Exponential Functions
Math Secondary IV
Topics
Calculation
 Growth & Decay
 Factor
 Graph
 Equation
 Point of Intersection
 Word Problems
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Calculators
This topic will involve a lot of calculator
use and specifically the exponential key
 Usually it is a button yx , xy or ^
 For example 25 = ?
 32
 177
 410338673
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Calculators (2nd Step)
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Add another step: 8 x 94
52488
16 x 724
429981696
42 x (2/3)8
Do 42 x open bracket 2/3 close bracket yx 8 = ??
1.64 (round off to two decimal places)
Exercises
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2.176 x 0.8158
0.42
273 x 4711
6.75 x 1020
0.82 x 0.006340
7.72 x 10-89
Work in class – Do #1
a-j
Growth & Decay
Consider the powers of 7 and 1/7
 7 = 70 = 1
 72 = 49
 73 = 343
 We note as the power increases value
increases; we call this…
 growth
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Growth & Decay
Consider 1/70 = 1
 (1/7)2 = 0.02
 (1/7)3 = 0.0029
 We note as the power increases the value
decreases; we call this…
 decay
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Growth & Decay
In general, if the base or factor is greater
than 1 we have growth.
 If the base or factor is between 0 and 1,
we have decay.
 We note we do not use 0 and 1 or
negative numbers
 Work in class / homework – do #2 a-e
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Factors
We constantly need to see what factor or
base we are using.
 Some are easy Double =
2
 Half =
½
 By ten =
 10
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Factors
Then there is per cent.
 If I have
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If I double it…
 I would multiply by 2
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Other Factors
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If I have a square and add 10% I would
multiply by?
1.1 (110%)
 We would get
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More Examples
10% increase is a factor of 1.1 (1+.1)
 33% increase is a factor of
 1.33 (1+.33)
 1% increase is a factor of
 1.01 (1+.01)
 4.75 increase is a factor
 1.0475 (1+0.0475).
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Let’s Go the Other Way
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We can use the same logic for a
10% decrease. 10% decrease is a
factor of 0.9 (1-.1)
39% decrease is a factor of
0.61 (1-.39)
Please note increase = up,
appreciation, interest and decrease
= down, depreciation
Work in class / homework do #3 a-j
Exponential Formula
The exponential formula is y = S ● Fx
 Formula Defined
 S Parameter: S is the starting value
Where does the function start
 In other words what is the value of y
when x=0
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Exponential Formula Defined
F Parameter –
 F stands for factor
 what is the function increasing or
decreasing by?
 i.e., what is the ratio between the value
when x = 1 and x = 0
 i.e. what is the value when x=1 divided by
the value when x = 0.
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Graphs
Graph the following. State the S and F parameters
and whether the curve is a growth or decay.
 X 0 1 2
3 4
 Y 2 6 18 54 162
 S: x = 0 y = 2; S = ?
 S=2
 F: x = 1 y = 6
 x = 0 y = 2; F = ?
 F=6/2=3
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Graph con’t
S=2
 F=3
 Graph it!
 Growth
 Hence y = 2 ● 3x
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Work in class / homework
do #4 a-e; 2 per page!
Homework Solutions
4a: S = 6
F=3
 Growth
 Graph it on graph paper; two per page
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Work in Class / Homework
#5: y = 5 (2)x
 Plot X and Y (use 1,2,3,&4)
 Quiz 
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Exercises
Give the formula to show the situation
where you invest $1000 at 7% annually.
 After 20 years, how much do you have?
 What is S and what is F?
 S = 1000 ; F = 1.07; Put in formula…
 Where y is the money the investment is
worth; x is the number of years
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Work in Class / Homework
#6 a – j
 State the Exponential Equation… 6a
 6a) Y = 14 (½)x
 6b) y = 1000 (1/100) x
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Solution
y = 1000 ● (1.07)x
 After 20 years your $1000 investment is
worth…
 Y = 1000 (1.07)20
 = $3869.68
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More Exercises
Radioactive elements decay (the atoms fall
apart) in a set formula. The element
Pingdanga has a half life of a year. Give
the formula if you start with 1000 kg
 1st step: Write down Exponential Formu.
 Y = 1000 (½) x
 Y is the amount of Pingdanga
 X is the number of years
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Solution Con’t
In 20 years we would have…
 Y = 1000 (½)20
 = 0.001 kg of Pingdanga
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Comparing Investments
If you have two investments $1000 @
20% and $2000 @ 5%, when will they be
worth the same? Plot points every two
years for eight years. What is the value of
each investment after 50 years?
 1st step… create two exponential formulas
(double the fun!)
 Y = 1000 (1.2)X
y = 2000 (1.05)X
 Create Table of Values for both…
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Investment Solutions
Investment A
X 0
2
4
6
8
Y $1000 $1440 $2073.60 $2985.98 $4299.82
 Investment B
X 0
2
4
6
8
Y 2000 2205 2431.01 2680.19 2954.91
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Investment Solutions
After 50 years?
 Investment A
 1000 (1.2)50
 = $9100438.15
 Investment B
 2000 (1.05)50
 = $22934.80
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Car Depreciation
Which car is worth more after five years
given the following values. You are given
the cost of the car and the depreciation
(amount your car goes down per year).
 Car A: $5000 @ 7%
 Car B: $6000 @ 8%
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Car Solution
Car A: 5000 (0.93)5
 = $3478.44
 Car B: $6000 (.92)5
 = $3954.49
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Work in Class / Homework
#7-15…
 Study Guide
 Test 
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