Introduction

Welcome back to AP Precalculus! We are on Unit 2. In the previous articles, we discussed exponential functions very extensively, talking about what they are, their form being $f(x)=a\cdot b^x$, and different ways to apply transformations such as shifting them up or down, or dilating them horizontally or vertically. Today we will be talking about Exponential Function Context and Data Modeling where we use these functions to express different scenarios that are applicable! With no further ado, let’s get started!

Modeling Data with Math

Let’s dive into the following scenario: A company is trying to track the total number of sales a product made over the first year that it was released. To visualize this data, the company took a snapshot of the total number of sales each week and graphed it in a chart, which is shown below:

Now the company wants to predict how many total sales they will have by the end of the next year. The issue is that, at the moment, all we have is data, and while data on its own is nice, it doesn’t tell us much besides the general trends on what is happening. This is where regressions come in. Regressions allow us to take a dataset, such as the one shown below, and try to fit a function to the data by changing the constants of the data. For example, if I were to use an exponential regression on the dataset, the calculator will try to find values of $a$ and $b$ that make the function $a{b}^x$ most closely fit the data. In this case, the optimal values of $a$ and $b$ were $0.520224$ and $1.10658$ respectively. Below is the graph of the function $0.520224\cdot 1.10658^x$.

The regression above is an exponential regression, since it attempts to fit the data to an exponential function. There are many different types of regressions you can use, such as a linear regression (which tries to match the function to a linear equation).

Calculating a regression

The below guide will be using Desmos, as you are given access to it on the Bluebook application. [Editor's Note: More details about Desmos Calculator Usage can be found in Exam Guides section of AP Precalculus]. Please refer to your specific calculator’s manual for advice regarding your calculator

1) Click on the $+$ icon above the equations:

2) Click on the table:

3) Fill in the data into the table

Click this icon next to the the table and select your desired regression type.

Modeling Data without a calculator

The AP exam is divided into two parts, one where you are allowed to use your calculator, and one where you aren’t. Using a regression tool on your calculator works for the calculator-allowed section of the test, but is not possible on the calculator-disallowed section of the test. The methods used to solve these problems have been mostly discussed in previous topics, so you should already be familiar with how to solve these. Below are a couple of question formats you might be given on a quiz or the AP exam along with how to solve them.

You have quite a bunch of options here! Let’s do an example so you can understand a few of these methods.

EXAMPLE 1:

A person is playing a game where the number of points is represented by $p(r)$ after each round $r$. Points are used to upgrade tools and armor and a player can earn more points in each round. The following table shows the player’s progress.

$\begin{array}{lccccc}\text{Rounds Played}(r)& 1& 2& 3& 4& 5\\ \text{Points}(p(r))& 10& 12& 16& 24& 40\end{array}$

Find an exponential function that best models this data.

Take some time to work it out and try to find the ideal function.

The first thing that you may notice is that the data is pretty interesting in the fact that if you try to verify if the data is exponential, the numbers just don’t add up. If we take the points from Round 2 to 1 and divide, we get $\frac{p(2)}{p(1)}=\frac{12}{10}=1.2$. However, if we do from Round 3 to 2 and divide, we get $\frac{p(3)}{p(2)}=\frac{16}{12}=1.33$. Not the biggest difference, but if we continue, we get $\frac{p(4)}{p(3)}=\frac{24}{16}=1.5$ and $\frac{p(5)}{p(4)}=\frac{40}{24}=1.67$. Something is not right. 

If you have a keen eye, then you may be able to spot that if we subtract $8$ from $p(r)$, we get something very close to a sort of exponential function.

$\begin{array}{lccccc}\text{Rounds Played}(r)& 1& 2& 3& 4& 5\\ \text{Points}-8(p(r)-8)& 2& 4& 8& 16& 32\end{array}$

Turns out, this does match an exponential function, specifically $2^x$. As such, we can say that $2^x$ models this well. We subtracted 8 from the table, so it only makes sense that our original table is modeled with the function $f(x)=2^x+8$.

Notice that we had to modify the table a little and realize that there is some constant that is added to the data that makes the data not match up perfectly to an exponential function. This can occur, and it is something to look out for.

EXAMPLE 2:

Jackson deposits $$100$ into a savings account represented by $s(m)$ that earns $4\%$ interest annually and is compounded. In terms of $m$, months, find a function that can predict the amount of money Jackson will have in the future, and use the function to find the amount of money he will have in exactly $40$ months.

Take some time to figure out this problem. Be a little innovative!

Alright. Firstly, this is a weird-ish problem, because we are technically given all we need, but it is not very clear, but let’s break down everything. If Jackson deposits $$100$ initially, that means that the savings account will start with $$100$. In an exponential function in the form $a\cdot b^x$, $a$ is the starting value. Since our starting value is $100$, we set $a=100$.

Next up, realize that Jackson earns four percent of what he has in his account every year, or plus $4\%.$ How do we make this into an exponential equation? Well, let’s take a quick example.

If we double something continuously, we can represent it as $2^x$. Now, it is not wrong to say that this is also $(1+1{)}^x$. If you don’t see the point yet, let me make it clearer by saying that this is also $(1.00+1.00{)}^x$. If I had two objects and they doubled, I would have four of those objects. I have just gained and added to my total 100% of my objects. This matters because $(1.00+1.00{)}^x$ is saying that the first 1.00 is the baseline, and the second 1.00 is how much I gain after each time. As such, wouldn’t $(1.00+0.04{)}^x$ mean that I gain $0.04$, or $4\%$ each time? As such, it only makes sense that $b=1.04$.

Our equation is now $f(x)=100\cdot (1.04{)}^x$, and this is correct, right? Well… not quite. You see, Jackson gains $4\%$ every year, but the question asks us to model what happens each month. How do we convert our years into months? Well, we know one year is twelve months (hopefully). If we increase $x$ by one (our placeholder variable for a year), then Jackson earns a nice 4%. However, how do we replace $x$ with $m$ in such a way that if we raise $m$ by 12 months, or one year, it will give Jackson that sweet $4\%$? Well, can’t we just replace $x$ with $\frac{m}{12}$? Think about it for a second.

Our equation is now $s(m)=100\cdot (1.04{)}^{\frac{m}{12}}$, and we are done with solving for the function. The only step now is to plug in $m=40$ and get an answer.

$s(40)=100\cdot (1.04{)}^{\frac{40}{12}}=113.997\approx 114$

This is quite a hefty problem, so I am going to walk you through another one of a similar style so that you can lock this type of problem into your mind.

EXAMPLE 3:

In a game, every five waves is one round, and a player begins with $200$ health represented by $h(w)$. A player knows that a deadly poison effect takes away a fifth of the player’s total health every round, but each wave causes the player to take damage that compounds. Find an exponential function that models a player’s health each wave, $w$, after receiving the deadly poison effect, and find how many waves it takes for the player’s health to go down to $30$ or less.

This one is an interesting scenario, so take your time.

Firstly, the exponential form is $a\cdot b^x$. Since the player begins with $200$ health, $a=200$. For the value of $b$, we do $(1.00+\_\_{)}^x$. What is the missing number? Well, if we are losing health, then the function must be an exponential decay, which means $b$ is from $0<b<1$. As such, the missing number must be negative and only makes sense that it is $-20$, or $-0.20$. As such, our base is $1.00+(-0.20)=0.80=b$. We have $200\cdot (0.80{)}^x$. We can also see this as the player conserving $80\%$ of his health after the damage.

As we know, $x$ represents each round, but we have to represent the function in terms of each wave. As we know, five waves is one round. If $x$ goes up by $1$, the player loses $20\%$ health. We need to make it so that if $w$ goes up by $5$, which is equivalent to one round, the player also loses $20\%$. As such, only $\frac{w}{5}$ works to replace $x$. Our new function is $h(w)=200\cdot (0.80{)}^{\frac{w}{5}}$.

Finally, let’s find out how many waves it takes for the player’s health to go down to $30$ or less by setting the function equal to $30$.

$30=200\cdot (0.80{)}^{\frac{w}{5}}\to w=42.509$

Finally, round up because waves are in whole numbers, not in decimals. As such, it will take at least $43$ waves to get down to $30$ health or less.

e

$\mathbf{e}$ is a constant that is approximately equal to $2.718$, and appears frequently in problems involving real-life scenarios. I mean, just know about this constant as it can appear.

Questions