Exponential Graphs
Warm Up
Solve:
f ( x)  3x  x  10 x
3

2

x 3x 2  x  10  0
x3x  5x  2  0
5
x  0, x   , x  2
3
Find the Vertex:
b
 24
x

3
2a
24
f ( x)  4 x  24 x  5
2
y  43  243  5  31
2
V  3,31
Definition

In an exponential function, the base is fixed
and the exponent is a variable.
Ex ponent
f x   b
x
base
Exploration

Using your GDC, graph the following
exponential functions on the same screen:
y1  3 , y 2  2 , y 3  1 .5 
x
x
At what POINT
do they all INTERSECT ?
All functions of the
form y  a pass
x
through the POINT 0,1.
x
Exploration

What do you observe about the function as
the base gets larger, and the exponent
remains positive?
Observe that for x  0,
the LARGER the base,
the faster the function
grows.
This is called a
Growth Model .
This is called a
Decay Model .
Exploration

Using your GDC, graph the following
exponential functions on the same screen:
y1  3 , y 2  2 , y 3  1 .5 
x
x
x
How do these graphs
differ from the previous
graphs and why ?
Negative Exponents
force the function to
rise on the left instead
of the right .
Exploration…

Using your GDC, graph the following
exponential functions on the same screen:
x
1
x
y1  3 , y 2   
How are the
3
 
graphs of each
pair related ?
The DECAY MODELS are
a reflection of the GROWTH
MODELS across the y  axis.
Continued….
1
y1  3 , y 2   
3
x
x
How are the bases of
the pairs related ?
They are reciprocal s
of each other.
x
1
NOTE : y    is the same as y  3  x .
3
y  2x
Graph:
x
y
-2
0.25
-1
0.5
0
1
1
2
2
4
HA: y = 0
Domain:
 , 
Range:
0, 
y  2 x
Graph:
Decreasing!
x
y
-2
4
-1
2
0
1
1
0.5
2
0.25
HA: y = 0
Domain:
 , 
Range:
0, 
Graph:
y  2 x 3
HA: y = 0
Domain:
Range:
 , 
0, 
The graph has been
moved 3 to the left
from the parent function.
Graph:
y  2x  2
HA: y = 2
Domain:
 , 
Range:
2, 
The graph has been
moved 2 up from the
parent function.
Graph:
y  2 x 4  3
HA: y = -3
Domain:
Range:
 , 
 3, 
Parent Function
Right 4
Down 3
Graph:
y2
 x 1
5
y  2  1  x 1   5
HA: y = -5
Parent Function
Domain:  , 
Left 1
Down 5 Range:  5, 
Re versed
Graph:
y  2 x4  2
HA: y = 2
Parent Function
Right 4
Up 2
Domain:
Range:
 , 
2, 
Natural exponential function

f ( x)  e

e  2.718281828...
x
Graph:
f ( x)  e x 1  3
Left 1
Down 3
Domain:
Range:
 , 
 3, 
Logarithmic Function


loga x  y

a x
y
It’s the inverse of the exponential function
Switch the x’s and the y’s!
Graph:
f ( x)  log 2 x
Is the inverse of
y  2x
x2

y  2x
Domain:  , 
Range:
0, 
Domain:
Range:
0, 
 , 
y
Graph:
f ( x)  3  log 2 x
Up 3 from previous example!


Domain:
0, 
Range:
 , 
Graph: f ( x)  log 2 ( x  4)
Left 4 from Original Example!

 4, 
Domain:
Range:  , 
Graph: f ( x)  log 2 ( x  2)
Right 2 from Original Example!

Domain:
Range:
2, 
 , 
Graph:
f ( x)  log 2 ( x)
Reflected over y-axis.

Domain:
Range:
 ,0
 , 
Graph:
f ( x)   log 2 x
Reflected over x-axis.

Domain:
Range:
0, 
 , 
Compound Interest
An infectious disease begins to spread in a small city of
population 10,000. After t days, the number of persons who
have succumbed to the virus is modeled by the function:
v(t ) 

10, 000
5  1245e 0.97 t
How many infected people are there initially?
10,000
10,000
v0  
8
  0.97 0  
5  1245e
1250

How many people are infected after five days?
10,000
v0  

678
.
1
  0.97 5 
5  1245e
Compound Interest
 r
A(t )  P 1  
 n
P = Principal
r = rate
t = time in years
n = number of times it’s compounded per year
Compounded:
annually
n=1
quarterly
n=4
monthly
n = 12
daily
n = 365
nt
Find the Final Amount:
$8000 at 6.5% compounded quarterly for 8 years

At   P 1 

r

n
nt
 .065 
At   80001 

4 

n4
 48 
 13400.09
Find the Final Amount:
$600 at 9% compounded daily for 20 years
n  365
 0.09 
At   6001 

365 

36520 
 3628.98
Find the Final Amount:
$300 at 6% compounded annually for 25 years

At   P 1 

r

n
n t
n 1
 0.06 
At   3001 

1 

125
 1287.56
Compounded Continuously:
A(t )  Pert
P = Principal
r = rate
t = time in years
E = 2.718281828…
Find the Final Amount:
$2500 at 4% compounded continuously for 25 years
At   Pe
A t   2500 e
rt
0 .04 25
 6795 .70
Suppose your are offered a job that lasts one month, and
you are to be very well paid. Which of the following methods
of payment is more profitable for you? How much will you
make?

One million dollars at the end of the month.

Two cents on the first day of the month, 4 cents
on the second day, 8 cents on the third day, and,
in general, 2n cents on the nth day.
A  2 cents
n
A  2  2147483648 cents
31
A  $21,474836.48
More Profitable