Mathematical Methods · Units 3 & 4

Exponential Functions

Master exponential functions the easy way, with plain English intuition, growth and decay, asymptotes, transformations, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Money left alone in a savings account does something quietly amazing. It does not grow by the same dollar amount each year. It grows by a slice of itself, so the bigger it gets, the faster it grows, and the line on the graph keeps tilting steeper forever. That runaway, snowballing behaviour is exactly what an exponential function captures, and the same shape describes spreading rumours, cooling coffee and decaying medicine in your bloodstream.

What makes a function exponential

In every function you have met so far the $x$ sat on the ground floor, as in $x^{2}$ or $3x$ . An exponential function flips that. Now the $x$ is up in the exponent, riding on top of a fixed number called the base.

y = a^{x}, \qquad a > 0, \ a \neq 1

Each time $x$ goes up by one, you multiply by another copy of $a$ . That is the whole personality of the function. Adding gives you straight lines. Repeated multiplying gives you exponentials, and multiplying compounds far faster than adding ever could.

The special base e

Out of all possible bases there is one the whole of calculus is built around, an irrational number close to $2.718$ written as $e$ . The function $y = e^{x}$ is special because its steepness at any point is exactly equal to its height there. Nothing else does that. For now just treat $e$ as a particular number a bit under $3$ , so $y = e^{x}$ is an ordinary growth curve that happens to be the one the exam uses most.

Growth, decay and the floor it never touches

Whether $y = a^{x}$ climbs or falls comes down to one thing.

If $a > 1$ the curve grows, sweeping upward to the right.
If $0 < a < 1$ the curve decays, sliding downward to the right.

A negative sign in the power flips growth into decay, because $a^{-x}$ is the same as $\left(\tfrac{1}{a}\right)^{x}$ . So $e^{-x}$ is a decay curve even though its base is bigger than one.

Here is the feature examiners love. As $x$ heads far to the left, $y = e^{x}$ dives toward the $x$ axis but never reaches it. That line it forever approaches is the horizontal asymptote, with equation $y = 0$ for the basic curve. The function is always positive, so the graph lives entirely above this floor.

As x heads left, y = eˣ (blue) flattens toward its asymptote y = 0, while y = eˣ + 3 (orange) flattens toward y = 3: shifting the curve up by 3 lifts its floor by 3.

Transforming the graph

Real exam graphs are rarely the plain $y = e^{x}$ . They are stretched, flipped and slid. The general form gathers every change into one expression.

y = A\,e^{n(x - h)} + k

Read each letter as a separate instruction, and apply them in the right order.

$A$ stretches the graph away from the $x$ axis. This is a dilation by a factor of $A$ from the $x$ axis. A negative $A$ also reflects the graph in the $x$ axis.
$n$ stretches or squeezes sideways. A negative $n$ reflects the graph in the $y$ axis.
$h$ slides the graph sideways. A value of $x - h$ moves it $h$ units to the right, which feels backwards to most students.
$k$ slides the graph up by $k$ and, crucially, carries the asymptote with it to $y = k$ .

When you write a dilation, the exam wants the precise wording, naming both the factor and the axis, as in dilation by a factor of $2$ from the $x$ axis. A common loss of marks is quoting the factor as its reciprocal, or naming the wrong axis. Order matters too. Reflections and dilations must be applied before you describe a horizontal translation, although the vertical translation $k$ can be stated at any stage.

Solving exponential equations

Many questions secretly hide a quadratic. An equation like $e^{2x} - 5e^{x} + 4 = 0$ looks frightening until you notice $e^{2x}$ is just $\left(e^{x}\right)^{2}$ . Let $a = e^{x}$ and it collapses into a friendly quadratic $a^{2} - 5a + 4 = 0$ that factorises in seconds.

Two traps wait at the end, and the examiner reports flag both every year. First, $a = e^{x}$ is never the final answer. You must finish by solving $e^{x} = a$ to recover $x$ . Second, because $e^{x}$ is always positive, any negative or zero value of $a$ is impossible and must be rejected. Throwing away a valid positive root, or keeping an invalid negative one, are the two errors that cost the most marks.

Learn it, then fade the help

1 · I do — fully worked

Solve the hidden quadratic Solve $e^{2x} - 5e^{x} + 4 = 0$ . Let $a = e^{x}$ , so $a^2 - 5a + 4 = 0$ , which factorises as $(a-1)(a-4) = 0$ , giving $a = 1$ or $a = 4$ . Both are positive, so $e^x = 1 \Rightarrow x = 0$ and $e^x = 4 \Rightarrow x = \log_e(4)$ .

2 · We do — your turn, guided

Same steps, your turn. Solve $e^{2x} + e^{x} - 6 = 0$ . Let $a = e^x$ , factorise $a^2 + a - 6 = (a+3)(a-2)$ , so $a = \dfrac{?}{}$ (keep only the valid value).

Check your step

$a = -3$ or $a = 2$ . Reject $a = -3$ since $e^x > 0$ , leaving $e^x = 2$ , so $x = \log_e(2)$ .

3 · You do — solo

No scaffolding now: solve $e^{2x} - 4e^{x} + 3 = 0$ , giving exact values.

Reveal solution

Let $a = e^x$ : $a^2 - 4a + 3 = (a-1)(a-3) = 0$ , so $a = 1$ or $a = 3$ . Both positive, so $x = \log_e(1) = 0$ or $x = \log_e(3)$ .

See these recipes in action in the Worked Examples tab, then test yourself in Try It.

Lock it in with active recall

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

For

y = a^x

, when does the curve grow and when does it decay?

Where is the horizontal asymptote of

y = e^x + k

Why is the

y

intercept of

y = 3e^x - 2

equal to

(0, 1)

How do you solve an equation containing both

e^{2x}

and

e^x

y = A\,e^{n(x-h)} + k

, what does each of

A

and

n

do?

What is

\displaystyle\lim_{x \to -\infty}(2^x + 5)

Recall · Logarithmic Functions

y = \log_e(x)

is the inverse of which function, and what does that swap?

Recall · Transformations of Graphs

Do changes outside the function act on

x

y

, and are they “obvious” or “backwards”?

Worked Example 1Recovering a factor from a product, from a real exam

Let $f(x) = e^{x-1}$ . Given that the product function $f(x) \times g(x) = e^{(x-1)^2}$ , find the rule for the function $g$ .

1
Divide both sides by $f(x)$ . Dividing powers of $e$ subtracts the exponents.
$g(x) = \frac{e^{(x-1)^2}}{e^{x-1}} = e^{(x-1)^2 - (x-1)}$
2
Factor the exponent by taking out the common factor $(x-1)$ .
$(x-1)^2 - (x-1) = (x-1)(x-1-1) = (x-1)(x-2)$
3
Write the rule with the factored exponent.
$g(x) = e^{(x-2)(x-1)}$

Answer

g(x) = e^{(x-2)(x-1)}

VCAA 2023 Mathematical Methods Exam 2, Section A Q16

Worked Example 2Reading a transformed exponential

The graph of $g(x) = 2e^{x} + 3$ is the graph of $y = e^{x}$ after some changes. State the dilation, the translation, the equation of the horizontal asymptote, and the $y$ intercept.

1
The factor $2$ multiplies the whole function, so it stretches the graph away from the $x$ axis.
$\text{Dilation by a factor of } 2 \text{ from the } x \text{ axis}$
2
The $+3$ lifts every point up. The asymptote moves up with it.
$\text{Translation of } 3 \text{ units up}$
3
$y = e^{x}$ sits above $y = 0$ . After lifting $3$ units the floor is now $y = 3$ .
$y = 3$
4
Find the $y$ intercept by putting $x = 0$ and using $e^{0} = 1$ .
$g(0) = 2e^{0} + 3 = 2(1) + 3 = 5$

Answer

\text{Dilation factor } 2 \text{ from the } x \text{ axis, up } 3, \text{ asymptote } y = 3, \text{ intercept } (0, 5)

Worked Example 3Solving a hidden quadratic

Solve $e^{2x} - 5e^{x} + 4 = 0$ for $x$ , giving exact values.

1
Let $a = e^{x}$ . Then $e^{2x} = (e^{x})^{2} = a^{2}$ , so the equation becomes a quadratic in $a$ .
$a^{2} - 5a + 4 = 0$
2
Factorise and solve for $a$ .
$(a - 1)(a - 4) = 0 \quad \Rightarrow \quad a = 1 \text{ or } a = 4$
3
Now undo the substitution. Both values are positive, so both are allowed because $e^{x} > 0$ always.
$e^{x} = 1 \quad \text{or} \quad e^{x} = 4$
4
Take the natural log of each side. Remember $a = e^{x}$ was never the final answer.
$x = \log_{e}(1) = 0 \quad \text{or} \quad x = \log_{e}(4)$

Answer

x = 0 \quad \text{or} \quad x = \log_{e}(4)

Worked Example 4Rejecting an impossible solution

Solve $e^{2x} + e^{x} - 6 = 0$ for $x$ , giving an exact value.

1
Let $a = e^{x}$ , so the equation is a quadratic in $a$ .
$a^{2} + a - 6 = 0$
2
Factorise.
$(a + 3)(a - 2) = 0 \quad \Rightarrow \quad a = -3 \text{ or } a = 2$
3
Reject $a = -3$ because $e^{x}$ is never negative. Only $a = 2$ survives.
$e^{x} = 2 \quad (\text{since } e^{x} > 0)$
4
Take the natural log.
$x = \log_{e}(2)$

Answer

x = \log_{e}(2)

Practice test

Try it yourself

7 questions, 9 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.Consider the function $g : R \to R$ , $g(x) = 2^{x} + 5$ . State the value of $\displaystyle\lim_{x \to -\infty} g(x)$ .

1mark

Work this on paper. The worked solution appears once you submit.

VCAA 2023 Mathematical Methods Exam 2, Section B Q3a

Q2.The horizontal asymptote of $y = e^{x} - 4$ has equation:

1mark

Need a hint?

The basic curve

y = e^{x}

has asymptote

y = 0

. Ask which way the

-4

slides the whole graph, and move the floor by the same amount.

Q3.The $y$ intercept of $y = 3e^{x} - 2$ is:

1mark

Need a hint?

Substitute

x = 0

, and remember that

e^{0} = 1

, not

0

, before you simplify.

Q4.The graph of $y = e^{-x}$ is obtained from the graph of $y = e^{x}$ by a:

1mark

Need a hint?

The minus sign is on the

x

, not on the whole function. Decide whether that flips the graph horizontally or vertically.

Q5.The solution to $e^{2x} - 3e^{x} - 4 = 0$ is:

1mark

Need a hint?

Let

a = e^{x}

to get a quadratic. After solving for

a

, discard any value that

e^{x}

can never equal.

Q6.A population modelled by $P(t) = 200 \times 2^{-t}$ , with $t$ in hours, is best described as:

1mark

Need a hint?

Put

t = 0

to find the starting value, then read the negative exponent to decide whether the base acts as a multiplier above or below

1

Q7.Solve $e^{2x} - 4e^{x} + 3 = 0$ for $x$ , giving exact values. Show every step.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

How do I find the horizontal asymptote of an exponential function?

Start from the fact that the basic curve sits on the line y equals zero, then follow the vertical shift. Whatever number is added or subtracted outside the exponential moves the asymptote by that same amount, so a curve like e to the x plus three has asymptote y equals three.

Why is e to the power of zero equal to one and not zero?

Any non zero number raised to the power of zero is one, and e is no exception. This is why you must always use one, not zero, when you substitute x equals zero to find the y intercept of an exponential graph.

How do I solve an equation with both e to the 2x and e to the x in it?

Substitute a single letter for e to the x, which turns the equation into an ordinary quadratic you can factorise. After solving, swap back and solve e to the x equals each value, and reject any answer that is negative or zero because e to the x is always positive.

Exponential Functions

Learn

What makes a function exponential

The special base e

Growth, decay and the floor it never touches

Transforming the graph

Solving exponential equations

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions