Unit 6 Introduction To Exponential Functions — Unit Plan

Here are two tables representing two different situations.

• A student runs errands for a neighbor every week. The table shows the pay he may receive, in dollars, in any given week.

  number of errands	pay in dollars	difference from previous week	factor from previous week
  0	10	-	-
  1	15	5	1.5
  2	20	5	1.33
  3	25	5	1.25
  4	30	5	1.2

• A student at a high school heard a rumor that a celebrity will be speaking at graduation. The table shows how the rumor is spreading over time, in days.

  day	people who have heard the rumor	difference from previous day	factor from previous day
  0	1	-	-
  1	5	4	5
  2	25	20	5
  3	125	100	5
  4	625	500	5

Once we recognize how these patterns change, we can describe them mathematically. This allows us to understand their behavior, extend the patterns, and make predictions.

Notice that in the situation with the student running errands, the difference is constant from week to week, while the factor changes. In the situation about a rumor spreading, the difference changes from day to day, but the factor is constant. This can give us clues to how we might write out the pattern in each situation.

In relationships where the change is exponential, a quantity is repeatedly multiplied by the same amount. The multiplier is called the growth factor.

Suppose a population of cells starts at 500 and triples every day. The number of cells each day can be calculated as follows:

We can see that the number of cells ($p$) is changing exponentially, and that $p$ can be found by multiplying 500 by 3 as many times as the number of days ($d$) since the 500 cells were observed. The growth factor is 3. To model this situation, we can write this equation: $\displaystyle p = 500 \boldcdot 3^d$.

The equation can be used to find the population on any day, including day 0, when the population was first measured. On day 0, the population is $500 \boldcdot 3^0$. Since $3^0 = 1$, this is $500 \boldcdot 1$ or 500.

Here is a graph of the daily cell population. The point $(0,500)$ on the graph means that on day 0, the population starts at 500.

Graph of an exponential function, origin O. Horizontal axis, number days, scale 0 to 4, by 1’s Vertical axis, cell population, scale 0 to 20,000, by 5,000’s. The function is discrete and has these points: (0 comma 500), (1 comma 1,500), (2 comma 4,500) and (3 comma 13,500).

Each point is 3 times higher on the graph than the previous point. $(1,1500)$ is 3 times higher than $(0,500)$, and $(2,4500)$ is 3 times higher than $(1,1500)$.

Here is a graph showing the luminescence of a glow-in-the-dark paint, measured in lumens, over a period of time, measured in hours. The luminescence of this glow-in-the-dark paint can be modeled by an exponential function.

Notice that the amounts are decreasing over time. The graph includes the point $(0, 12)$. This means that when the glow-in-the-dark paint started glowing, its glow measured 12 lumens. The point $(1, 6)$ tells us the glow measured 6 lumens 1 hour later. Between 3 and 4 hours after the glow-in-the-dark paint began to glow, the luminescence fell below 1 lumen.

We can use the graph to find out what fraction of luminescence stays each hour. Notice that $\frac{6}{12}=\frac{1}{2}$ and $\frac{3}{6}=\frac{1}{2}$. As each hour passes, the luminescence that stays is multiplied by a factor of $\frac{1}{2}$.

If $y$ is the luminescence, in lumens, and $t$ is time, in hours, then this situation is modeled by the equation:

$y=12 \boldcdot (\frac{1}{2})^t$

We can confirm that the data is changing exponentially because it is multiplied by the same value each time. When the growth factor is between 0 and 1, the quantity being multiplied decreases, the situation is sometimes called “exponential decay,” and the growth factor may be called a “decay factor.”

Graphs are useful for comparing relationships. Here are two graphs representing the amount of caffeine in Person A and Person B, in milligrams, at different times, measured hourly, after an initial measurement.

Graph of an exponential function, origin O. Horizontal axis, time (hours), scale 0 to 10, by 1’s Vertical axis, caffeine (mg), scale 0 to 200, by 100’s. The function is discrete and has these approximate points: (0 comma 200), (1 comma 160), (2 comma 125), (3 comma 105), (4 comma 80), (5 comma 65), (6 comma 50), (7 comma 45), (8 comma 35), (9 comma 25), (10 comma 20), (11 comma 18) and (12 comma 12).

Graph of an exponential function, origin O. Horizontal axis, time (hours), scale 0 to 10, by 1’s. Vertical axis, caffeine (mg), scale 0 to 200, by 100’s. The function is discrete and has these approximate points: (0 comma 100), (1 comma 90), (2 comma 80), (3 comma 70 ), (4 comma 65), (5 comma 60), (6 comma 57), (7 comma 50), (8 comma 45), (9 comma 40), (10 comma 38), (11 comma 35) and (12 comma 30).

The graphs reveal interesting information about the caffeine in each person over time:

• At the initial measurement, Person A has more caffeine (200 milligrams) than Person B (100 milligrams).
• The caffeine in Person A's body decreases faster. It went from 200 to 160 milligrams in an hour. Because 160 is $\frac{8}{10}$ (or $\frac45$) of 200, the growth factor is $\frac45$.
• The caffeine in Person B's body went from 100 to about 90 milligrams, so that growth factor is about $\frac{9}{10}$. This means that after each hour, a larger fraction of caffeine stays in Person B than in Person A.
• Even though Person A started out with twice as much caffeine, because of the growth factor, Person A had less caffeine than Person B after 6 hours.

Equations are useful not only for representing relationships that change exponentially, but also for answering questions about these situations.

Suppose a bacteria population of 1,000,000 has been increasing by a factor of 2 every hour. What was the size of the population 5 hours ago? How many hours ago was the population less than 1,000?

We could go backward and calculate the population of bacteria 1 hour ago, 2 hours ago, and so on. For example, if the population doubled each hour and was 1,000,000 when first observed, an hour before then it must have been 500,000, and two hours before then it must have been 250,000, and so on.

Another way to reason through these questions is by representing the situation with an equation. If $t$ measures time in hours since the population was 1,000,000, then the bacteria population can be described by the equation:

$\displaystyle p = 1,000,000 \boldcdot 2^t$

The population is 1,000,000 when $t$ is 0, so 5 hours earlier, $t$ would be -5 and here is a way to calculate the population:

$\displaystyle \begin{aligned} 1,000,000 \boldcdot 2^{\text-5} &= 1,000,000 \boldcdot \frac{1}{2^5} \\ &= 1,000,000 \boldcdot \frac{1}{32} \\ &= 31,250 \end{aligned}$

Likewise, substituting -10 for $t$ gives us $1,000,000 \boldcdot 2^{\text-10}$ (or $1,000,000 \boldcdot \frac{1}{2^{10}}$), which is a little less than 1,000. This means that 10 hours before the initial measurement the bacteria population was less than 1,000.

The situations we have looked at that are characterized by exponential change can be seen as functions. In each situation, there is a quantity—an independent variable—that determines another quantity—a dependent variable. They are functions because any value of the independent variable that makes sense corresponds to only one value of the dependent variable. Functions that describe exponential change are called exponential functions.

For example, suppose $t$ represents time in hours, and $p$ is a bacteria population $t$ hours after the bacteria population was measured. For each time $t$, there is only one value for the corresponding number of bacteria, so we can say that $p$ is a function of $t$ and we can write this as $p = f(t)$.

If there were 100,000 bacteria at the time it was initially measured and the population decreases so that $\frac{1}{5}$ of it remains after each passing hour, we can use function notation to model the bacteria population:

$\displaystyle f(t) = 100,000 \boldcdot \left(\frac{1}{5}\right)^t$

Notice the expression in the form of $a \boldcdot b^t$ (on the right side of the equation) is the same as in previous equations that we wrote to represent situations characterized by exponential change.

When we calculate the average rate of change for a linear function, no matter what interval we pick, the value of the rate of change is the same. A constant rate of change is an important feature of linear functions! When a linear function is represented by a graph, the slope of the line is the rate of change of the function.

Exponential functions also have important features. We've learned about exponential growth and exponential decay, both of which are characterized by a constant quotient over equal intervals. But what does this mean for the value of the average rate of change for an exponential function over a specific interval?

The average rate of change of $A$ from the start of treatment to Week 2 is about -107 square yards per week because $\dfrac{A(2)-A(0)}{2-0} \approx \text-107$. The average rate of change of $A$ from Week 2 to Week 4, however, is only about -12 square yards per week because $\dfrac{A(4)-A(2)}{4-2} \approx \text-12$.

The negative average rates of change show that $A$ is decreasing over both intervals, but the average rate of change for the time during Weeks 0 to 2 indicates that the values are decreasing more rapidly than during Weeks 2 to 4 due to the effect of the decay factor. For an exponential function with a growth factor greater than 1, the values for the average rate of change of each interval are positive, with the second interval increasing more quickly due to the effect of the growth factor.

The height, $h$, of the ball, in cm, after $n$ bounces can be modeled by the equation: 

$\displaystyle h = 142 \boldcdot \left(\frac{2}{3}\right)^{n}$

Here is a graph of the equation.

Graph of a function on grid, origin O. Horizontal axis, number of bounces, from 0 to 12, by 1's. Vertical axis, height in centimeters, from 0 to 150, by 25's. Line of given equation, h equals 142 time 2 thirds to the n, is graphed, passing through 0 comma 142, 1 comma 94 and two thirds, 2 comma 63 and 1 ninth. 3 comma 42 and 2 twenty-sevenths, 4 comma 28 and 4 eighty-firsts. Data points of 1 comma 95, 2 comma 61, 3 comma 39, 4 comma 26 also plotted.  

This graph shows both the points from the data and the points generated by the equation, which can give us new insights. For example, the height from which the ball was dropped is not given but can be determined. If $\frac23$ of the initial height is about 95 centimeters, then that initial height is about 142.5 centimeters, because $95 \div \frac23 = 142.5$. For a second example, we can see that it will take 7 bounces before the rebound height is less than 10 centimeters.

An exponential function can give us information about a graph that represents it.

For example, suppose that function $q$ represents a bacteria population $t$ hours after it is first measured, and $q(t) = 5,000 \boldcdot (1.5)^t$. The number 5,000 is the bacteria population measured, when $t$ is 0. The number 1.5 indicates that the bacteria population increases by a factor of 1.5 each hour.

A graph can help us see how the starting population (5,000) and growth factor (1.5) influence the population. Suppose functions $p$ and $r$ represent two other bacteria populations and are given by $p(t) = 5,000 \boldcdot 2^t$ and  $r(t) =5,000 \boldcdot (1.2)^t$. Here are the graphs of $p$, $q$, and $r$.

Graph of 3 functions labeled, p, q, and r on grid, origin O. Horizontal axis, time in hours, from 0 to 10, by 2's. Vertical axis, number of bacteria, from 0 to 40,000 by 20,000's. All three functions start at 0 comma 5,000 and trend upward and to the right. Function p passes through 2 comma 20,000 and trends rapidly upward and right. Function q passes through 4 comma11,250 and trends steadily upward and right. Function r passes through 6 comma 7,200 and trends more slowly upward and right.

All three graphs start at $5,000$, but the graph of $r$ grows more slowly than does the graph of $q$, while the graph of $p$ grows more quickly. This makes sense because a population that doubles every hour is growing more quickly than one that increases by a factor of 1.5 each hour, and both grow more quickly than a population that increases by a factor of 1.2 each hour.

 If we have enough information about a graph representing an exponential function $f$, we can write a corresponding equation.

Here is a graph of $y = f(x)$.

An equation defining an exponential function has the form $f(x) = a \boldcdot b^x$. The value of $a$ is the starting value or $f(0)$, so it is the $y$-intercept of the graph. We can see that $f(0)$ is 500 and that the function is decreasing.

The value of $b$ is the growth factor. It is the number by which we multiply the function’s output at $x$ to get the output at $x+1$. To find this growth factor for $f$, we can calculate $\frac{f(1)}{f(0)}$, which is $\frac{300}{500}$ (or $\frac35$).

Graph of a function on grid, origin O. Horizontal axis, x, from 0 to 4, by 2's. Vertical axis, y. from 0 to 600, by 200's. Plotted points as follows: 0 comma 500, 1 comma 300. Line is drawn through the points, moving downward and to the right.  

So an equation that defines $f$ is: $f(x) = 500 \boldcdot \left(\frac{3}{5}\right)^x$

We can also use graphs to compare functions. Here are graphs representing two different exponential functions, labeled $g$ and $h$. Each one represents the area of algae (in square meters) in a pond, $x$ days after certain fish were introduced.

• Pond A had 40 square meters of algae. Its area shrinks to $\frac{8}{10}$ of the area on the previous day.
• Pond B had 50 square meters of algae. Its area shrinks to $\frac 25$ of the area on the previous day.

Graph of 2 functions on grid, origin O. Horizontal axis, x, Vertical axis, y. Function h, blue line, starts on the vertical axis,  moves downward and to the right, almost linearly.. Function g, green line, starts above function h, moves downward and to the right, exponentially.  

Can you tell which graph corresponds to which algae population?

We can see that the $y$-intercept of $g$'s graph is greater than the $y$-intercept of $h$'s graph. We can also see that $g$ has a smaller growth factor than $h$ because as $x$ increases by the same amount, $g$ is retaining a smaller fraction of its value compared to $h$. This suggests that $g$ corresponds to Pond B, and $h$ corresponds to Pond A.

When we borrow money from a lender, the lender usually charges interest, a percentage of the borrowed amount as payment for allowing us to use the money. The interest is usually calculated at a regular interval of time (for example, daily, monthly, or yearly).

Suppose you received a loan of $500 and the interest rate is 15%, calculated at the end of each year. If you make no other purchases or payments, the amount owed after one year would be $500 + (0.15) \boldcdot 500$, or $500 \boldcdot(1+0.15)$. If you continue to make no payments or other purchases in the second year, the amount owed would increase by another 15%. The table shows the calculation of the amount owed for the first three years.

The pattern here continues. Each additional year means multiplication by another factor of $(1+0.15)$. With no further purchases or payments, after $t$ years the debt in dollars is given by the expression:

$\displaystyle 500\boldcdot(1+0.15)^t$

In this representation, we might leave the growth factor as $(1 + 0.15)$ rather than combining it to 1.15 so that the percentage increase is easier to see. In other situations, it may make sense to write it as 1.15, depending on what is being emphasized. Because exponential functions eventually grow very quickly, leaving a debt unpaid can be very costly.

Expressions can be written in different ways to highlight different aspects of a situation or to help us better understand what is happening. A growth rate tells us the percent change. As always, in percent change situations, it is important to know if the change is an increase or decrease. For example:

• A population is increasing by 20% each year. The growth rate is 20%, so after one year, 0.2 times the population at the beginning of that year is being added.  If the initial population is $p$, the new population is $p + 0.2p$, which equals $(1 + 0.2)p$, or $1.2p$.

• A population is decreasing by 20% each year. The growth rate is -20%, so after one year, 0.2 times the population at the beginning of that year is being lost.  If the initial population is $p$, the new population is $p – 0.2p$, which equals $(1 – 0.2)p$, or $0.8p$.

Suppose the area, $a$, covered by a forest is currently 50 square miles, and it is growing by 0.2% each year. If $t$ represents time, from now, in years, we can express the area of the forest as:

$\displaystyle a = 50 \boldcdot (1+0.002)^t$

$\displaystyle a = 50 \boldcdot (1.002)^t$

In this situation, the growth rate is 0.002, and the growth factor is 1.002. Because 0.002 is such a small number, however, it may be difficult to tell from this function how quickly the forest is growing. We may find it more meaningful to measure the growth every decade or every century. There are 10 years in a decade, so to find the growth rate in decades, we can use the expression $(1.002)^{10}$, which is approximately 1.02. This means a growth rate of about 2% per decade. Using $d$ for time, in decades, the area of the forest can be expressed as:

$\displaystyle a=50 \boldcdot \left((1+0.002)^{10}\right)^d$

$\displaystyle a \approx50 \boldcdot (1.02)^d$

If we measure time in centuries, the growth rate is about 22% per century because $1.002^{100} \approx 1.22$. Using $c$ to measure time, in centuries, our equation for area becomes:

$\displaystyle a = 50 \boldcdot \left((1+0.002)^{100}\right)^c$

$\displaystyle a \approx 50 \boldcdot (1.22)^c$

Suppose that you won the top prize from a game show and are given two options. The first option is a cash gift of $10,000 and $1,000 is added per day for the next 7 days. The second option is a cash gift of 1 cent (or $0.01) that grows tenfold each day for 7 days. You must wait the entire time and get all of the prize money at the end of the week. Which option would you choose?

In the first option, the amount of money increases by the same amount ($1,000) each day, so we can represent it with a linear function. In the second option, the money grows by multiples of 10, so we can represent it with an exponential function. Let $f$ represent the amount of money $x$ days after winning with the first option, and let $g$ represent the amount of money $x$ days after winning with the second option.

For the first few days, the second option trails far behind the first. Because of the repeated multiplication by 10, however, after 7 days it surges past the amount in the first option.

What if the factor of growth is much smaller than 10? Suppose we have a third option, represented by a function $h$. The starting amount is still $0.01 and it grows by a factor of 1.5 times each day.

If we graph the function $h(x)=(0.01) \boldcdot (1.5)^x$, we see that it takes many, many more days before we see rapid growth. But given time to continue growing, the amount in this exponential option will eventually also outpace that in the linear option. If the prize rules are changed so that both prizes can grow for more than 38 days, this new exponential prize may be worth more than the linear option, but if the prizes can grow for only a shorter amount of time, the linear option is worth more.

Graph of 2 functions on grid, origin O. Horizontal axis, number of days, from .0 to 40, by 2's. Vertical axis, amount in dollars, from 0 to 90,000 by 10,000's. Linear function, f of x equals 10,000 plus 1,000 x represented by red dots. Exponential function, h of x equals 0 point 01 times 1 point 5 to the x, represented by blue dots. Red dots well above blue dots from 0 days to 37 days. Dots intersect around 38 comma 48,000. Blue dot exceeds red dot at 40 days.