1. 7-6 & 7-7 Exponential Functions
Evaluate and graph exponential functions

2. A function in the form of
Exponential function
A function in the form of
y = 𝑎∙𝑏 𝑥
Examples:

3. 𝑦=𝑎∙ 𝑏 𝑥 The base & when b>1, called the Growth factor
Exponential Growth, modeled by the following
y = a∙ 𝑏 𝑥
Initial amount
(this is when x = 0)
𝑦=𝑎∙ 𝑏 𝑥
exponent
The base & when b>1, called the Growth factor
(1 + the percent rate written as a decimal)

4. 𝑦=𝑎∙ 𝑏 𝑥 Initial amount (this is when x = 0)
Exponential Decay
Initial amount
(this is when x = 0)
𝑦=𝑎∙ 𝑏 𝑥
exponent
The base is the decay factor
(1 – percent rate written as a decimal)

5. x What is the graph of y = 3∙ 2 𝑥 ? y = 3∙ 2 𝑥 y = 3∙ 2 −2 y = 3∙ 2 −1
(x, y) -2 -1 1 2
y = 3∙ 2 −2
(-2, 3 4 )
y = 3∙ 2 −1
(-1, ) cc
y = 3∙ 2 0
(0, 3)
y = 3∙ 2 1
(1, 6)
y = 3∙ 2 2
(2, 12)

6. ANSWER: EXPONENTIAL FUNCTION. B. y = 3x ANSWER: LINEAR FUNCTION.
Does the table or rule represent a linear or an exponential function? A.
ANSWER:
EXPONENTIAL FUNCTION.
B. y = 3x
ANSWER:
LINEAR FUNCTION.

7. Suppose 30 flour beetles are left undisturbed in a warehouse bin.
The beetle population doubles each week. The function
f(x) = 30∙ 2 𝑥 gives the population after x weeks. How many beetles
will there be after 56 days?
f(x) = 30∙ 2 𝑥
What does x represent?
= 30∙ 2 8
= 30∙256
= 7680
Answer: after 56 days, there will be 7,680 beetles.

8. Evaluate the function for the given value.
𝑦= 3∙4 𝑥 for x = 3
𝑦= 3∙4 ( ) 3
𝑦= 3∙64
𝑦=192

9. 𝑦=𝑎∙ 𝑏 𝑥 𝑦=360∙ 1.07 10 𝑦=708.174488 𝑦=$708 billion Let y =
Since 2005, the amount of money spent at restaurants in the US has increased about 7% each year. In 2005, about $360 billion was spent at restaurants. If the trend continues, about how much will be spent at restaurants in 2015?
𝑦=𝑎∙ 𝑏 𝑥
Let y =
The annual amount spent in restaurants (in billions of dollars)
Let a =
The initial amount:
360
Let b =
The growth factor:
(1 + %) or = 1.07
Let x =
The number of years since 2005: 10
𝑦=360∙ 𝑦=
𝑦=$708 billion

10. When a bank pays interest on both the principal and the interest an account has earned. (it uses the following formula)
Compound interest:
r = the annual interest rate----convert from % to a decimal—(move 2 places to the left)
A = The balance
A=P ( 1+ r n ) nt
t= the time in years
P = the principal
(the initial deposit)
n = the number of times interest is compounded per year

11. A=P ( 1+ r n ) nt A=12,000 ( 1+ .048 1 ) 1(7) A = $16,661.35
Find the balance in the account after the given period:
$12,000 principal earning 4.8% compounded annually, after 7 years
P =
r =
n =
t =
12,000
A=P ( 1+ r n ) nt
.048 1 7
A=12,000 ( ) 1(7)
A = $16,661.35

12. A=P ( 1+ r n ) nt A=20,000 ( 1+ .035 12 ) 12(10) A = $28,366.90
Find the balance in the account after the given period:
$20,000 principal earning 3.5% compounded monthly, after 10 years
P =
r =
n =
t =
20,000
A=P ( 1+ r n ) nt
.035 12 10
A=20,000 ( ) 12(10)
A = $28,366.90

13. 𝑦=𝑎∙ 𝑏 𝑥 𝑦=101∙ .885 3 𝑦=70.0085 𝑦=70 kilopascals Let y =
The kilopascal is unit of measure for atmospheric pressure. The atmospheric pressure at sea level is about 101 kilopascals. For every 1000-m increase in altitude, the pressure decreases about 11.5%. What is the approximate pressure at an altitude of 3000 m?
𝑦=𝑎∙ 𝑏 𝑥
Let y =
The atmospheric pressure (in kilopascals)
Let a =
The initial amount:
101
Let b =
The decay factor:
(1 - %) or = .885
Let x =
The altitude (in thousands of meters) 3
𝑦=101∙ 𝑦=
𝑦=70 kilopascals

14. Pg 457: odd & 20
pg 464: 9-21 odd (skip 13)