Exponential Functions

A function that changes by a constant multiplier is called an exponential function. For example, these functions can be used to describe population changes in a country and how the value of a used car decreases over time.

Concept

Exponential Function

Many functions containing a variable exponent, are called exponential functions. Formally, any function that can be written in the following form is an exponential function.

$y = a \cdot b^{x}$

Here, the coefficient, $a,$ is the $y$ -intercept, which is sometimes referred to as the initial value. The base, $b,$ can be interpreted as the constant multiplier. To ensure that $y$ is an exponential function, there are restrictions on $a$ and $b .$

Concept

$a \neq = 0$

If the coefficient $a$ is $0,$ the function becomes a horizontal line. $y = 0 \cdot b^{x} \Rightarrow y = 0$ This is a line along $y = 0,$ and thus, a linear relationship. Therefore, if $a = 0$ the function is not exponential.

Concept

$b > 0$ and $b \neq = 1$

If the base,

b,

is negative, the function gives undefined results for certain

x

-values. For example, since

b^{1 / 2} = b,

a negative

b

would yield non-real values for

x = \frac{1}{2} .

Then, a condition is needed.

b \geq 0

Furthermore, if

b = 0

or

b = 1

, the function becomes a horizontal line.

a \cdot 0^{x} = 0 and a \cdot 1^{x} = a

Therefore,

b

can not equal

0

or

1 .

Therefore, for all exponential functions

y = a \cdot b^{x},

a \neq = 0

and

b > 0, b \neq = 1 .

Method

Graphing an Exponential Function using the Function Rule

For an exponential function

y = a \cdot b^{x},

a

represents the initial value and

b

represents the constant multiplier. These values can be used to graph the function. Consider

y = 10000 \cdot 0 . 8^{x} .

1

Identify

a

and

b

The initial value, $a,$ of an exponential function is the number without an exponent. In this case, $a = 10000 .$ The constant multiplier, $b$ is the number with the exponent. Here, $b = 0.8 .$

2

Plot the initial value

The initial value is the $y$ -value when $x = 0 .$ It can also be thought of as the $y$ -intercept of the function. Here, the initial value is $10000$ so $(0, 10000)$ is a point on the graph.

3

Use the constant multiplier to find more points

When the $x$ -value increases by $1,$ the $y$ -value is multiplied by $b .$ Since $b = 0.8,$ the $y$ -value when $x = 1$ is $10000 \cdot 0.8 = 8000 .$ Thus, $(1, 8000)$ also lies on the graph of the function. Similarly, the point $(2, 6400)$ lies on the graph because $8000 \cdot 0.8 = 6400 .$ These points are shown on the graph.

This process can be repeated until a general form of the graph emerges.

4

Draw the curve

Lastly, the graph can be drawn by connecting the points with a smooth curve.

Graph the exponential function and describe its key features

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Exercise

Graph the exponential function using the function rule and describe its key features. $y = 3 \cdot 2^{x}$

Show Solution

Solution

The function $y = 3 \cdot 2^{x}$ has the initial value $a = 3$ and the constant multiplier $b = 2 .$ Let us use these values to mark four points on the function's graph.

The function can now be graphed by connecting the points with a smooth curve.

Example

Key features

Let us now describe the function's key features.

First the graph shows a $y$ -intercept at $(0, 3) .$
The function $y = 3 \cdot 2^{x}$ is greater than $0$ for all $x .$ Although the left-end of the graph approaches the $x$ -axis it never intersects it. Thus, there is no $x$ -intercept.
As $x$ approaches $- \infty$ the function continues to approach, but never becomes parallel with, the $x$ -axis. Thus, the function increases for all $x .$

Looking at the graph, we can see that the left end approaches $y = 0$ and the right end extends upward. Thus, the end behavior of $y = 3 \cdot 2^{x}$ can be written as follows.

$As x \to - As x \to + \infty, \infty, y \to 0 y \to + \infty$ Let us show this in the graph.

Interpret the initial value and constant multiplier

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Exercise

In $1976,$ scientists discovered a rare population of Flemish Giant rabbits in a secluded forest. Since then, they've been monitoring the population. During the five years of the study, the number of rabbits could be modeled with the exponential function shown.

Use the graph to write the rule for the function, then interpret its initial value and constant multiplier.

Show Solution

Solution

To write an exponential function rule, we need the initial value of the function, $a,$ and the constant multiplier, $b .$ $y = a \cdot b^{x}$ Notice that the graph starts at $(0, 80) .$ This means that $80$ is the initial value.

Since $a = 80,$ we can write the following incomplete function rule. $y = 80 \cdot b^{x}$ To determine $b,$ we can use another point on the graph.

The point

(1, 100)

lies on the graph. Thus, we can susbtitute

x = 1

and

y = 100

into the rule above and solve for

b .

y = 80 \cdot b^{x}

SubstituteII

x = 1

,

y = 100

100 = 80 \cdot b^{1}

Solve for

b

ExponentOne

a^{1} = a

100 = 80 \cdot b

RearrangeEqnRearrange equation

80 \cdot b = 100

DivEqn

LHS / 80 = RHS / 80

b = \frac{1 0 0}{8 0}

CalcQuotCalculate quotient

b = 1.25

The constant multiplier is

b = 1.25 .

Thus, the function rule can be written as follows.

y = 80 \cdot 1.2 5^{x}

Next, we can interpret the values of

a

and

b

we found above. The initial value,

a = 80

, means that the initial population when the rabbits were discovered was

80 .

Additionally, a constant multiplier of

1.25

means that each year the population is

1.25

times more than the previous year.

Rule

Exponential Growth and Decay

Concept

Exponential Growth

Exponential growth occurs when a quantity increases by the same factor over equal intervals of time. This leads to an exponential function, where the independent variable in the exponent, $t,$ is time. $y = a \cdot b^{t}$ Since the quantity increases over time, the constant multiplier $b$ has to be greater than $1 .$ Thus, the growth factor $b$ can be expressed as $b = 1 + r,$ where $r$ is some positive number. The resulting function is called an exponential growth function.

The constant $r$ can then be interpreted as the rate of growth, in decimal form. A value of $0.06,$ for instance, means that the quantity increases by $6 %$ over every unit of time. As is the case with all exponential functions, $a$ is the $y$ -coordinate of the $y$ -intercept.

Since the growth factor is greater than $1,$ the quantity grows faster and faster, without bound.

Concept

Exponential Decay

The counterpart of exponential growth is exponential decay; when a quantity decreases by the same factor over equal intervals of time, the constant multiplier of the exponential decay function is less than $1 .$ This factor can be expressed as $(1 - r)$ and is known as the decay factor.

The constant $r$ can then be interpreted as the rate of decay, in decimal form. A value of $0.12,$ for instance, would mean that the quantity decreases by $12 %$ over every unit of time.

Since the decay factor is smaller than

1,

the quantity decays toward

0

over time.

Find the rule of the exponential growth function

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Exercise

In an ideal environment, bacteria populations grow exponentially and can be modeled with an exponential growth function. The bacteria Lactobacillus acidophilus duplicates about once every hour. A single bacteria cell is placed in an ideal environment. State and interpret the constants $a$ and $r$ for the growth that will occur. Then, write a function rule describing this growth.

Show Solution

Solution

The constant $a$ is the initial value of the quantity, in this case the number of bacteria. There was only one bacteria placed in the environment, so $a$ is $1 .$ The constant $r$ is the rate of growth. Since the bacteria duplicate every hour, the amount of bacteria doubles every hour. This corresponds to an increase by $100 % .$ Thus, $r$ is $1 .$ Substituting this into the rule of an exponential growth function gives $P (t) = 1 \cdot (1 + 1)^{t},$ which can be simplified as $P (t) = 2^{t} .$ Since this is an exponential growth function, population will grow faster and faster, without bound. In a real environment, this would not happen, since the available space and nutrition would have to be infinite. At some point, the environment would no longer be ideal, so the growth would slow down or stop.

Graph the exponential growth or decay function

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Exercise

During a time period, the number of carps in a small lake can be modeled by the function $H (t) = 800 \cdot 0.8 8^{t},$ where $t$ is the time in years. State whether the function shows growth or a decay, and then find the rate of growth or decay, $r .$ Finally, graph the function.

Show Solution

Solution

To begin, let's analyze the given function rule. It's written in the form $y = a \cdot b^{x},$ where $a$ is the initial value and $b$ is constant multiplier/growth factor. The constant multiplier, $0.88,$ is less than $1,$ so it is a decay factor. Therefore, the function shows decay. Since the decay factor is always equal to $1 - r,$ we can write the equation. $0.88 = 1 - r,$ which can be solved for $r .$

0.88 = 1 - r

AddEqn

LHS + r = RHS + r

0.88 + r = 1

SubEqn

LHS - 0.88 = RHS - 0.88

r = 0.12

Thus, the rate of decay is $0.12,$ or $12 %,$ per year. The initial value is $800,$ and the constant multiplier is $0.88 .$ Using this information, we can graph the exponential decay function by plotting some points that lie on $H$ and connecting them with a smooth curve.

Concept

Compound Interest

When money is deposited to a savings account, interest is accrued, often yearly. Different types of interest work in different ways. When the interest earned is then added to the original amount, future interest accrues for a larger amount. This is called compound interest. To calculate the balance on the account at a specific time, an exponential growth function can be used. When the interest is compounded yearly, the balance can be modeled with a function.

$y = P (1 + r)^{t}$

In this context, $P$ stands for the principal, which is the initial amount of money, and $r$ is the interest rate in decimal form. If the interest is not compounded yearly, the function looks a little different.

$y = P (1 + \frac{r}{n})^{n t}$

The constant $n$ is the number of times the interest is compounded per year, while $r$ is still the annual interest rate. For an account with the principal $ $100$ and an annual interest of $15 %$ compounded twice a year, the growth function is:

$B (t) = 100 (1 + \frac{0 . 1 5}{2})^{2 t}$

Notice that this function grows continuously, whereas, in reality, the account balance only increases at the times of compound. Graphing the function together with the actual balance of the account will highlight how it can be used in practice.

Every time the interest is compounded, in this case every half year, the value of

B

is equal to the account balance. However, at all other times, it is not. To find, for instance, the account balance after

1.75

years,

B (1.5)

should then be evaluated, since that was the last time interest compounded.

Find the monthly interest rate

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Exercise

One savings account, with a principal of $ $100,$ offers an annual interest rate of $15 %$ compounded twice a year. Find the balance in the account after $5$ years. Another savings account with the same principal will have the same balance after $5$ years. However, the interest is compounded monthly. Find the interest rate of the second account.

Show Solution

Solution

First, we'll find the function rule describing the growth of the first account. It is given that $P = 100, r = 0.15,$ and $n = 2 .$ Substituting these values in the compound interest formula gives $B (t) = 100 (1 + \frac{0 . 1 5}{2})^{2 t} .$ Let's simplify this function before continuing.

B (t) = 100 (1 + \frac{0 . 1 5}{2})^{2 t}

CalcQuotCalculate quotient

B (t) = 100 (1 + 0.075)^{2 t}

AddTermsAdd terms

B (t) = 100 \cdot 1.07 5^{2 t}

Since the interest will accrute for

5

years,

t = 5 .

Therefore, we can find the account balance by evaluating

B (5) .

B (t) = 100 \cdot 1.07 5^{2 t}

Substitute

t = 5

B (5) = 100 \cdot 1.07 5^{2 \cdot 5}

MultiplyMultiply

B (5) = 100 \cdot 1.07 5^{10}

UseCalcUse a calculator

B (5) = 206.10315 \dots

RoundDecRound to

B (5) \approx 206.10

The account balance is $ $206.10$ after $5$ years. Now we can consider the second account. We know the balance of both accounts is equal, at least when both just had their interest compounded. This means that $B (t)$ also describes the growth in the second account. However, since the interest in the second account accrues monthly, or $12$ times a year, $n = 12 .$ Thus, the exponent in the rule should be $12 t .$ By using the equality $2 t = 1 / 6 \cdot 12 t$ and the power of a power property, we can rewrite $B (t)$ so that it's possible to find the monthly interest rate. Having the exponent $12 t$ means that the base of the power is equal to the monthly growth factor, which we can then use to find the monthly interest rate.

B (t) = 100 \cdot 1.07 5^{2 t}

RewriteRewrite

2 t

as

1 / 6 \cdot 12 t

B (t) = 100 \cdot 1.07 5^{1 / 6 \cdot 12 t}

ProdInExponent

a^{m \cdot n} = (a^{m})^{n}

B (t) = 100 \cdot (1.07 5^{1 / 6})^{12 t}

UseCalcUse a calculator

B (t) = 100 \cdot (1.01212 \dots)^{12 t}

RoundDecRound to

B (t) \approx 100 \cdot 1.01 2^{12 t}

We find an approximate monthly growth factor $1.012,$ which corresponds to a rate of growth that is $0.012 .$ Thus, the monthly interest rate is roughly $1.2 % .$