2012, the population had grown to 180 deer.
The population was growing exponentially. Write an
exponential function N(t) representing the population (N)
of deer over time t.
If we let t be the number of years after 2006, we can write
the information in the problem as two ordered pairs:
(0, 80) and (6, 180). We also have an initial value, so a = 80,
and we can use the process in step 1.
15. Writing Exponential Functions
Set up the initial equation (y = a(b)x) and substitute a and
the second set of values into it.
Thus, the function becomes
( )
6
6
1
6
180 80
180 9
80 4
9
1.1447
4
b
b
b
=
= =
=
( ) ( )
80 1.447
t
N t =
16. Writing Exponential Functions
Example: Find an exponential function that passes through
the points (‒2, 6) and (2, 1).
Since we don’t have an initial value, we will need to set up
and solve a system. It will usually be simplest to use the
first equation with the first set of values to solve for a, and
then substitute that into the second equation with the
second set of values to solve for b.
17. Writing Exponential Functions
Example: Find an exponential function that passes through
the points (‒2, 6) and (2, 1).
Thus, the function is
2
2
2
6
6
6
ab
a
b
a b
−
=
=
=
( )
2 2
4
4
1
4
1 6
1 6
1
6
1
0.6389
6
b b
b
b
b
=
=
=
=
( )
2
6 0.6389
2.4492
a =
=
( ) ( )
2.4492 0.6389
x
f x =
18. Writing Exponential Functions
Example: Find an exponential function that passes through
the points (‒2, 6) and (2, 1).
It may (or may not) surprise you to
learn that Desmos also offers us a
shortcut for this.
( ) ( )
2.4495 0.6389
x
f x =
19. Compound Interest
⚫ The formula for compound interest (interest paid on
both principal and interest) is an important application
of exponential functions.
⚫ Recall that the formula for simple interest, I = Prt, where
P is principal (amount deposited), r is annual rate of
interest, and t is time in years.
20. Compound Interest (cont.)
⚫ Now, suppose we deposit $1000 at 10% annual interest.
At the end of the first year, we have
so our account now has 1000 + .1(1000) = $1100.
⚫ At the end of the second year, we have
so our account now has 1100 + .1(1100) = $1210.
( )( )
= =
1000 0.1 100
I
( )( )
= =
1100 .1 110
I
21. Compound Interest (cont.)
⚫ Another way to write 1000 + .1(1000) is
⚫ After the second year, this gives us
( ) ( )
( )
+ + +
1000 1 .1 .1 1000 1 .1 ( )( )
= + +
1000 1 .1 1 .1
( )
= +
2
1000 1 .1
( )
+
1000 1 .1
22. Compound Interest (cont.)
⚫ If we continue, we end up with
This leads us to the general formula.
Year Account
1 $1100 1000(1 + .1)
2 $1210 1000(1 + .1)2
3 $1331 1000(1 + .1)3
4 $1464.10 1000(1 + .1)4
t 1000(1 + .1)t
23. Compound Interest Formulas
⚫ For interest compounded n times per year:
⚫ For interest compounded continuously:
where e is the irrational constant 2.718281…
= +
1
tn
r
A P
n
= rt
A Pe
24. Examples
1. If $2500 is deposited in an account paying 6% per year
compounded twice per year, how much is the account
worth after 10 years with no withdrawals?
2. What amount deposited today at 4.8% compounded
quarterly will give $15,000 in 8 years?
25. Examples
1. If $2500 is deposited in an account paying 6% per year
compounded twice per year, how much is the account
worth after 10 years with no withdrawals?
2. What amount deposited today at 4.8% compounded
quarterly will give $15,000 in 8 years?
( )
= +
2 10
.06
2500 1
2
A P = 2500, r = .06,
n = 2, t = 10
26. Examples
1. If $2500 is deposited in an account paying 6% per year
compounded twice per year, how much is the account
worth after 10 years with no withdrawals?
2. What amount deposited today at 4.8% compounded
quarterly will give $15,000 in 8 years?
( )
= +
2 10
.06
2500 1
2
A
27. Examples
1. If $2500 is deposited in an account paying 6% per year
compounded twice per year, how much is the account
worth after 10 years with no withdrawals?
2. What amount deposited today at 4.8% compounded
quarterly will give $15,000 in 8 years?
( )
= +
2 10
.06
2500 1
2
A = $4515.28
28. Examples
1. If $2500 is deposited in an account paying 6% per year
compounded twice per year, how much is the account
worth after 10 years with no withdrawals?
2. What amount deposited today at 4.8% compounded
quarterly will give $15,000 in 8 years?
( )
= +
2 10
.06
2500 1
2
A
( )
= +
4 8
.048
15000 1
4
P
A = 15000, r = .048,
n = 4, t = 8
= $4515.28
29. Examples
1. If $2500 is deposited in an account paying 6% per year
compounded twice per year, how much is the account
worth after 10 years with no withdrawals?
2. What amount deposited today at 4.8% compounded
quarterly will give $15,000 in 8 years?
( )
= +
2 10
.06
2500 1
2
A
( )
= +
4 8
.048
15000 1
4
P
( )
15000 1.4648
P
= $4515.28
30. Examples
1. If $2500 is deposited in an account paying 6% per year
compounded twice per year, how much is the account
worth after 10 years with no withdrawals?
2. What amount deposited today at 4.8% compounded
quarterly will give $15,000 in 8 years?
( )
= +
2 10
.06
2500 1
2
A
( )
= +
4 8
.048
15000 1
4
P
( )
15000 1.4648
P
= $4515.28
31. Examples
1. If $2500 is deposited in an account paying 6% per year
compounded twice per year, how much is the account
worth after 10 years with no withdrawals?
2. What amount deposited today at 4.8% compounded
quarterly will give $15,000 in 8 years?
( )
= +
2 10
.06
2500 1
2
A
( )
= +
4 8
.048
15000 1
4
P
( )
15000 1.4648
P
= $10,240.35
P
= $4515.28
32. Examples
3. If $8000 is deposited in an account paying 5% interest
compounded continuously, how much is the account
worth at the end of 6 years?
4. Which is a better deal, depositing $7000 at 6.25%
compounded every month for 5 years or 5.75%
compounded continuously for 6 years?
33. Examples
3. If $8000 is deposited in an account paying 5% interest
compounded continuously, how much is the account
worth at the end of 6 years?
4. Which is a better deal, depositing $7000 at 6.25%
compounded every month for 5 years or 5.75%
compounded continuously for 6 years?
( )( )
=
.05 6
8000
A e
= $10,798.87
A
( )
= +
=
12 5
.0625
7000 1
12
$9560.11
A
( )( )
=
=
.0575 6
7000
$9883.93
A e