Unit 6 Introduction To Exponential Functions — Unit Plan

TitleAssessment
Lesson 1
Growing and Growing
Fish Wish

While on a different beach, your friend discovers another magical fish. This fish also offers two purses to choose from, and it gives them this following graph to show how the money in each purse will grow. The set of triangular marks that lie on a line represent values in Purse A, and the set of square marks that make a curve represent those in Purse B.

<p>Graph of 2 functions on grid.</p>
Graph of 2 functions on grid, origin O. Horizontal axis, number of days, from 0 to 15 by 1's. Vertical axis, amount in dollars from 0 to 4,500 by 500's. Approximate plotted coordinates as follows: For Purse A: 0 comma 300, 1 comma 500, 2 comma 700, 3 comma 900, 4 comma 1,100, 5 comma 1,300, 6 comma 1,500, 7 comma 1,700, 8 comma 1,900, 9 comma 2,100, 10 comma 2,300, 11 comma 2,500, 12 comma 2,700, 13 comma 2,900, 14 comma 3,100, 15 comma 3,300. Approximate coordinates for Purse B: 0 comma 100, 1 comma 125, 2 comma 180, 3 comma 250, 4 comma 350, 5 comma 500, 6 comma 650, 7 comma 900, 8 comma 1,250, 9 comma 1,700, 10 comma 2,300, 11 comma 3,200, 12 comma 4,300.

The fish is still deciding how many days it will let the money in the purses grow. Help your friend plan a strategy for picking the purse with the most money while the fish thinks.

Show Solution

The vertical intercept for Purse B is lower, so Purse B starts off with less money than Purse A. The values in Purse B stay lower than the values in Purse A until day 10, when they appear to be the same. So Purse A is better up to day 10, and after that point Purse B becomes a better choice.

Section A Check
Section A Checkpoint
Problem 1

Consider the two patterns. For each pattern: 

  1. Determine whether they grow by a constant difference, constant factor, or neither.
  2. If they grow by a constant difference or factor, find its value. Show your reasoning.

Pattern 1

xx 1 2 3 4 5
yy 8 16 32 64 128

Pattern 2

aa 0 1 2 3 4
bb 8 16 24 32 40

Show Solution
Pattern 1 
  1. It grows by a constant factor.
  2. The factor is 2 because 168=2\frac{16}{8} = 2, 3216=2\frac{32}{16}=2, and so on.
Pattern 2
  1. It grows by a constant difference.
  2. The difference is 8 because 16−8=816 - 8 = 8, 24−16=824-16=8, and so on.
Problem 2

You are given 2 options for claiming prize money that must not be touched until you are a certain age. 

  • Option 1: You are given $50,000 every year. 
  • Option 2: You are given $10,000 initially, and its value is doubled every year.

Will Option 2 ever be worth more than Option 1? Show your reasoning.

Show Solution
Yes. Sample response: After 5 years, Option 1 will be worth $250,000 (50,000⋅5=250,00050,000 \boldcdot 5 = 250,000), and Option 2 will be worth $320,000 (10,000⋅25=320,00010,000 \boldcdot 2^5 = 320,000).
Lesson 3
Representing Exponential Growth
Mice in the Forest

A group of biologists is surveying the mice population in a forest. The equation n=75⋅3tn=75 \boldcdot 3^t gives the total number of mice, nn, tt years since the survey began. Explain what the numbers 75 and 3 mean in this situation.

Show Solution

75 is the initial number of mice in the forest when the survey started, or the number of mice when tt is 0. The 3 is the growth factor, meaning that each year, the population of mice in the forest is 3 times the previous year's population.

Lesson 4
Representing Exponential Decay
Freezing Soup

A soup is placed in a freezer to save. Here is a graph showing the temperature of the soup at different times after being placed in the freezer.

Graph of points. Horizontal axis time in hours.  Vertical axis, temperature in degrees Celsius.

  1. What is the vertical intercept? What does it mean in this situation?
  2. What fraction of the temperature remained after one hour?
  3. Write an equation that represents the temperature of the soup, tt, after hh hours.
Show Solution
  1. 60. The soup is 60∘C60^\circ \text{C} when it is placed in the freezer.
  2. 610\frac{6}{10} of the temperature remained one hour after the soup was placed in the freezer.
  3. t=60(610)ht=60(\frac{6}{10})^h
Lesson 5
Understanding Decay
The Depreciating Phone

Suppose that a phone that originally sold for $800 loses 35\frac35 of its value each year after it is released.

  1. After 2 years, how much is the phone worth?
  2. Write an equation for the value of the phone, pp, tt years after it is released.
Show Solution
  1. $128, because 800⋅25⋅25=800⋅(25)2=128800 \boldcdot \frac25\boldcdot \frac25 = 800\boldcdot \left(\frac25\right)^2=128.
  2. p=800⋅(25)tp=800 \boldcdot\left(\frac25\right)^t
Lesson 6
Analyzing Graphs
A Phone, a Company, a Camera

<p>Graph of function on grid.</p>
Graph of a function on grid, origin O. Horizontal axis, time, years, from 0 to 14 by 0 point 5's. Vertical axis, value, dollars, from 0 to 1,400 by 100's. Approximate plotted coordinates as follows:  0 comma 1,200, 1 comma 720, 2 comma 430, 3 comma 260, 4 comma 155, 5 comma 90, 6 comma 55, 7 comma 33, 8 comma 20, 9 comma 12, 10 comma 7, 11 comma 4, 12 comma 2, 13 comma 1.  

  1. This graph represents one of the following descriptions. Which one?
    1. A phone loses 45\frac{4}{5} of its value every year after purchase: the relationship​ between​ ​the​ ​number​ ​of​ ​years​ ​since purchasing​ ​the​ ​phone​ ​and​ ​the​ ​value​ ​of​ ​the​ ​phone.
    2. The number of stores that a company has triples approximately every 5 years: the relationship ​between​ ​the number of years and the​ ​number​ ​of​ ​stores​.
    3. A camera loses 25\frac{2}{5} of its value every year after purchase: the relationship​ between​ ​the​ ​number​ ​of​ ​years​ ​since purchasing​ ​the​ ​camera​ ​and​ ​the​ ​value​ ​of​ ​the​ ​camera.
  2. Explain how you know the graph represents the description you chose.
Show Solution
  1. C
  2. Sample responses:
    • The graph cannot represent Description A because the phone is retaining only 15\frac15 of its value, which is less than half, and the vertical coordinate of the second point on the graph is more than half of the vertical intercept.
    • If the camera loses 25\frac{2}{5} of its value each year, then its value is 35\frac{3}{5} that of the previous year. The vertical intercept seems to be 1,200, and 35\frac35 of 1,200 is about 700, which is roughly the vertical coordinate of the second point.
Lesson 7
Using Negative Exponents
Invasive Fish

The equation p=5,000⋅2tp =5,000 \boldcdot 2^t represents the population of an invasive fish species in a large lake, tt years since 2005, when the fish population in the lake was first surveyed.

  1. What was the population in 2005?
  2. For this model, what does it mean when tt is -2?
  3. For t=-2t = \text-2, is the fish population more or less than 1,000? How do you know?
Show Solution
  1. 5,000, because 5,000⋅20=5,0005,000 \boldcdot2^0 =5,000.
  2. It means 2 years before 2005, which is 2003.
  3. More than 1,000. If t=-2t=\text-2, then 5,000⋅2-2=1,2505,000 \boldcdot 2^{\text-2}=1,250.
Section B Check
Section B Checkpoint
Problem 1

Scientists in a laboratory have a sample of a material that contains many types of atoms. One of the atoms is iodine-131, a radioactive isotope used to treat some forms of cancer. Every day, about 110\frac{1}{10} of the iodine-131 present releases some radioactive material and turns into another type of atom.

  1. Scientists measure a particular sample and find that it has 5 grams of iodine-131 in it. Write an equation for AA, the amount of the iodine-131 left tt days after it was measured.
  2. What is the value of AA when t=-1t=\text{-}1? Explain what it means in this situation.
  3. Sketch a graph that shows the general shape of this equation. Include the coordinates of the vertical intercept.

blank coordinate grid from -10 to 10 on both axes

Show Solution
  1. A=5⋅(910)tA = 5 \boldcdot \left( \frac{9}{10} \right)^t or equivalent.
  2. About 5.56 grams. Sample response: It is the amount of iodine-131 one day before the scientist measured the amount in the sample.
  3. A graph that shows exponential decay and includes the point (0,5)(0,5).
Lesson 8
Exponential Situations as Functions
Beaver Population

The graph shows the population of beavers in a forest for different numbers of years after 1995. The beaver population is growing exponentially. 

<p>Graph of function.</p>
Graph of function, origin O. Horizontal axis, from 0 to 5, by 1's, labeled time in years. Vertical axis, from 0 to 1000, by 200s, labeled number of beavers. Approximate Points at 0 comma 50, 1 comma 100, 2 comma 200, 3 comma 400, Point Q at 3 point 5 comma 500, 4 comma 800.   

  1. Explain why we can think of the beaver population as a function of time in years.
  2. Estimate the coordinate for point QQ. What is the meaning of these values in this context?

  3. Write an equation using function notation to represent this situation.
Show Solution
  1. The number of beavers depends on time in years. In any given year, there is a particular number of beavers in the forest.
  2. Sample response: About (3.5,550)(3.5, 550). This means that 3.5 years after 1995, the beaver population was about 550.
  3. p(t)=50⋅2tp(t) =50 \boldcdot 2^t where tt is years after 1995 and p(t)p(t) is the beaver population at that time.
Lesson 9
Interpreting Exponential Functions
Bacteria Population

The graph shows the bacteria population on a petri dish as a function, ff, the days, dd, since an antibiotic is introduced.

<p>Graph of function.</p>
Graph of function, origin O. Horizontal axis, from 0 to 10, by 2's, labeled time, days. Vertical axis, from 0 to 1,000,000, by 500,000's, labeled number of bacteria. Line starts at 0 comma 1,000,000, passes through 1 comma 900,000 and 2 comma 810,000.

  1. What is the approximate value of f(4.5)f(4.5)?
  2. Approximately what is dd when f(d)=400,000f(d)=400,000?
  3. Explain what you would do, using your usual graphing technology, to be able to see f(15)f(15) on the graph.
Show Solution
  1. The exact value is 622,431, but reasonable approximations should be greater than 600,000 and less than 650,000.
  2. The value to one decimal place is 8.7, but reasonable approximations should be greater than 8.5 and less than 9.
  3. Answers vary based on what graphing technology is used.
Lesson 10
Looking at Rates of Change
An Average Rate of Change

Here is the function ff for Clare's moldy bread that you saw earlier.

dd, time since mold spotting (days)

f(d)f(d), area covered by mold (square millimeters)
0 1
1 2
2 4
3 8
4 16
5 32
6 64
  1. What is the average rate of change for the mold over the 6 days?
  2. How well does the average rate of change describe how the mold changes for these 6 days?
Show Solution

  1. 10.5 square millimeters per day: f(6)−f(0)6−0=10.5\displaystyle \dfrac{f(6)-f(0)}{6-0} =10.5

  2. The average rate of change does not accurately describe how the mold changes over the 6 day period. The first day it only grows by 1 square millimeter while on the fifth day it grows by 32 square millimeters.
Lesson 11
Modeling Exponential Behavior
Drop Height

A ball is dropped from a certain height. The table shows the rebound heights of the ball after a series of bounces.

bounce number height in centimeters
1 30
2 6
3 1
4 0

From what height, approximately, do you think the ball was dropped? Explain your reasoning.

Show Solution

Sample response: Between 150 cm and 180 cm. The rebound factors are 15\frac{1}{5}, 16\frac{1}{6} and 0 (this last measurement is probably not reliable because it could have been a very small bounce height, difficult to measure). Because 15\frac{1}{5} of 150 is 30 and 16\frac{1}{6} of 180 is 30 the ball was probably dropped from between 150 cm and 180 cm.

Lesson 12
Reasoning about Exponential Graphs (Part 1)
A Possible Equation

Here are three graphs representing three exponential functions, ff, gg, and hh.

<p>Graph of 3 functions on a x y plane. H. G. F.</p>
Graph of 3 functions on grid, origin O. Horizontal axis, x, vertical axis, y. Function h, g and f drawn. Function h and g cross the vertical axis at the same location, all other values for function h above function g. Function g crosses the vertical axis below functions h and g. Function g grows more steeply than function f.  

The functions ff and hh are given by f(x)=10⋅2xf(x) = 10 \boldcdot 2^x and h(x)=20⋅4xh(x) = 20 \boldcdot 4^x. Which of the following could define the function gg? Explain your reasoning.

  • Equation A: g(x)=20⋅(1.5)xg(x) = 20 \boldcdot (1.5)^x
  • Equation B: g(x)=20⋅(2.5)xg(x) = 20 \boldcdot (2.5)^x
  • Equation C: g(x)=10⋅(3.5)xg(x) = 10\boldcdot (3.5)^x
  • Equation D: g(x)=20⋅(4.5)xg(x) = 20 \boldcdot (4.5)^x
Show Solution

B. The graph of gg has the same yy-intercept as the graph of hh, which is 20. It grows more quickly than ff but more slowly than hh so the growth factor must be greater than 2 but less than 4.

Section C Check
Section C Checkpoint
Problem 1
A group of 10 students form a Good Deed Club. Their mission is to each perform a good deed for 3 people on a day, convince those 3 people to perform good deeds for 3 other people each, and continue the process. Assume that everyone in the chain does their part.
  1. Let the number of days be the independent variable and the number of good deeds done caused by this chain be the dependent variable. Is the relationship a function? Explain your reasoning.
  2. Write a function that represents this relationship.
Show Solution
  1. Yes, because each day there is a certain number of good deeds done that were caused by this chain.
  2. G(t)=10⋅3tG(t) = 10 \boldcdot 3^t
Problem 2

Match the function to the graph that represents the function. Explain your reasoning for each.

  1. f(t)=2⋅3tf(t) = 2 \boldcdot 3^t

  2. g(t)=2⋅(1.5)tg(t) = 2 \boldcdot (1.5)^t

  3. h(t)=3⋅(32)th(t) = 3 \boldcdot \left( \frac{3}{2} \right)^t

  4. j(t)=3⋅(23)tj(t) = 3 \boldcdot \left( \frac{2}{3} \right)^t

graph of 4 exponential functions labeled A, B, C, and D

Show Solution
  1. A. Sample reasoning: It goes through the point (0,2)(0,2), is the steepest, and has the greatest growth factor.
  2. C. Sample reasoning: It goes through the point (0,2)(0,2), but grows slower than the graph for f(t)f(t).
  3. B. Sample reasoning: It goes through the point (0,3)(0,3) and shows exponential growth because the growth factor is greater than 1.
  4. D. Sample reasoning: It goes through the point (0,3)(0,3) and shows exponential decay because the growth factor is between 0 and 1.
Lesson 14
Recalling Percent Change
A Book and a Present
  1. A book costs $45. Sales tax on the book is 7%. Write two different expressions that represent the final cost of the book in dollars. One of the expressions should use only a single multiplication and no other operations.
  2. A present that costs $32 is on sale for 15% off, and there are no taxes. Explain why the expression (0.85)⋅32(0.85) \boldcdot 32 represents the final price of the present in dollars.
Show Solution
  1. Sample response: 45⋅(1+0.07)45 \boldcdot (1 +0.07) (or 45+(0.07)⋅4545 +(0.07)\boldcdot 45, or equivalent) and 45⋅(1.07)45 \boldcdot (1.07)
  2. 15% less than 32 can be written as 32−(0.15)⋅3232 - (0.15)\boldcdot32, which equals 32⋅(0.85)32 \boldcdot (0.85).
Lesson 15
Functions Involving Percent Change
Delayed Payments

A business owner receives a $5,000 loan with 13% interest, charged at the end of each year.

  1. Write an expression to represent the amount owed, in dollars, after the given number of years of making no payments:
    1. after 1 year
    2. after 2 years
    3. after tt years
  2. Explain how to convince someone that your expression for tt years is correct.
Show Solution
    1. 5,000⋅(1.13)5,000 \boldcdot (1.13)
    2. 5,000⋅(1.13)25,000 \boldcdot (1.13)^2
    3. 5,000⋅(1.13)t5,000 \boldcdot (1.13)^t
  2. Sample response: After one year that the loan is not repaid, the amount gets multiplied by a factor of 1.13, or (1+0.13)(1 + 0.13). After 2 years, the balance from the first year, 5,000⋅(1.13)5,000 \boldcdot (1.13), is multiplied by 1.13 again, which gives 5,000⋅(1.13)25,000 \boldcdot (1.13)^2. After tt years, the $5,000 will be multiplied by 1.13 tt times, or by (1.13)t(1.13)^t.
Lesson 16
Compounding Interest
Two Months Later

A bank account that earns 1% interest each month has a balance of $1,500. Any interest is deposited into the account, and no further deposits or withdrawals are made.

Noah thinks that after two months the balance will be $1,530 because 2% of 1,500 is 30.

Do you agree with Noah? Explain your reasoning.

Show Solution

Disagree. Sample explanations:

  • After one month, the account balance will be 1,500⋅(1+0.01)1,500 \boldcdot (1+0.01) or $1,515. After two months, it will be 1,500⋅(1.01)21,500 \boldcdot (1.01)^2 or $1,530.15.
  • After one month, the account will grow by $15, which is 1% of the initial 1,500. After two months, the 1% interest applies not only to the initial $1,500 but also to the earned $15 from the first month. 1% of $15 is $0.15, so the interest earned that month is $15.15, bringing the new balance to $1,530.15. Noah's answer is short by $0.15.
Lesson 17
Different Compounding Intervals
How Often Is It Calculated?

A savings account pays a 3% nominal annual interest rate and has a balance of $1,000. Any interest earned is deposited into the account and no further deposits or withdrawals are made.

  1. Write an expression that represents the balance in one year if interest is compounded annually.
  2. If interest is compounded semi-annually (every six months), what interest rate would be used for each calculation?
  3. If interest is compounded semi-annually, which expression represents the account balance in tt years?
    • 1,000⋅(1+0.015)t1,000 \boldcdot (1 + 0.015)^t
    • 1,000⋅((1+0.015)2)t1,000 \boldcdot \left((1 + 0.015)^{2}\right)^t
    • 1,000⋅((1+0.015)6)t1,000 \boldcdot \left((1 + 0.015)^{6}\right)^t
    • 1,000⋅((1+0.03)2)t1,000 \boldcdot \left((1 + 0.03)^{2}\right)^t
Show Solution
  1. 1,000⋅(1+0.03)1,000 \boldcdot (1 + 0.03), or 1,000⋅(1+3100)1,000 \boldcdot \left(1 + \frac{3}{100}\right)
  2. 1.5%, or half of 3%
  3. 1,000⋅((1+0.015)2)t1,000 \boldcdot \left((1 + 0.015)^{2}\right)^t
Section D Check
Section D Checkpoint
Problem 1

A large field used to be sprayed so that only grass would grow. A new owner wants wildflowers to grow, so they stop spraying the field. The first year after not spraying there are 1,000 wildflowers in the field. Each month, the number of wildflowers increases by 10%.

If this pattern holds for 6 months, will there be 60% more wildflowers in the field? Explain your reasoning.

Show Solution
No, there will be more than 60% more wildflowers. Sample reasoning: The increase is compounding, so the increase each month comes from the original 1,000 wildflowers as well as any flowers that grew in the previous months.
Problem 2

An investor has two options for a savings account. To determine which is the better deal, the investor imagines investing $2,000 and making no deposits or withdrawals for 5 years.

  • Bank 1: 3% nominal annual interest compounded monthly
  • Bank 2: 3% nominal annual interest compounded quarterly
  1. How much would be in each account at the end of the 5 years? Show your reasoning.

  2. Which bank should the investor choose?
Show Solution
Sample responses:
  1. Bank 1: $2,323.23 because 0.0312=0.0025\frac{0.03}{12} = 0.0025, and 2000⋅(1+0.0025)60=2323.232000 \boldcdot (1 + 0.0025)^{60} = 2323.23.
    Bank 2: $2,322.37 because 0.034=0.0075\frac{0.03}{4} = 0.0075, and 2000⋅(1+0.0075)20=2322.372000 \boldcdot (1 + 0.0075)^{20} = 2322.37
  2. The investor should choose Bank 1 because with the same initial deposit and interest, the account that compounds more often will result in more money.
Lesson 20
Changes over Equal Intervals
Increasing Input by One
  1. The linear function gg is defined by g(x)=4xg(x) = 4x.
    1. Show that g(3)g(3) and g(2)g(2) have a difference of 4.
    2. Show that g(x+1)−g(x)=4g(x+1) - g(x) = 4.
  2. The exponential function hh is defined by ​h(x)=4xh(x) = 4^x.
    1. Show that h(3)h(3) and h(2)h(2) have a quotient of 4.
    2. Show that h(x+1)h(x)=4\frac{h(x+1)}{h(x)} = 4.
Show Solution
  1.  
    1. Sample responses:
      • g(3)=4⋅3=12g(3)=4 \boldcdot 3 = 12 and g(2)=4⋅2=8g(2) = 4 \boldcdot 2=8. The difference between 12 and 8 is 4.
      • g(3)−g(2)=4⋅(3)−4⋅(2)=4⋅(3−2)=4g(3)-g(2)=4\boldcdot(3) - 4\boldcdot(2)= 4\boldcdot (3-2) = 4
    2. g(x+1)−g(x)=4⋅(x+1)−4x=4⋅(x+1−x)=4⋅(1)g(x+1)-g(x)=4\boldcdot(x+1) - 4x= 4\boldcdot (x+1-x) = 4\boldcdot(1), which equals 4.
  2.  
    1. Sample responses:
      • h(3)=43=64h(3)=4^3 = 64 and h(2)=42=16h(2) = 4^2=16. The quotient is 6416\frac{64}{16}, which is 4.
      • h(3)h(2)=4342=43−2=4\frac{h(3)}{h(2)} = \frac{4^3}{4^2}=4^{3-2}=4
    2. h(x+1)h(x)=4x+14x=4x+1−x=41\frac{h(x+1)}{h(x)} = \frac{4^{x+1}}{4^x}=4^{x+1-x}=4^1, which equals 4.
Section E Check
Section E Checkpoint
Problem 1

A function is given by f(x)=2xf(x) = 2^x. 

  1. Write an exponential expression of the form a⋅bxa \boldcdot b^x equal to f(x+1)f(x+1). Show your reasoning.

  2. Show that f(x+1)f(x)\frac{f(x+1)}{f(x)} is a constant.
Show Solution
  1. 2⋅2x2 \boldcdot 2^x because f(x+1)=2x+1f(x+1) = 2^{x+1} which is equal to 2⋅2x2 \boldcdot 2^x
  2. Sample response: f(x+1)f(x)=2x+12x=2x+1−x=21\frac{f(x+1)}{f(x)} = \frac{2^{x+1}}{2^x} = 2^{x+1-x} = 2^1 which is 2.
Problem 2

The function gg is given by g(x)=3xg(x) = 3^x, and the function hh is given by h(x)=1,000+3xh(x) = 1,000 + 3x.

  1. Show that h(0)>g(0)h(0) > g(0)

  2. For the same xx-value, will g(x)>h(x)g(x) > h(x)? If so, give an example.
Show Solution
  1. h(0)=1000h(0) = 1000 and g(0)=1g(0) = 1, so h(0)>g(0)h(0) > g(0)
  2. Yes. Sample response: For any value greater than 7 this is true. For example, g(7)=2,187g(7) = 2,187 and h(7)=1,021h(7) = 1,021.
Lesson 21
Predicting Populations
No cool-down