Unit 6 Plan - Algebra 1

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Fish Wish (1 problem)

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.

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.

Lesson 2

Patterns of Growth

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Meow Island and Purr Island (1 problem)

The tables show the cat population on two islands over several years. Describe mathematically, as precisely as you can, how the cat population on each island is changing.

year	0	1	2	3	4
number of cats on Meow Island	2	6	18	54	162

year	0	1	2	3	4
number of cats on Purr Island	2	6	10	14	18

Show Solution

Sample responses:

The cat population on Meow Island is:

Tripling each year
Growing by a common factor each year

The cat population on Purr Island is:

Adding 4 cats each year
Characterized by a common difference each year

Section A Check

Section A Checkpoint

Problem 1

Consider the two patterns. For each pattern:

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

Pattern 1

$x$	1	2	3	4	5
$y$	8	16	32	64	128

Pattern 2

$a$	0	1	2	3	4
$b$	8	16	24	32	40

Show Solution

Pattern 1

It grows by a constant factor.
The factor is 2 because $\frac{16}{8} = 2$ , $\frac{32}{16}=2$ , and so on.

Pattern 2

It grows by a constant difference.
The difference is 8 because $16 - 8 = 8$ , $24-16=8$ , 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 \boldcdot 5 = 250,000

), and Option 2 will be worth $320,000 (

10,000 \boldcdot 2^5 = 320,000

).

Lesson 3

Representing Exponential Growth

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Mice in the Forest (1 problem)

A group of biologists is surveying the mice population in a forest. The equation $n=75 \boldcdot 3^t$ gives the total number of mice, $n$ , $t$ 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 $t$ 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

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Freezing Soup (1 problem)

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.

What is the vertical intercept? What does it mean in this situation?
What fraction of the temperature remained after one hour?
Write an equation that represents the temperature of the soup, $t$ , after $h$ hours.

Show Solution

60. The soup is $60^\circ \text{C}$ when it is placed in the freezer.
$\frac{6}{10}$ of the temperature remained one hour after the soup was placed in the freezer.
$t=60(\frac{6}{10})^h$

Lesson 5

Understanding Decay

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The Depreciating Phone (1 problem)

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

After 2 years, how much is the phone worth?
Write an equation for the value of the phone, $p$ , $t$ years after it is released.

Show Solution

$128, because $800 \boldcdot \frac25\boldcdot \frac25 = 800\boldcdot \left(\frac25\right)^2=128$ .
$p=800 \boldcdot\left(\frac25\right)^t$

Lesson 6

Analyzing Graphs

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A Phone, a Company, a Camera (1 problem)

This graph represents one of the following descriptions. Which one?
1. A phone loses $\frac{4}{5}$ 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 $\frac{2}{5}$ of its value every year after purchase: the relationship between the number of years since purchasing the camera and the value of the camera.
Explain how you know the graph represents the description you chose.

Show Solution

C
Sample responses:
- The graph cannot represent Description A because the phone is retaining only $\frac15$ 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 $\frac{2}{5}$ of its value each year, then its value is $\frac{3}{5}$ that of the previous year. The vertical intercept seems to be 1,200, and $\frac35$ of 1,200 is about 700, which is roughly the vertical coordinate of the second point.

Lesson 7

Using Negative Exponents

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Invasive Fish (1 problem)

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

What was the population in 2005?
For this model, what does it mean when $t$ is -2?
For $t = \text-2$ , is the fish population more or less than 1,000? How do you know?

Show Solution

5,000, because $5,000 \boldcdot2^0 =5,000$ .
It means 2 years before 2005, which is 2003.
More than 1,000. If $t=\text-2$ , then $5,000 \boldcdot 2^{\text-2}=1,250$ .

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 $\frac{1}{10}$ of the iodine-131 present releases some radioactive material and turns into another type of atom.

Scientists measure a particular sample and find that it has 5 grams of iodine-131 in it. Write an equation for $A$ , the amount of the iodine-131 left $t$ days after it was measured.
What is the value of $A$ when $t=\text{-}1$ ? Explain what it means in this situation.
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

$A = 5 \boldcdot \left( \frac{9}{10} \right)^t$ or equivalent.
About 5.56 grams. Sample response: It is the amount of iodine-131 one day before the scientist measured the amount in the sample.
A graph that shows exponential decay and includes the point $(0,5)$ .

Lesson 8

Exponential Situations as Functions

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Beaver Population (1 problem)

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

Explain why we can think of the beaver population as a function of time in years.
Estimate the coordinate for point $Q$ . What is the meaning of these values in this context?
Write an equation using function notation to represent this situation.

Show Solution

The number of beavers depends on time in years. In any given year, there is a particular number of beavers in the forest.
Sample response: About $(3.5, 550)$ . This means that 3.5 years after 1995, the beaver population was about 550.
$p(t) =50 \boldcdot 2^t$ where $t$ is years after 1995 and $p(t)$ is the beaver population at that time.

Lesson 9

Interpreting Exponential Functions

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Bacteria Population (1 problem)

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

What is the approximate value of $f(4.5)$ ?
Approximately what is $d$ when $f(d)=400,000$ ?
Explain what you would do, using your usual graphing technology, to be able to see $f(15)$ on the graph.

Show Solution

The exact value is 622,431, but reasonable approximations should be greater than 600,000 and less than 650,000.
The value to one decimal place is 8.7, but reasonable approximations should be greater than 8.5 and less than 9.
Answers vary based on what graphing technology is used.

Lesson 10

Looking at Rates of Change

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An Average Rate of Change (1 problem)

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

$d$ , time since mold spotting (days)	$f(d)$ , area covered by mold (square millimeters)
0	1
1	2
2	4
3	8
4	16
5	32
6	64

What is the average rate of change for the mold over the 6 days?
How well does the average rate of change describe how the mold changes for these 6 days?

Show Solution

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

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Drop Height (1 problem)

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

Lesson 12

Reasoning about Exponential Graphs (Part 1)

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A Possible Equation (1 problem)

Here are three graphs representing three exponential functions, $f$ , $g$ , and $h$ .

<p>Graph of 3 functions on a x y plane. H. G. F.</p>

The functions $f$ and $h$ are given by $f(x) = 10 \boldcdot 2^x$ and $h(x) = 20 \boldcdot 4^x$ . Which of the following could define the function $g$ ? Explain your reasoning.

Equation A: $g(x) = 20 \boldcdot (1.5)^x$
Equation B: $g(x) = 20 \boldcdot (2.5)^x$
Equation C: $g(x) = 10\boldcdot (3.5)^x$
Equation D: $g(x) = 20 \boldcdot (4.5)^x$

Show Solution

B. The graph of $g$ has the same $y$ -intercept as the graph of $h$ , which is 20. It grows more quickly than $f$ but more slowly than $h$ so the growth factor must be greater than 2 but less than 4.

Lesson 13

Reasoning about Exponential Graphs (Part 2)

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Two Graphs (1 problem)

Here are two graphs representing the function $f$ given by $f(x) = 10 \boldcdot 2^x$ and the function $g$ defined by $g(x) = a \boldcdot b^x$ .

Is $b$ greater than or less than 2? Explain how you know.
Write an equation that defines $g$ . Show your reasoning.
$f$ and $g$ represent the number, in thousands, of social media followers of two organizations as a function of years since 2010. What does the intersection of $f$ and $g$ mean in this context?

Show Solution

$b$ is greater than 2. Sample responses:
- The graph of $g$ starts off being below the graph of $f$ , but later it surpasses the graph of $f$ . Since $g$ grows more quickly than $f$ , the growth factor $b$ for $g$ must be greater than the growth factor for $f$ , which is 2.
- The growth factor from the first point $(0,3)$ to the second point $(1,12)$ is 4, because $\frac{12}{3}= 4$ , and 4 is greater than 2.
$g(x)=3 \boldcdot 4^x$ because the initial value is 3 and the $y$ -value is multiplied by 4 when the $x$ -value increases by 1, so the growth factor is 4.
It represents the time, in years since 2010, when the two organizations have the same number of followers.

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.

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.
Write a function that represents this relationship.

Show Solution

Yes, because each day there is a certain number of good deeds done that were caused by this chain.
$G(t) = 10 \boldcdot 3^t$

Problem 2

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

$f(t) = 2 \boldcdot 3^t$
$g(t) = 2 \boldcdot (1.5)^t$
$h(t) = 3 \boldcdot \left( \frac{3}{2} \right)^t$
$j(t) = 3 \boldcdot \left( \frac{2}{3} \right)^t$

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

Show Solution

A. Sample reasoning: It goes through the point $(0,2)$ , is the steepest, and has the greatest growth factor.
C. Sample reasoning: It goes through the point $(0,2)$ , but grows slower than the graph for $f(t)$ .
B. Sample reasoning: It goes through the point $(0,3)$ and shows exponential growth because the growth factor is greater than 1.
D. Sample reasoning: It goes through the point $(0,3)$ and shows exponential decay because the growth factor is between 0 and 1.

Lesson 14

Recalling Percent Change

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A Book and a Present (1 problem)

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.
A present that costs $32 is on sale for 15% off, and there are no taxes. Explain why the expression $(0.85) \boldcdot 32$ represents the final price of the present in dollars.

Show Solution

Sample response: $45 \boldcdot (1 +0.07)$ (or $45 +(0.07)\boldcdot 45$ , or equivalent) and $45 \boldcdot (1.07)$
15% less than 32 can be written as $32 - (0.15)\boldcdot32$ , which equals $32 \boldcdot (0.85)$ .

Lesson 15

Functions Involving Percent Change

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Delayed Payments (1 problem)

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

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 $t$ years
Explain how to convince someone that your expression for $t$ years is correct.

Show Solution

1. $5,000 \boldcdot (1.13)$
2. $5,000 \boldcdot (1.13)^2$
3. $5,000 \boldcdot (1.13)^t$
Sample response: After one year that the loan is not repaid, the amount gets multiplied by a factor of 1.13, or $(1 + 0.13)$ . After 2 years, the balance from the first year, $5,000 \boldcdot (1.13)$ , is multiplied by 1.13 again, which gives $5,000 \boldcdot (1.13)^2$ . After $t$ years, the $5,000 will be multiplied by 1.13 $t$ times, or by $(1.13)^t$ .

Lesson 16

Compounding Interest

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Two Months Later (1 problem)

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 $1,500 \boldcdot (1+0.01)$ or $1,515. After two months, it will be $1,500 \boldcdot (1.01)^2$ 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

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How Often Is It Calculated? (1 problem)

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.

Write an expression that represents the balance in one year if interest is compounded annually.
If interest is compounded semi-annually (every six months), what interest rate would be used for each calculation?
If interest is compounded semi-annually, which expression represents the account balance in $t$ years?
- $1,000 \boldcdot (1 + 0.015)^t$
- $1,000 \boldcdot \left((1 + 0.015)^{2}\right)^t$
- $1,000 \boldcdot \left((1 + 0.015)^{6}\right)^t$
- $1,000 \boldcdot \left((1 + 0.03)^{2}\right)^t$

Show Solution

$1,000 \boldcdot (1 + 0.03)$ , or $1,000 \boldcdot \left(1 + \frac{3}{100}\right)$
1.5%, or half of 3%
$1,000 \boldcdot \left((1 + 0.015)^{2}\right)^t$

Lesson 18

Expressed in Different Ways

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Printing Business (1 problem)

A small printing company launched an online ordering system to expand its business. The equation $c = 400 \boldcdot (1.03)^m$ represents the number of customers, $c$ , it has in terms of the number of months, $m$ , since it launched the ordering system.

By what factor does the number of customers grow in one year? Write your answer as an expression and a numerical value.
If $y$ is time in years, write an equation for the number of customers, $c$ , as a function of the number of years, $y$ , after the introduction of the ordering system.
If this model continues to apply for a decade, by what factor will the number of customers grow in one decade?

Show Solution

$(1.03)^{12}$ , or about 1.43 (or 1.426)
$c = 400 \left((1.03)^{12}\right)^y$ , or $c\approx 400 \boldcdot (1.43)^y$
$\left((1.03)^{12}\right)^{10}$ , or $(1.03)^{120}$ , or about $(1.43)^{10}$ , or approximately 35.76

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

How much would be in each account at the end of the 5 years? Show your reasoning.
Which bank should the investor choose?

Show Solution

Sample responses:

Bank 1: $2,323.23 because $\frac{0.03}{12} = 0.0025$ , and $2000 \boldcdot (1 + 0.0025)^{60} = 2323.23$ .
Bank 2: $2,322.37 because $\frac{0.03}{4} = 0.0075$ , and $2000 \boldcdot (1 + 0.0075)^{20} = 2322.37$
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 19

Which One Changes Faster?

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Which One Gets There First? (1 problem)

The function $f$ is given by $f(x) = 10x + 3$ , and the function $g$ is given by $g(x) =2^x$ . For each question, show your reasoning.

Which function reaches 50 first?
Which function reaches 100 first?

Show Solution

$f(5) = 53$ and $g(5) = 32$ , so $f$ reaches 50 first.
$f(7) = 73$ and $g(7) = 128$ , so $g$ reaches 100 first.

Lesson 20

Changes over Equal Intervals

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Increasing Input by One (1 problem)

The linear function $g$ is defined by $g(x) = 4x$ .
1. Show that $g(3)$ and $g(2)$ have a difference of 4.
2. Show that $g(x+1) - g(x) = 4$ .
The exponential function $h$ is defined by $h(x) = 4^x$ .
1. Show that $h(3)$ and $h(2)$ have a quotient of 4.
2. Show that $\frac{h(x+1)}{h(x)} = 4$ .

Show Solution

1. Sample responses:
  - $g(3)=4 \boldcdot 3 = 12$ and $g(2) = 4 \boldcdot 2=8$ . The difference between 12 and 8 is 4.
  - $g(3)-g(2)=4\boldcdot(3) - 4\boldcdot(2)= 4\boldcdot (3-2) = 4$
2. $g(x+1)-g(x)=4\boldcdot(x+1) - 4x= 4\boldcdot (x+1-x) = 4\boldcdot(1)$ , which equals 4.
1. Sample responses:
  - $h(3)=4^3 = 64$ and $h(2) = 4^2=16$ . The quotient is $\frac{64}{16}$ , which is 4.
  - $\frac{h(3)}{h(2)} = \frac{4^3}{4^2}=4^{3-2}=4$
2. $\frac{h(x+1)}{h(x)} = \frac{4^{x+1}}{4^x}=4^{x+1-x}=4^1$ , which equals 4.

Section E Check

Section E Checkpoint

Problem 1

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

Write an exponential expression of the form $a \boldcdot b^x$ equal to $f(x+1)$ . Show your reasoning.
Show that $\frac{f(x+1)}{f(x)}$ is a constant.

Show Solution

$2 \boldcdot 2^x$ because $f(x+1) = 2^{x+1}$ which is equal to $2 \boldcdot 2^x$
Sample response: $\frac{f(x+1)}{f(x)} = \frac{2^{x+1}}{2^x} = 2^{x+1-x} = 2^1$ which is 2.

Problem 2

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

Show that $h(0) > g(0)$
For the same $x$ -value, will $g(x) > h(x)$ ? If so, give an example.

Show Solution

$h(0) = 1000$ and $g(0) = 1$ , so $h(0) > g(0)$
Yes. Sample response: For any value greater than 7 this is true. For example, $g(7) = 2,187$ and $h(7) = 1,021$ .

Lesson 21

Predicting Populations

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No cool-down

Unit 6 Introduction To Exponential Functions — Unit Plan