IB Applications and Interpretation SL · Functions · Section 5 of 7

Exponential functions and models of growth and decay

3 worked examples · By Álvaro López

Key idea

An exponential function multiplies by the same factor each time the variable increases by one unit. It is used to model populations, money with interest, medicine in the bloodstream or a cup of coffee cooling down. In the exam you will have to find constants, evaluate the model, solve equations using the GDC and interpret the asymptote.

The basic exponential function

An exponential function has the form $f (x) = a^{x}$ , where the base $a$ is a positive number other than $1$ . The variable is in the exponent, which distinguishes it from a power such as $x^{2}$ .

If $a > 1$ , the function increases: the larger $x$ is, the faster it rises. This is exponential growth.
If $0 < a < 1$ , the function decreases: it approaches zero as $x$ increases. This is exponential decay.

In both cases the graph passes through the point $(0, 1)$ , because $a^{0} = 1$ , and it never touches the $x$ -axis. The line that the curve approaches without ever reaching it is called a horizontal asymptote. For $f (x) = a^{x}$ the asymptote is $y = 0$ . The domain is $R$ and the range is $] 0, + \infty [$ .

The IB uses the special base $e \approx 2.718$ a lot. The function $f (x) = e^{x}$ increases and $f (x) = e^{- x}$ decreases. The GDC has it built in; you only need to recognise it in the model.

To see what each constant does, evaluate the function at $x = 0$ : $f (0) = k + c$ . The $y$ -intercept is always $(0, k + c)$ . For the $x$ -intercept you must solve $k \cdot a^{x} + c = 0$ , and it exists only if $- c / k > 0$ .

Graph of f(x) = 3·2^x − 12, increasing, with horizontal asymptote y = −12, intercepts at (0; −9) and (2; 0), and point (3; 12)

Exponential models

A quantity that changes by a fixed percentage in each period follows an exponential model. A common model is $P (t) = A \cdot a^{t} or P (t) = c + k e^{- r t}$ where $t$ is time, $A = P (0)$ is the initial value if there is no added constant, and $r > 0$ measures the rate of decay.

In cooling, the temperature of an object approaches that of its surroundings without ever reaching it in finite time. That is why the asymptote of the model is the ambient temperature. In a population model that grows without limit there is no added constant, and the asymptote is $y = 0$ .

Worked examples

Example 1 Basic

Intercepts and asymptote of an exponential

Let $f (x) = 3 \cdot 2^{x} - 12$ .

a) Find the intersection of the graph with the $y$ -axis.

b) Find the value of $x$ for which $f (x) = 0$ .

c) Write down the equation of the horizontal asymptote and state the value $c$ of the line $y = c$ .

Solution

a) Substitute $x = 0$ : $f (0) = 3 \cdot 2^{0} - 12 = 3 - 12 = - 9$ The $y$ -intercept is $(0, - 9)$ .

b) Solve $3 \cdot 2^{x} - 12 = 0$ : $3 \cdot 2^{x} = 12 \Rightarrow 2^{x} = 4 \Rightarrow x = 2$

c) In $f (x) = k \cdot a^{x} + c$ the horizontal asymptote is $y = c$ . Here $c = - 12$ , so the asymptote is $y = - 12$ . As $x$ takes very negative values, $2^{x}$ approaches $0$ and $f (x)$ approaches $- 12$ .

Answer: a) $- 9$ ; b) $x = 2$ ; c) $y = - 12$ .

Example 2 Medium

Yeast culture

The number of cells in a yeast culture, in thousands, is modelled by $P (t) = A \cdot 3^{t}$ , where $t$ is the time in hours since the start of the experiment, with $t \geq 0$ . After $2$ hours there are $450$ thousand cells.

a) Find the value of $A$ .

b) Find how many thousand cells there will be after $4$ hours.

c) Find the time at which there will be $12150$ thousand cells.

Solution

a) Substitute $t = 2$ and $P = 450$ : $A \cdot 3^{2} = 450 \Rightarrow 9 A = 450 \Rightarrow A = 50$ The culture started with $50$ thousand cells.

b) With $A = 50$ : $P (4) = 50 \cdot 3^{4} = 50 \cdot 81 = 4050$ After $4$ hours there are $4050$ thousand cells.

c) Solve $P (t) = 12150$ : $50 \cdot 3^{t} = 12150 \Rightarrow 3^{t} = 243 \Rightarrow 3^{t} = 3^{5} \Rightarrow t = 5$ You can also draw $P (t)$ on the GDC and find its intersection with the line $y = 12150$ .

Answer: $A = 50$ ; $4050$ thousand cells; after $5$ hours.

Example 3 Hard

A cup of coffee cooling down

The temperature of a cup of coffee, in degrees Celsius, $t$ minutes after it is served, is given by $T (t) = 22 + 73 e^{- 0.05 t}, t \geq 0$

a) Find the temperature of the coffee when it is served.

b) Find the temperature after $10$ minutes. Give your answer to one decimal place.

c) Find how long it takes for the coffee to reach $40^{\circ} C$ . Give your answer to one decimal place.

d) Write down the equation of the horizontal asymptote of $T$ and explain what it means in this context.

Solution

a) Serving the coffee is the instant $t = 0$ : $T (0) = 22 + 73 e^{0} = 22 + 73 = 95$ The coffee is at $95^{\circ} C$ .

b) Substitute $t = 10$ : $T (10) = 22 + 73 e^{- 0.5} \approx 66.3$ After $10$ minutes it is at $66.3^{\circ} C$ .

c) Set up $T (t) = 40$ : $22 + 73 e^{- 0.05 t} = 40 \Rightarrow e^{- 0.05 t} = \frac{18}{73}$ This equation is solved with the GDC, for example by finding the point where the graph of $T$ meets the line $y = 40$ . We get $t \approx 28.0$ . It takes about $28$ minutes.

d) As $t$ increases, $e^{- 0.05 t}$ approaches $0$ and $T (t)$ approaches $22$ . The horizontal asymptote is $T = 22$ . It is the temperature of the room: the coffee cools towards it, but never falls below that value.

In the exam: for the initial temperature, expect to see the substitution $t = 0$ . If you enter a number into the GDC, write down the equation you set up and give the result to the precision asked for. The interpretation of the asymptote is marked on the wording and the context.

Common mistakes

Summary

An exponential function is $f (x) = k \cdot a^{x} + c$ with $a > 0$ and $a \neq = 1$ ; it increases if $a > 1$ and decreases if $0 < a < 1$ .
Its horizontal asymptote is $y = c$ , and the curve never touches it.
The $y$ -intercept is $(0, k + c)$ ; the $x$ -intercept requires isolating the power and checking that it exists.
In a model, $t$ is time, the constant is found by substituting one piece of data, and the asymptote has an interpretation in context.
Equations with $t$ in the exponent are solved with the GDC or with logarithms; always write down the equation you solve.

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