Exponential Functions: math lesson, Grade 11

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Math lessons Grade 11 : Exponential Functions — Zyro the alien explorer of Planète Maths

A viral video doubles its audience every day. A new car loses a slice of its value every year. A savings account earns interest on its own interest. All three stories share one pattern: the quantity is multiplied by the same factor again and again. In this chapter you will learn the functions that describe that pattern, graph them, shift them, and use them to model money, populations and decay.

Sommaire

1. What makes a function exponential?

Exponential function

An exponential function has the form \( f(x) = a \cdot b^{x} \), where \( a \neq 0 \) is the initial value, and \( b > 0 \), \( b \neq 1 \) is the base (also called the growth or decay factor). The variable \( x \) is in the exponent.

The base cannot be \( 1 \) (then \( 1^x = 1 \) and the function is just a constant), and it cannot be negative (then \( b^x \) would fail to exist for many real numbers \( x \), such as \( x = \tfrac{1}{2} \)).

Example 1: a table of values

Let \( f(x) = 3 \cdot 2^{x} \). Then \( f(0) = 3 \cdot 1 = 3 \), \( f(1) = 6 \), \( f(2) = 12 \), \( f(3) = 24 \) and \( f(-1) = 3 \cdot 2^{-1} = 1.5 \).

\( x \)	\( -1 \)	\( 0 \)	\( 1 \)	\( 2 \)	\( 3 \)
\( f(x) \)	\( 1.5 \)	\( 3 \)	\( 6 \)	\( 12 \)	\( 24 \)

Each time \( x \) goes up by 1, the output is multiplied by 2. That constant ratio is the signature of an exponential function; a linear function would add the same amount instead.

Watch out

\( x^{2} \) and \( 2^{x} \) are very different. In \( x^{2} \) the variable is the base (a power function); in \( 2^{x} \) the variable is the exponent. At \( x = 5 \), \( x^{2} = 25 \) but \( 2^{x} = 32 \), and the gap explodes after that.

2. Exponential growth and decay

Growth or decay?

For \( f(x) = a \cdot b^{x} \) with \( a > 0 \):
If \( b > 1 \), the function shows exponential growth. If \( 0 < b < 1 \), it shows exponential decay.
A change of \( r \) per period (written as a decimal) gives the factor \( b = 1 + r \) for growth and \( b = 1 - r \) for decay.

From a percent to a factor

Write the percent as a decimal: 20% becomes \( 0.20 \).
Add it for growth (\( 1 + 0.20 = 1.2 \)) or subtract it for decay (\( 1 - 0.20 = 0.8 \)).
Use that factor as the base and the starting amount as \( a \).

Example 2: growth

A culture starts with 500 bacteria and grows 20% every hour. The model is \( P(t) = 500(1.2)^{t} \). After 5 hours: \( P(5) = 500 \cdot 1.2^{5} = 500 \cdot 2.48832 = 1244.16 \), so about 1,244 bacteria.

Example 3: decay

A van is worth 24,000 dollars and loses 15% of its value each year. The model is \( V(t) = 24000(0.85)^{t} \). After 4 years: \( V(4) = 24000 \cdot 0.85^{4} = 24000 \cdot 0.52200625 = 12528.15 \), so about 12,528 dollars.

3. Graphing exponential functions

Here are the graphs of \( y = 2^{x} \) (growth) and \( y = \left(\tfrac{1}{2}\right)^{x} \) (decay). They are mirror images of each other across the y-axis.

Key features of \( f(x) = a \cdot b^{x} \)

Domain: all real numbers. Range: \( y > 0 \) when \( a > 0 \).
The y-intercept is \( (0, a) \), because \( b^{0} = 1 \).
The x-axis, \( y = 0 \), is a horizontal asymptote: the curve gets closer and closer to it but never touches it.
There is no x-intercept. The function is increasing if \( b > 1 \) and decreasing if \( 0 < b < 1 \).

Zyro’s tip

To sketch fast, plot the y-intercept \( (0, a) \), then the point at \( x = 1 \), which is \( (1, ab) \). Draw the curve hugging the asymptote on one side and shooting up on the other.

4. Transformations of exponential functions

Exponentials move like any other function. For \( g(x) = a \cdot b^{\,x-h} + k \):

\( h \) shifts the graph right by \( h \) units (left if \( h \) is negative).
\( k \) shifts the graph up by \( k \) units, and the asymptote moves to \( y = k \).
\( a < 0 \) reflects the graph across the x-axis; a larger \( |a| \) stretches it vertically.

Example 4: shifting

Graph \( g(x) = 2^{\,x-1} + 3 \). Start from \( y = 2^{x} \): move it 1 unit right and 3 units up. The point \( (0, 1) \) goes to \( (1, 4) \), the asymptote goes from \( y = 0 \) to \( y = 3 \), and the range becomes \( y > 3 \). The y-intercept is \( g(0) = 2^{-1} + 3 = 3.5 \).

5. The number e

Suppose you invest 1 dollar at 100% interest per year. If the interest is added once, you end with 2 dollars. If it is added \( n \) times a year, you end with \( \left(1 + \tfrac{1}{n}\right)^{n} \). More frequent additions help, but the total does not grow forever.

\( n \)	1	2	12	100	1,000	1,000,000
\( \left(1 + \tfrac{1}{n}\right)^{n} \)	\( 2 \)	\( 2.25 \)	\( 2.6130 \)	\( 2.7048 \)	\( 2.7169 \)	\( 2.71828 \)

The number e

The values settle toward an irrational number called \( e \approx 2.71828 \). The function \( f(x) = e^{x} \) is the natural exponential function. Its graph lies between those of \( 2^{x} \) and \( 3^{x} \), and it passes through \( (0, 1) \) and \( (1, e) \).

The inverse of \( e^{x} \) is the natural logarithm: if \( e^{t} = c \), then \( t = \ln c \). Your calculator has both keys, and you will need \( \ln \) to solve for an unknown exponent.

6. Compound interest

Compound interest formula

If \( P \) dollars earn an annual rate \( r \) (as a decimal), compounded \( n \) times per year for \( t \) years, the balance is \[ A = P\left(1 + \frac{r}{n}\right)^{nt}. \]

Example 5: quarterly compounding

Invest 2,000 dollars at 4% compounded quarterly for 6 years. Here \( P = 2000 \), \( r = 0.04 \), \( n = 4 \), \( t = 6 \), so \( A = 2000(1 + 0.01)^{24} = 2000 \cdot 1.26973\ldots \approx 2539.47 \). The balance is about 2,539.47 dollars. With annual compounding you would get \( 2000 \cdot 1.04^{6} \approx 2530.64 \) dollars.

The orange curve (compound interest) bends upward because each year’s interest earns interest. The violet line (simple interest) adds the same 80 dollars every year.

7. Continuous compounding

Push the compounding frequency as high as possible, and the formula turns into one that uses \( e \).

Continuous compounding

If \( P \) dollars earn rate \( r \) compounded continuously for \( t \) years, then \[ A = P\,e^{rt}. \]

Example 6: continuous

Same account as before: \( A = 2000\,e^{0.04 \cdot 6} = 2000\,e^{0.24} \approx 2000 \cdot 1.27125 \approx 2542.50 \) dollars. That is only about 3 dollars more than quarterly compounding, which shows how quickly the gains level off.

8. Modeling with exponential functions

Finding an exponential model from two points

Use the point with \( x = 0 \) to get \( a \) (the y-intercept). If there is no such point, divide the two equations.
Divide the outputs and take the right root to find \( b \).
Write \( f(x) = a \cdot b^{x} \) and check both points.

Example 7: two points

A streaming channel had 150 subscribers at \( t = 0 \) months and 1,200 at \( t = 3 \). Then \( a = 150 \) and \( 150\,b^{3} = 1200 \), so \( b^{3} = 8 \) and \( b = 2 \). The model is \( S(t) = 150 \cdot 2^{t} \): the channel doubles every month.

Solving for the exponent. To find when 5,000 dollars at 6% compounded yearly doubles, solve \( 1.06^{t} = 2 \). Taking logarithms gives \( t = \dfrac{\ln 2}{\ln 1.06} \approx 11.9 \) years. For continuous growth, \( e^{rt} = 2 \) gives \( t = \dfrac{\ln 2}{r} \).

Key takeaways

An exponential function is \( f(x) = a \cdot b^{x} \) with \( b > 0 \) and \( b \neq 1 \); equal steps in \( x \) multiply \( y \) by the same factor.
\( b = 1 + r \) means growth, \( b = 1 - r \) means decay; \( b > 1 \) increases and \( 0 < b < 1 \) decreases.
The graph has y-intercept \( (0, a) \), a horizontal asymptote, domain all reals and range \( y > 0 \) (for \( a > 0 \)).
In \( a \cdot b^{x-h} + k \), the shifts move the graph \( h \) right and \( k \) up, and the asymptote becomes \( y = k \).
\( e \approx 2.71828 \); compound interest is \( A = P(1 + r/n)^{nt} \), continuous is \( A = Pe^{rt} \).
To model, find \( a \) and \( b \) from data; to find a time, use logarithms.