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Understanding the *growth* or *decay* of many real-life situations leads to making correct decisions and getting the best results. This lesson will explore situations that can be modeled using linear or exponential functions. Analyzing the rate of change will determine which of these models best fits each situation. ### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

During the holidays, Ignacio's family plans to visit the theme park Mondo Marino in California. While looking at the park's website to buy tickets, Ignacio found information about a brand new aquarium inside the park.

Wow! Ignacio notices that the park has a great contest. The person who solves a series of problems about the fish population in the aquarium will win an annual family fun pass. The website provides a function that models the fish population's growth.$f(x)=100b_{x} $

In this function, $x$ represents the weeks after the fish were introduced into the aquarium, and $b$ is an unknown positive growth factor. To win the contest, these values need to be determined. a What is the number of fish initially released into the aquarium?

b If the aquarium contains $850$ fish after $25$ weeks, what is the value of $b?$ Round to $2$ decimal places.

c Suppose that the growth factor is $2.$ What is the percentage growth rate per week of the fish?

Functions can be used to model many real-life situations and solve many real-life problems. For this reason, it is essential to identify which type of function is the most appropriate for any given situation. Two of the most common types of functions are linear and exponential functions.
### Answer

### Hint

### Solution

*less than* $1,$ it describes a decay factor. Sorry for Marky, but the given situation represents an exponential decay. This means that this is modeled by an exponential function.
*less than* $1,$ this situation describes an exponential decay. This means that the situation can be modeled by an exponential function.
*greater than* $1,$ this is a growth factor. Therefore, this situation describes an exponential growth and is modeled by an exponential function.

$Linear Function f(x)=ax+b Exponential Function g(x)=a⋅b_{x} $

While linear functions grow by equal differences over equal intervals, exponential functions grow by equal factors. The following figure shows the graph of example linear and exponential functions.
In the following situations, identify and explain whether it can be modeled by a linear or an exponential function.

a Marky just got paid after releasing his first pop single

$$$ Be Growing,so he opens a bank account. He deposits an initial amount of $$850$ and plans to deposit the same quantity every month. The account earns no interest.

b The value of Marky's collection of dodgeball trading cards decreases at a rate of $12%$ per year.

c Marky takes a specific medicine to treat a fever. Every hour, $87 $ of the medicine's concentration in the bloodstream remains.

d A certain rabbit colony's population, started by Marky releasing his pet bunnies into the

wild, increases by $2.5%$ every year.

e Marky's small town had a population of $1150$ people in $1890.$ Each year, the town's population increased by $150$ people. It even continued increasing despite the pop song

$$$ Be Growingbeing played non-stop on the local radio.

a Linear

b Exponential

c Exponential

d Exponential

e Linear

a The balance of the account after $x$ months of the initial deposit can be determined by adding the product of $x$ and the deposit to the initial amount.

b Write an expression for the new price of the dodgeball trading card collection after one year.

c Write an expression for the medicine's concentration found in the bloodstream after one hour.

d If $P$ is the current population of rabbits, what will be the population after one year?

e The town's population after $x$ years from $1890$ can be calculated by adding the product of $x$ and $150$ to the initial population size.

a It is given that the person plans to deposit the quantity of $$850$ every month. The balance $B(x)$ of the account after $x$ months of the initial deposit can be calculated by adding the product of $x$ and $850$ to the initial amount $850.$

$Balance AfterxMonths B(x)=850+850x $

This shows that the situation can be modeled by a linear function.
b The value of the dodgeball trading cards collection decreases by $12%$ each year. Let $C$ represent the current price of the collection. Then, in the next year, the price of the collection will be replaced by $C$ minus $0.12C.$

$New Price of the Car C−0.12C=0.88C $

It can be seen that the value of the dodgeball cards will be multiplied every year by a constant factor of $0.88.$ Because this constant factor is c If $D$ represents the current concentration of the medicine, in the next hour, $87 D$ will remain in the bloodstream.

$New Concentration of the Drug: 87 D $

Similar to Part B, because this constant factor is d Suppose that $P$ represents the current size of the population of rabbits. The next year, this population will increase by $2.5%.$ Therefore, the new population size will be substituted by the sum of $P$ and $0.025P.$

$New Population Size: P+0.025P=1.025P $

Note that the population is multiplied every year by a constant factor of $1.025.$ Because this constant factor is e It is given that the town's population in $1890$ was $1150$ people. Also, each year, the population increased by $150$ people. Therefore, the town's population $P(x)$ after $x$ years from $1890$ can be calculated by adding the product of $x$ and $150$ to $1150.$

$Population AfterxYears P(x)=1150+150x $

It can be concluded that this situation is modeled by a linear function. Marky's music had nothing to do with the function. Marky and his friend Tifanniqua are playing a video game in which they each have to build a city. The game gives them $15000$ v-coins for buying supplies. Since this virtual money is not enough, they must invest this initial amount. After some exploration, they have discovered two banks in the game that can help them increase their v-coins.

Considering that one year in the game is about $12$ minutes in the real world, the following are the offers of each virtual bank in the game.

**Bank A**— a simple interest of $3%$ per year, meaning that the balance will increase by $3%$ of the initial investment each year.**Bank B**— a compound interest of $2.5%$ per year, compounded monthly, meaning that the balance increases by one-twelfth of $2.5%$ of the previous balance each month.

a If Marky and Tiffaniqua plan to maintain their investment for $10$ years, who made the better choice?

b Instead of $10$ years, they decided to keep their investment for $20$ years. Does the better option change?

c Write an expression for $A(t)$ representing the balance in Bank A after $t$ years of the initial investment. Similarly, write an expression for $B(m)$ representing the balance in Bank B after $m$ months of the initial investment.

d Make a table of values showing the balances of the two banks from the first to the twentieth year. Interpret the results.

a Marky

b Tiffaniqua

c **Bank A: ** $A(t)=15000+450t$

**Bank B: ** $B(m)=15000(1+120.025 )_{m}$

d **Table of Values: **

Investment Period | Bank A | Bank B |
---|---|---|

$A(t)=15000+450t$ | $B(m)=15000(1+120.025 )_{m}$ | |

$1$ Year ($12$ Months) | $15450$ | $≈15379$ |

$2$ Years | $15900$ | $≈15768$ |

$3$ Years | $16350$ | $≈16167$ |

$4$ Years | $16800$ | $≈16576$ |

$5$ Years | $17250$ | $≈16995$ |

$6$ Years | $17700$ | $≈17425$ |

$7$ Years | $18150$ | $≈17865$ |

$8$ Years | $18600$ | $≈18317$ |

$9$ Years | $19050$ | $≈18780$ |

$10$ Years | $19500$ | $≈19255$ |

$11$ Years | $19950$ | $≈19742$ |

$12$ Years | $20400$ | $≈20241$ |

$13$ Years | $20850$ | $≈20753$ |

$14$ Years | $21300$ | $≈21278$ |

$15$ Years | $21750$ | $≈21816$ |

$16$ Years | $22200$ | $≈22368$ |

$17$ Years | $22650$ | $≈22934$ |

$18$ Years | $23100$ | $≈23514$ |

$19$ Years | $23550$ | $≈24108$ |

$20$ Years | $24000$ | $≈24718$ |

**Interpretation: ** Bank B generates higher income after the fourteenth year. However, for the previous years, Bank B is more profitable.

a Begin by finding how much Marky will receive per year. Then determine the growth factor of Tiffaniqua's investment and compare the results.

c Marky's balance after $t$ years will be given by $t$ times the earnings per year added to the initial investment. Furthermore, Tiffaniqua's balance after $m$ months will be given by the product of the initial investment and the growth factor to the $m_{th}$ power.

d Use the formulas obtained in Part C. In which year does one balance surpasses the other?

a Both Marky and Tiffaniqua plan to keep their investment for $10$ years. The options will be analyzed one at a time to find the amounts Marky and Tiffaniqua will receive after $10$ years.

$Annual Profit 0.03⋅15000=450 $

Bank A will give $450$ v-coins to Marky for each year he keeps his investment in the bank. Now, because Marky plans to keep his investment for $10$ years, $450$ will be multiplied by $10$ to get the total profit. $Total Profit 450⋅10=4500Final Balance 15000+4500=19500 $

Therefore, Marky will be able to withdraw $19500$ v-coins from the bank after ten years. $Growth Factor 1+120.025 $

The balance after $10$ years, which is equal to $120$ months, will be given by the product of the initial amount and the growth factor raised to $120_{th}$ power. Use a calculator to evaluate the final balance.
$Final Balance 15000(1+120.025 )_{120}≈19255 $

Therefore, Tiffaniqua will receive about $19255$ v-coins after ten years. It has been obtained that the balance of Marky's investment after $10$ years will be $19500$ v-coins and Tiffaniqua's balance will be about $19255.$ Therefore, the offer of Bank A is better for the period of $10$ years.

b Following a similar reasoning as in Part A, the balance after $20$ years for each investment can be determined. In this case, Marky will have a total profit of $20⋅450=9000.$ This will be added to his initial investment to get the final balance.

$Marky’s Final Balance 15000+9000=24000 $

Now, because there are $240$ months in $20$ years, the growth factor will be raised to $240_{th}$ power and multiplied by the initial investment to obtain Tiffaniqua's final balance. $Tiffaniqua’s Final Balance 15000(1+120.025 )_{240}≈24718 $

It can be seen that for a $20-$year period, the Tiffaniqua's profit will be greater. This now means that the better option is offered by Bank B.
c To find an expression for the balance $A(t)$ after $t$ years, the product of $t$ and $450$ should be added to the initial investment of $15000$ v-coins.

$Balance in Bank A AftertYears A(t)=15000+450t $

In case of Bank B, $B(m)$ will be given by the product of the growth factor $1+120.025 $ raised to the $m_{th}$ power and the initial investment.
$Balance in Bank B AftermMonths B(m)=15000(1+120.025 )_{m} $

d Using the formulas found in Part C, a table of values can be generated showing the balance of Bank A and Bank B from the first to the twentieth year. Keep in mind that $t$ is equal to $12$ times $m.$

Investment Period in Years | Bank A | Bank B |
---|---|---|

$A(t)=15000+450t$ | $B(m)=15000(1+120.025 )_{m}$ | |

$1$ | $15450$ | $≈15379$ |

$2$ | $15900$ | $≈15768$ |

$3$ | $16350$ | $≈16167$ |

$4$ | $16800$ | $≈16576$ |

$5$ | $17250$ | $≈16995$ |

$6$ | $17700$ | $≈17425$ |

$7$ | $18150$ | $≈17865$ |

$8$ | $18600$ | $≈18317$ |

$9$ | $19050$ | $≈18780$ |

$10$ | $19500$ | $≈19255$ |

$11$ | $19950$ | $≈19742$ |

$12$ | $20400$ | $≈20241$ |

$13$ | $20850$ | $≈20753$ |

$14$ | $21300$ | $≈21278$ |

$15$ | $21750$ | $≈21816$ |

$16$ | $22200$ | $≈22368$ |

$17$ | $22650$ | $≈22934$ |

$18$ | $23100$ | $≈23514$ |

$19$ | $23550$ | $≈24108$ |

$20$ | $24000$ | $≈24718$ |

It can be seen that Bank B generates a higher income after the fourteenth year. However, for the previous years, Bank A is more profitable. The friends have been having such a great time playing video games. In the end, all was just fake money.

Besides singing, Marky is really into fishing. A small company in town called SeaBase Fish specializes in shrimp farming. They currently only farm shrimp, but they want to expand and introduce two types of fish next year, Catfish and Tilapia. The good news for Marky is that they will allow local kids the chance to fish recreationally.

Marky has never caught either type of fish in his life. He is interested in finding information about both types and then sharing this information with his friends. Online, Marky found information from another study about the growth model of each species' population after $x$ weeks of the initial population being introduced into the sea.$Catfish y_{C}=55x+10 Tilapia y_{T}=4⋅2_{x} $

Marky is a bit unsure of how to model the graphs. He wonders what the growth rates of the fish will actually be. a Graph both functions in the same coordinate plane.

b How much time after introducing the initial fish population is needed to expect that there will be the same amount of Catfish and Tilapia?

c Compare the growth rates of the two fish types.

a

b About six and a half weeks

c The Catfish population will increase by $55$ every week. The Tilapia population will double each week.

a Begin by making a table of values for each function.

b The size of the populations will be the same when the graphs of their functions intersect.

c Linear functions grow by equal differences over equal intervals and exponential functions grow by equal factors.

a The given functions can be graphed by making a table of values. Then, each data point will be plotted in the coordinate plane. Begin with a table corresponding to Catfish's model. Because $x$ is the number of weeks, only non-negative values of $x$ will be used.

$x$ | $55x+10$ | $y_{C}=55x+10$ | $(x,y_{C})$ |
---|---|---|---|

$0$ | $55(0)+10$ | $10$ | $(0,10)$ |

$2$ | $55(2)+10$ | $120$ | $(2,120)$ |

$4$ | $55(4)+10$ | $230$ | $(4,230)$ |

$6$ | $55(6)+10$ | $340$ | $(6,340)$ |

$8$ | $55(8)+10$ | $450$ | $(8,450)$ |

The data plots can be plotted and connected in a coordinate plane.

Similarly, a table for Tilapia's model will be created.

$x$ | $4⋅2_{x}$ | $y_{T}=4⋅2_{x}$ | $(x,y_{T})$ |
---|---|---|---|

$0$ | $4⋅2_{0}$ | $4$ | $(0,4)$ |

$2$ | $4⋅2_{2}$ | $16$ | $(2,16)$ |

$4$ | $4⋅2_{4}$ | $64$ | $(4,64)$ |

$6$ | $4⋅2_{6}$ | $256$ | $(6,256)$ |

$8$ | $4⋅2_{8}$ | $1024$ | $(8,1024)$ |

Now, the graph for Tilapia fish can be added by plotting and connecting the data points obtained in the second table.

b The graphs of the functions represent the population size of fish after $x$ weeks they are introduced into the sea. Therefore, the $x-$value where the graphs intersect represents the number of weeks after there is approximately the same amount of Catfish and Tilapia.

It can be seen that the functions intersect at $x≈6.5.$ Therefore, it is expected that after about six and a half weeks, the sizes of both populations will be approximately equal. This can be verified by evaluating each function when $x=6.5.$

Catfish | Tilapia | |
---|---|---|

Growth Function | $y_{C}=55x+10$ | $y_{T}=4⋅2_{x}$ |

Growth After $621 $ Weeks | $y_{C}=55(6.5)+10$ | $y_{T}=4⋅2_{6.5}$ |

Evaluate and Approximate | $y_{C}≈368$ | $y_{T}≈362$ |

It can be noted although $x=6.5$ is not the exact answer, population sizes are very close.

$Catfish y_{C}=55x+10⇒m=55 $

This means that the Catfish population will have a constantly increase of $55$ every week. Next, the growth rate of the exponential growth is the growth factor given by the base $b$ of the exponent.
$Tilapia y_{T}=4⋅2_{x}⇒b=2 $

It can be concluded that Tilapia´s population will double every week.
A possible interpretation that Marky can make from his findings is that after six-and-a-half weeks after the fish are introduced, he can focus on catching Tilapia. That interpretation can be made assuming that Marky wants to catch the fish species that has such a high population that it is easier to catch.

Ignacio's family is going to have some large expenses for a holiday trip, so Ignacio's mom asked him for a hand. She wants to post some ads on social media to reach more clients to generate sales for her a client. She represents an up-and-coming musician named Marky. There is one problem, Ignacio has no idea which social media site is the best option.

For this reason, Ignacio collected some data about how the number of users of these social media sites increases on a weekly basis.

User Growth per Week on Each Social Media (in Thousands) | ||
---|---|---|

Time (in Weeks) | Option 1 | Option 2 |

$0$ | $18$ | $15$ |

$1$ | $21$ | $18$ |

$2$ | $24$ | $21.5$ |

$3$ | $27$ | $26$ |

$4$ | $30$ | $31$ |

$5$ | $33$ | $37.5$ |

Ignacio wonders if this data can help him make the best decision. Find the following information to help him decide where they should post the ads.

a Using the data collected in the table, determine the trend of user growth of each social media.

b Write a function that models the data of each social media.

c Predict the number of users of each social media after $1$ year. Which option should Ignacio choose to post his ads?

a **Option $1$: ** Linear

**Option $2$: ** Exponential

b **Option $1$: ** $f(x)=3x+18$

**Option $2$: ** $g(x)=15⋅1.2_{x}$

c **Option $1$: ** $54000$ users

**Which Option Should Ignacio Choose?** Option $2$

**Option $2$: ** $≈133700$ users

a Linear functions grow by equal differences over equal intervals and exponential functions grow by equal factors.

b A linear function has the form $f(x)=mx+b$ and an exponential function is given by the equation $g(x)=a⋅b_{x}.$

c Evaluate each function at $x=12.$ Which option is expected to attract more users?

a By finding the patterns in the data sets, the model that describes each user growth can be determined. Linear functions grow by equal differences over equal intervals and exponential functions grow by equal factors. Using this information, the pattern for Option $1$ will be determined.

The number of users of Option $1$ increases by a constant of $3$ thousands every week. This data follows a linear model. Using the same process, the pattern for Option $2$ will be found.

Each week, the number of users of Option $2$ grows by a factor of about $1.2.$ That means this data follows an exponential model.

b To write the function that describes each option, the data sets will be analyzed one at a time.

$f(x)=mx+b $

In Part A, it was found that the number of users of Option $1$ increases by $3$ thousands. Since the data is given in thousands, $3$ represents the slope of the function. This value will be substituted for $m$ in the general form.
$f(x)=mx+b⇓f(x)=3x+b $

Now, the value of $b$ can be determined by substituting one of the data points into the above equation and solving it for $b.$ In this case, $(0,18)$ will be used.
$f(x)=3x+b$

SubstituteII

$x=0$, $f(x)=18$

$18=3(0)+b$

ZeroPropMult

Zero Property of Multiplication

$18=b$

RearrangeEqn

Rearrange equation

$b=18$

$f(x)=3x+18 $

$g(x)=a⋅b_{x} $

In this function, $b$ represents the growth factor. Because the data corresponding to Option $2$ increases by a factor of about $1.2,$ this value will be substituted for $b$ into the general form.
$g(x)=a⋅b_{x}⇓g(x)=a⋅1.2_{x} $

Similar to the linear function, this function rule will be completed using the first data point $(0,15)$ corresponding to Option $2$ to determine the value of $a.$
$g(x)=a⋅1.2_{x}$

SubstituteII

$x=0$, $g(x)=15$

$15=a⋅1.2_{0}$

Solve for $a$

ExponentZero

$a_{0}=1$

$15=a⋅1$

IdPropMult

Identity Property of Multiplication

$15=a$

RearrangeEqn

Rearrange equation

$a=15$

$g(x)=15⋅1.2_{x} $

c To predict the number of users of each social media after $1$ year, each function will be evaluated at $x=12.$ Begin by evaluating the function for Option $1$ at this value.

$f(x)=3x+18$

Substitute

$x=12$

$f(x)=3(12)+18$

Multiply

Multiply

$f(x)=36+18$

AddTerms

Add terms

$f(x)=54$

$g(x)=15⋅1.2_{x}$

Substitute

$x=12$

$g(x)=15⋅1.2_{12}$

CalcPow

Calculate power

$g(x)≈15(8.9)$

Multiply

Multiply

$g(x)≈133.5$

Sells are doing great. Ignacio's family can now have an even more fantastic holiday at the theme park.

The challenge presented at the beginning of the lesson can now be solved. The theme park's website provides a function that models fish population growth into the aquarium.
b If the aquarium contains $850$ fish after $25$ weeks, what is the value of $b?$ Round to $2$ decimal places.
### Hint

### Solution

Therefore, the initial number of fish that was released into the Aquarium is $100.$

$f(x)=100⋅b_{x} $

In this function, $x$ represents the weeks after the fish was introduced into the aquarium and $b$ is an unknown positive growth factor. Since Ignacio wants to win the contest, he needs to determine this information. a What is the number of initial fish released into the aquarium?

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c Suppose that the growth factor is $2.$ What is the percentage growth rate per week of the fish?

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a Calculate the value of the exponential function at $x=0.$

b Substitute $25$ for $x$ and $850$ for $f(x).$ Then, solve the resulting equation for $b.$

c Substitute $x+1$ for $x.$ Then, rewrite the resulting equation in terms of $f(x).$

a Consider the given function.

$f(x)=100⋅b_{x} $

This function models the population size of the fish after $x$ weeks. Evaluate this function when $x=0$ to determine the size of the initial population. $f(x)=100⋅b_{x}$

Substitute

$x=0$

$f(0)=100⋅b_{0}$

ExponentZero

$a_{0}=1$

$f(0)=100⋅1$

IdPropMult

Identity Property of Multiplication

$f(0)=100$

b It is given that after $25$ weeks of the release, there are $850$ fish. Substitute these values into the function rule. Then, solve the resulting equation for $b.$

$f(x)=100⋅b_{x}$

SubstituteII

$f(x)=850$, $x=25$

$850=100⋅b_{25}$

Solve for $b$

DivEqn

$LHS/100=RHS/100$

$100850 =b_{25}$

CalcQuot

Calculate quotient

$8.5=b_{25}$

RaiseEqn

$LHS_{251}=RHS_{251}$

$8.5_{251}=(b_{25})_{251}$

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$8.5_{251}=b_{25⋅251}$

DenomMultFracToNumber

$25⋅25a =a$

$8.5_{251}=b_{1}$

ExponentOne

$a_{1}=a$

$8.5_{251}=b$

RearrangeEqn

Rearrange equation

$b=8.5_{251}$

UseCalc

Use a calculator

$b=1.089373…$

RoundDec

Round to $2$ decimal place(s)

$b≈1.09$

c In the given case, the fish's percentage growth rate per week will be determined if $b=2.$

$f(x)=100⋅b_{x}⇓f(x)=$