Math Core

Polynomials and rational functions change by adding and multiplying the input. Exponential functions put the input in the exponent, and the result is change that compounds: money earning interest, bacteria doubling, a drug leaving the bloodstream. This lesson sets up the family, its graphs, and the special base $e$ that the rest of the unit (and all of calculus) depends on.

The exponential family

Definition

Exponential function

An exponential function has the form

f(x) = a \cdot b^{x}, \qquad a \ne 0,\quad b > 0,\quad b \ne 1.

The number $a = f(0)$ is the initial value and $b$ is the base or growth factor.

The restrictions on $b$ are there for good reasons. If $b = 1$ , then $b^x = 1$ for every $x$ and the function is constant. If $b$ were negative, expressions like $(-4)^{1/2}$ would not be real numbers, so the function would not be defined on an interval.

The defining feature of an exponential function is a constant ratio. When $x$ increases by $1$ , the output is multiplied by $b$ :

\frac{f(x + 1)}{f(x)} = \frac{a b^{x+1}}{a b^{x}} = b.

Compare this with a linear function, where increasing $x$ by $1$ always adds the slope. So a table with equally spaced inputs is linear if the outputs have a common difference and exponential if they have a common ratio.

$x$	$0$	$1$	$2$	$3$	$4$
linear $g(x) = 3 + 6x$	$3$	$9$	$15$	$21$	$27$
exponential $f(x) = 3 \cdot 2^x$	$3$	$6$	$12$	$24$	$48$

The linear function starts ahead, but at $x = 3$ the exponential one passes it ( $24$ vs. $21$ ) and never looks back: by $x = 5$ it is $96$ vs. $33$ . Any exponential function with $b > 1$ eventually outgrows every polynomial.

Growth and decay

With $a > 0$ the base decides the shape:

If $b > 1$ , the function shows exponential growth: it increases, rising faster and faster.
If $0 < b < 1$ , it shows exponential decay: it decreases, leveling off toward $0$ .

A percent change converts to a base directly. Growing by $r$ per period multiplies by $1 + r$ ; shrinking by $r$ per period multiplies by $1 - r$ . So growing $6\%$ per year means $b = 1.06$ , and losing $15\%$ per year means $b = 0.85$ .

Growth with b = 2 and b = 1.3; decay with b = 1/2. All pass through (0, 1).Open in grapher →

Notice that $\left(\tfrac{1}{2}\right)^x = 2^{-x}$ , so the decay curve is the reflection of $y = 2^x$ across the $y$ -axis.

Features of y = b^x

For $y = b^x$ with $b > 0$ and $b \ne 1$ :

Domain: all real numbers. Range: $y > 0$ .
$y$ -intercept: $(0, 1)$ . There is no $x$ -intercept.
Horizontal asymptote: $y = 0$ (on the left for growth, on the right for decay).
The function is one-to-one, so it has an inverse. That inverse is the logarithm, the subject of the next lesson.

Transformations

Everything you know about transformations applies. In

f(x) = a \cdot b^{x - h} + k,

$h$ shifts the graph right, $k$ shifts it up, and $a$ stretches it vertically (and reflects it across the asymptote if $a$ is negative). The key idea is to track the asymptote: it starts at $y = 0$ and moves up or down with $k$ , so the asymptote is $y = k$ .

Worked example: Graph a transformed exponential

Describe the graph of $f(x) = -2 \cdot 3^{x+1} + 5$ . Give its asymptote, range and $y$ -intercept.

Solution. Start with $y = 3^x$ . Shift left $1$ , stretch by $2$ , reflect across the $x$ -axis, then shift up $5$ .

The asymptote moves with the vertical shift: $y = 5$ .
Since $a = -2$ is negative, the graph lies below its asymptote. The range is $y < 5$ .
$y$ -intercept: $f(0) = -2 \cdot 3^{1} + 5 = -1$ , so $(0, -1)$ .

As $x \to -\infty$ , $3^{x+1} \to 0$ and $f(x) \to 5$ . As $x \to \infty$ , $f(x) \to -\infty$ .

f(x) = -2·3^(x+1) + 5 approaches its asymptote y = 5 from below.Open in grapher →

Common mistake

The asymptote of $y = a \cdot b^{x-h} + k$ is $y = k$ , not $y = 0$ . And a horizontal shift $h$ does not move the asymptote at all. Find $k$ first and the range follows: $y > k$ if $a > 0$ , $y < k$ if $a < 0$ .

Finding an exponential function from two points

Two points determine an exponential function $y = ab^x$ . Divide one equation by the other: the $a$ 's cancel and you are left with a power of $b$ .

Worked example: Through two points

Find the exponential function $y = ab^x$ whose graph passes through $(1, 6)$ and $(3, 54)$ .

Solution. The points give $ab = 6$ and $ab^3 = 54$ . Divide:

\frac{ab^3}{ab} = \frac{54}{6} \quad\Longrightarrow\quad b^2 = 9 \quad\Longrightarrow\quad b = 3,

taking the positive root because $b > 0$ . Then $a = \dfrac{6}{3} = 2$ . The function is $y = 2 \cdot 3^x$ .

Check: $2 \cdot 3^3 = 54$ . ✓

The natural base e

Suppose you deposit $1 at an annual interest rate of $100\%$ . Compounded once a year, you have $2 after a year. Compounded $n$ times a year at rate $\tfrac{1}{n}$ each time, you have $\left(1 + \tfrac{1}{n}\right)^n$ dollars.

$n$	$1$	$12$	$365$	$10{,}000$	$1{,}000{,}000$
$\left(1 + \frac{1}{n}\right)^n$	$2$	$2.613\ldots$	$2.7145\ldots$	$2.71814\ldots$	$2.718280\ldots$

More frequent compounding helps, but less and less. The values approach a limit, an irrational number called $e$ :

e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n \approx 2.718281828.

The function $f(x) = e^x$ is the natural exponential function. Its graph sits between $y = 2^x$ and $y = 3^x$ . In calculus you will find that $e^x$ is the only exponential whose slope at every point equals its own height, which is why it is called natural.

Compound and continuous interest

Interest formulas

A principal $P$ at annual rate $r$ (as a decimal) for $t$ years grows to

A = P\left(1 + \frac{r}{n}\right)^{nt} \quad\text{compounded } n \text{ times per year,} \qquad A = Pe^{rt} \quad\text{compounded continuously.}

Continuous compounding is the limit of compounding more and more often, and it gives the largest balance for a given rate.

Worked example: Quarterly vs. continuous

You invest $5,000 at $4\%$ annual interest for $10$ years. Find the balance if interest is compounded (a) quarterly and (b) continuously.

Solution.

(a) $n = 4$ , so

A = 5000\left(1 + \frac{0.04}{4}\right)^{40} = 5000(1.01)^{40} \approx 7444.32.

(b) $A = 5000e^{0.04 \cdot 10} = 5000e^{0.4} \approx 7459.12$ .

Continuous compounding earns about $14.80 more.

Tip

Round only at the end. Rounding $(1.01)^{40}$ to $1.49$ before multiplying by $5000$ gives $7,450, off by about $6. Keep full calculator precision until the final step.

Practice

Practice 1

Let $f(x) = 5 \cdot 2^x$ . Find $f(3)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Which function models exponential decay?

Practice 3

Find the range of $g(x) = 3 \cdot 2^{x-5} - 4$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 4

Find the $y$ -intercept of $f(x) = 3 \cdot 2^{x-2} + 1$ . Give the $y$ -value.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

The table shows values of an exponential function. Write the function in the form $y = ab^x$ .

$x$	$0$	$1$	$2$	$3$
$y$	$4$	$12$	$36$	$108$

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 6

Find the exponential function $y = ab^x$ whose graph passes through $(2, 20)$ and $(4, 80)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 7

You deposit $2,000 in an account paying $6\%$ annual interest compounded monthly. What is the balance after $5$ years? Round to the nearest cent.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

The same $2,000 is instead invested at $6\%$ compounded continuously for $5$ years. What is the balance? Round to the nearest cent.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.