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Math 2 Precalculus Algebra
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Exponential Functions — 6.3
Exponential Function
f ( x ) = ax ,
a > 0 and a ≠ 1
Domain:
Horizontal Asymptote at y = 0 . The horizontal asymptote will move up or down depending on a horizontal shift.
The range depends on the global behavior which can be determined by the horizontal asymptote and the sign of vertical
stretch/compression factor, denoted by the letter b in your book..
1

All “basic” exponential functions contain the points ( 0 , 1) , (1 , a ) and  − 1 ,  .
a

Exponential Growth Functions
f ( x ) = ax ,
a >1
Domain:
Graph:
If a > 1 , the graph is a “
shape : right side up” . Range:
In mathematical terms, this means: as x → ∞, f ( x ) → ∞
and as x → − ∞, f ( x ) → 0+
Exponential Decay Functions
f ( x ) = ax ,
0 < a <1
Domain:
Graph:
If 0 < a < 1 , the graph is a “
shape : left side up” . Range:
In mathematical terms, this means: as x → ∞, f ( x ) → 0+
1.
Graph the equation.
f ( x ) = 3x
Domain:
Value of base:
Horizontal Asymptote:
Range:
Points:
and as x → − ∞, f ( x ) → ∞
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
2.
Graph the equation.
f ( x) =
−3
x
−2
Domain:
Value of base:
Horizontal Asymptote:
Transformations:
Range:
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
-4
-5
-6
-7
-8
Points:
3.
Graph the equation.
1
f ( x) =  
2
x −1
Domain:
Value of base:
Horizontal Asymptote:
Transformations:
Range:
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
-4
-5
-6
-7
-8
Points:
4.
Graph the equation.
1
=
f ( x)  
2
−x
+3
Domain:
Value of base:
Horizontal Asymptote:
Transformations:
Range:
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
-4
-5
-6
-7
-8
Points:
Many applications of exponential functions rely on a certain irrational base. Let’s use the compound interest formula to fill in
the following table, if $1 is invested at an annual interest rate of 100% for 1 year for the various numbers of compounding
periods per year.
Compound Interest Formula
r

=
A P 1 + 
n

P = principal
r = annual interest rate expressed as a decimal
n = number of interest periods per year
t = number of years P is invested
A = amount after t years
nt
,
where
)n ⋅1
or A
=
1
(1 + 1)1
=
4
(1 + 1 4 )4
12
(1 + 11 2 )1 2
=
56
(1 + 15 6 )5 6
=
365
(1 + 13 6 5 )3 6 5
n
1,000
10,000
100,000
1,000,000
=
A 1 (1 +
1
n
(1 + 1 n )n
=
=
(1 + 11,0 0 0 )
1,0 0 0
=
(1 + 11 0,0 0 0 )
1 0,0 0 0
=
(1 + 11 0 0,0 0 0 )
1 0 0,0 0 0
=
(1 + 11,0 0 0,0 0 0 )
1,0 0 0,0 0 0
=
What can you conclude about the value of the expression (1 +
In mathematical terms, as n → ∞ , (1 +
1
n
)n
1
n
)n as
n gets really large?
→
The Number e
If n is a positive integer, then as n → ∞ , (1 +
In calculus, this is written as l i m (1 +
n →∞
1
n
)n
1
n
)n
→ e ≈ 2.71828.
=≈
e 2.71828
Natural Exponential Function
f ( x ) = ex
( Note: e ≈ 2 . 7 1 8 2 8 )
Domain:
Horizontal Asymptote at y = 0 .
Range:
The natural exponential function contains
1

the points ( 0 , 1) , (1 , e ) and  − 1 ,  .
e

5.
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0 1 2 3 4 5 6 7 8 9 10
-2
0
-3
Graph the equation.
f (=
x ) e− x + 2
Domain:
Transformations:
Points:
Range:
Find an exponential function of the form f ( x ) = b a x or f (=
x ) b a x + c that has the given graph or conditions.
6.
y-intercept ( 0 , 4 ) ; passes through point P ( 2 , 1 6 ) and has a horizontal asymptote at y = − 1
7.
10
9
8
7
6
5
4
3
2
1
0
-5 -4 -3 -2 -1-1 0
-2
1
2
3
4
5
Exponential Functions Are One-to-One
The exponential function f given by f ( x ) = a x for 0 < a < 1 or a > 1 is one-to-one. Thus, the following equivalent
conditions are satisfied for real numbers x1 and x2
1.
If x1 ≠ x2 , then a x1 ≠ a x2 .
2.
If a x1 = a x2 , then x1 = x2
Solve each equation.
2
8.
4x =
9.
2 5x ⋅
( 12 )
3x
( 15 )
10. e − x e 2 =
4−2x
( 1e )
( )
=5 x
4 x +1 2
3
⋅125