4: Exponential Functions

A progressive guide to simplifying exponents, recognizing exponential behavior, building growth and decay models, calculating compound interest, fitting data, and interpreting transformed exponential graphs.

Recognizing Exponential Behavior

An exponential relationship describes repeated multiplication rather than repeated addition. If equal increases in the input produce equal multiplicative changes in the output, an exponential model may be appropriate. Common applications include population change, depreciation, , and radioactive decay.

A linear model adds a constant amount over equal intervals. An exponential model multiplies by a constant factor over equal intervals. For example, the values $3,6,12,24$ double each time, so they show multiplicative rather than additive behavior.

The central form is

f(x)=ab^x,

where $a\ne0$ , $b>0$ , and $b\ne1$ . The coefficient $a$ is the initial value because $f(0)=a$ , and the base $b$ is the multiplicative factor for each increase of $1$ in $x$ .

Takeaway: Look for a constant ratio, not merely a constant difference, when deciding whether exponential behavior is present.

Exponent Properties and Rational Exponents

Exponent rules provide equivalent ways to simplify powers and roots. For appropriate values, the main rules are

\begin{aligned} a^ma^n&=a^{m+n},\\ \frac{a^m}{a^n}&=a^{m-n},\\ (a^m)^n&=a^{mn},\\ (ab)^n&=a^nb^n,\\ \left(\frac{a}{b}\right)^n&=\frac{a^n}{b^n},\\ a^0&=1,\\ a^{-n}&=\frac{1}{a^n}. \end{aligned}

A represents a root:

a^{1/n}=\sqrt[n]{a},\qquad a^{m/n}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m.

For example,

16^{3/4}=\left(\sqrt[4]{16}\right)^3=2^3=8.

Apply these rules to factors and products while preserving the original structure. Thus,

(x^3y^2)^2=x^6y^4,

but a sum cannot be separated in the same way:

(x+y)^2\ne x^2+y^2.

Takeaway: Exponent rules operate on multiplication, division, powers, and roots; they do not permit distributing an exponent across addition.

Graphing and Interpreting Exponential Functions

For the parent function $f(x)=b^x$ , the base controls the direction of change. If $b>1$ , the graph increases and represents growth. If $0<b<1$ , the graph decreases and represents decay.

For the general $f(x)=ab^x$ with positive $a$ , the domain is all real numbers, the range is $(0,\infty)$ , the initial value is $f(0)=a$ , and the is $y=0$ when there is no vertical shift.

For example, in

f(x)=3(2)^x,

the initial value is $3$ , and each increase of $1$ in $x$ doubles the output:

f(0)=3,\quad f(1)=6,\quad f(2)=12,\quad f(3)=24.

A is a line the graph approaches. For an unshifted , the graph approaches $y=0$ but does not cross it when the coefficient is positive.

Takeaway: The coefficient sets the starting value, the base determines growth or decay, and the asymptote describes long-term behavior.

Exponential Growth and Decay Models

A constant percentage change becomes a multiplicative factor. For a growth rate $r$ written as a decimal, the is $1+r$ , so a $7\%$ increase uses factor $1.07$ . For a decrease, the is $1-r$ , so a $12\%$ decrease uses factor $0.88$ .

The discrete growth and decay model is

A(t)=A_0(1+r)^t

for growth, and

A(t)=A_0(1-r)^t

for decay.

If a town begins with $18{,}000$ people and grows by $2.5\%$ per year, then

P(t)=18{,}000(1.025)^t.

After $6$ years,

P(6)=18{,}000(1.025)^6\approx20{,}874.

If a machine worth $$24{,}000$ loses $15\%$ of its value each year, then

V(t)=24{,}000(0.85)^t.

After $4$ years,

V(4)=24{,}000(0.85)^4\approx\$12{,}528.

Exponential decay approaches zero but does not reach zero in this mathematical model. When change occurs continuously, use

A(t)=A_0e^{kt},

where $k>0$ indicates continuous growth and $k<0$ indicates continuous decay.

Takeaway: Convert the percentage to a decimal, choose the correct factor, identify the time units, and interpret the result in context.

and

applies interest to both the original principal and previously accumulated interest. With principal $P$ , annual rate $r$ as a decimal, $n$ compounding periods per year, and $t$ years, use

A=P\left(1+\frac{r}{n}\right)^{nt}.

For , use

A=Pe^{rt}.

Suppose $$2{,}500$ is invested at an annual rate of $4.8\%$ , compounded monthly, for $3$ years. Substitute $P=2500$ , $r=0.048$ , $n=12$ , and $t=3$ :

A=2500\left(1+\frac{0.048}{12}\right)^{12(3)}\approx\$2{,}887.10.

The interest earned is approximately

\$2{,}887.10-\$2{,}500=\$387.10.

The rate must be written as a decimal, and the exponent $nt$ must count the total number of compounding periods.

Takeaway: Match the rate, compounding frequency, and time units carefully before substituting into the formula.

Building and Evaluating Exponential Models

An exponential model can be built from a constant ratio or estimated from data. In the data pattern $80,100,125,156.25$ , each successive ratio is $1.25$ :

\frac{100}{80}=\frac{125}{100}=\frac{156.25}{125}=1.25.

Therefore, a model is

y=80(1.25)^x,

which represents a $25\%$ increase for each one-unit increase in $x$ .

If two points $(x_1,y_1)$ and $(x_2,y_2)$ are known for $y=ab^x$ , divide the equations to eliminate $a$ :

\frac{y_2}{y_1}=b^{x_2-x_1}.

Then

b=\left(\frac{y_2}{y_1}\right)^{1/(x_2-x_1)},\qquad a=\frac{y_1}{b^{x_1}}.

For a quantity equal to $40$ when $x=0$ and $90$ when $x=3$ , the initial value is $a=40$ . Solving $90=40b^3$ gives

b=\sqrt[3]{2.25}\approx1.31,

so an approximate model is

y=40(1.31)^x.

For larger data sets, estimates the parameters in $y=ab^x$ . A sound modeling process is:

Identify the input and output variables and their units.
Create a scatter plot.
Compare linear and exponential patterns.
Use when the data follow a curved growth or decay pattern.
Examine residuals or prediction errors.
Interpret the coefficient and base in context.
Avoid extrapolating far beyond the observed data unless continued exponential behavior is justified.

Takeaway: A good model reflects the data, explains its parameters, and is not treated as reliable beyond the context that supports it.

Transformations of Exponential Graphs

Transformations of the parent function $f(x)=b^x$ can be represented by

g(x)=Ab^{B(x-h)}+k.

The parameter $A$ controls vertical stretch or compression; if $A<0$ , it also reflects the graph across the $x$ -axis. The parameter $B$ controls horizontal scaling; a negative $B$ reflects the graph across the $y$ -axis. The value $h$ shifts the graph horizontally, and $k$ shifts it vertically. The becomes

y=k.

For

g(x)=2\cdot3^{x-4}+5,

the graph of $3^x$ is stretched vertically by a factor of $2$ , shifted $4$ units right, and shifted $5$ units up. Its is $y=5$ .

For

g(x)=-3\cdot2^{x+1}-4,

the graph is shifted $1$ unit left, stretched vertically by a factor of $3$ , reflected across the $x$ -axis, and shifted $4$ units down. Its is $y=-4$ . The negative coefficient makes the graph decrease even though the base $2$ is greater than $1$ .

Takeaway: Read the exponent for horizontal changes, the coefficient for vertical changes and reflection, and the final constant for the asymptote.

Interpreting Models in Context

Before using an exponential model, identify what each quantity means. In

P(t)=12{,}000(1.04)^t,

$12{,}000$ is the initial population, $1.04$ is a $4\%$ annual , and $t$ is measured in years.

Use these questions to evaluate a model:

What does the initial value represent?
Is the factor greater than $1$ , or between $0$ and $1$ ?
What percentage change does the factor represent?
What are the units of the input and output?
Is the prediction interpolation within the observed data range or extrapolation beyond it?
Does the context support continued growth or decay?

A model is an approximation. Even when the algebra is correct, unlimited growth or decay may not be realistic in every situation. Predictions should therefore be compared with observed data and limited to ranges where the constant-percentage assumption is reasonable.

Final takeaway: Interpret every parameter, check the model against the context, and distinguish mathematical behavior from realistic long-term assumptions.