Open Math›Year 11›Mathematics Advanced›Exponential and logarithmic functions›Exponential functions

Exponential functions

Evaluate exponential expressions for positive and negative integer inputs, identify exponential growth and decay, and explore the unique properties of

y = e^x

Worked examples

Evaluating an exponential function and identifying key features

Straightforward

Problem

For

y = 2^x

, evaluate

y

x = -2, -1, 0, 1, 2

. Hence state the y-intercept and the equation of the horizontal asymptote.

Evaluate

y = 2^x

at each negative

x

-value using the rule

a^{-n} = \dfrac{1}{a^n}

2^{-2} = \frac{1}{2^2} = \frac{1}{4}, \qquad 2^{-1} = \frac{1}{2}

Evaluate at

x = 0

using

a^0 = 1

2^0 = 1

Evaluate at each positive

x

-value.

2^1 = 2, \qquad 2^2 = 4

Identify the y-intercept (the value of

y

when

x = 0

) and the horizontal asymptote (the value

y

approaches as

x \to -\infty

\text{y-intercept: } y = 2^0 = 1 \implies (0,\, 1)

\text{As } x \to -\infty,\; 2^x \to 0 \implies \text{asymptote: } y = 0

Answer

The y-intercept is

(0, 1)

and the horizontal asymptote is

y = 0

Identifying exponential growth or decay and stating key features

Moderate

Problem

For

y = 5 \cdot (0.5)^x

, determine whether the function shows exponential growth or decay. State its domain, range and the equation of its horizontal asymptote.

Identify the base

a

and the coefficient

k

y = k \cdot a^x \text{ where } k = 5 \text{ and } a = 0.5

Since

0 < a < 1

, classify the function as exponential decay.

0 < 0.5 < 1 \implies \text{exponential decay}

State the domain. Exponential functions are defined for all real

x

\text{Domain: } x \in (-\infty,\, \infty)

Determine the range. Since

k = 5 > 0

and

(0.5)^x > 0

for all

x

, the function is always positive. As

x \to \infty

y \to 0

; as

x \to -\infty

y \to \infty

\text{Range: } y \in (0,\, \infty)

Identify the horizontal asymptote: the value

y

approaches but never reaches.

\text{As } x \to \infty,\; y \to 0 \implies \text{asymptote: } y = 0

Answer

Exponential decay. Domain: all real numbers. Range:

y > 0

. Horizontal asymptote:

y = 0

Exploring the unique property of $y = e^x$ and sketching key features

Challenging

Problem

The exponential function

y = e^x

has the unique property that its gradient at every point equals its y-value at that point. State

\dfrac{d}{dx}(e^x)

and hence describe the key features of both

y = e^x

and

y = e^{-x}

, including y-intercept, asymptote and behaviour.

State the derivative using the unique property of

e^x

\frac{d}{dx}(e^x) = e^x

Find the y-intercept of

y = e^x

by substituting

x = 0

y = e^0 = 1 \implies \text{y-intercept: } (0,\, 1)

State the behaviour and asymptote of

y = e^x

. The gradient is always positive (since

e^x > 0

), so the function is always increasing.

\text{As } x \to -\infty,\; e^x \to 0 \implies \text{asymptote: } y = 0

\text{As } x \to \infty,\; e^x \to \infty

Find the y-intercept of

y = e^{-x}

and determine its behaviour. Note that

e^{-x}

is the reflection of

e^x

in the y-axis.

y = e^{-0} = 1 \implies \text{y-intercept: } (0,\, 1)

\text{As } x \to \infty,\; e^{-x} \to 0 \implies \text{asymptote: } y = 0

\text{As } x \to -\infty,\; e^{-x} \to \infty

Summarise the relationship between the two curves.

y = e^{-x} \text{ is the reflection of } y = e^x \text{ in the y-axis.}

\text{Both pass through } (0, 1) \text{ and have asymptote } y = 0.

y = e^x \text{ increases; } y = e^{-x} \text{ decreases.}

Answer

\dfrac{d}{dx}(e^x) = e^x

. Both

y = e^x

and

y = e^{-x}

have y-intercept

(0, 1)

and horizontal asymptote

y = 0

. They are reflections of each other in the y-axis:

y = e^x

increases and

y = e^{-x}

decreases as

x

increases.

Practise

Q1·Straightforward

Evaluate

y = 2^x

when

x = 4

. What is the value of

y

Explanation

2^4 = 2 \times 2 \times 2 \times 2 = 16

Q2·Straightforward

Evaluate

y = 2^x

when

x = -3

. Write your answer as a fraction.

Explanation

2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}

Q3·Straightforward

What is the y-intercept of the graph of

y = 3^x

Explanation

The y-intercept occurs at

x = 0

y = 3^0 = 1

. This is true for any exponential function

y = a^x

, since

a^0 = 1

Q4·Straightforward

x \to -\infty

, the function

y = 4^x

approaches a horizontal asymptote. What is the equation of this asymptote?

$y = -4$$y = 0$$y = 1$$y = 4$

Explanation

x \to -\infty

4^x \to 0

(the values get smaller and smaller but never reach zero). The graph approaches the x-axis from above, so the horizontal asymptote is

y = 0

Q5·Moderate

Evaluate

y = 3 \cdot 2^x

when

x = 4

Explanation

y = 3 \cdot 2^4 = 3 \cdot 16 = 48

Q6·Moderate

Evaluate

y = 80 \cdot (0.5)^x

when

x = 4

Explanation

(0.5)^4 = \dfrac{1}{16}

, so

y = 80 \times \dfrac{1}{16} = 5

Q7·Moderate

Which of the following exponential functions shows exponential decay?

$y = 2^x$$y = e^x$$y = (0.25)^x$$y = 5 \cdot 3^x$

Explanation

For

y = a^x

, exponential decay occurs when the base

a

satisfies

0 < a < 1

. Here,

a = 0.25

, which lies in this range, so

y = (0.25)^x

decreases as

x

increases. The other options all have bases greater than

1

(since

e \approx 2.718

), so they show exponential growth.

Q8·Moderate

What is the y-intercept of

y = 7 \cdot 5^x

Explanation

x = 0

y = 7 \cdot 5^0 = 7 \cdot 1 = 7

. For any function of the form

y = k \cdot a^x

, the y-intercept is always

k

Q9·Challenging

The unique property of

y = e^x

is that the gradient of the curve at every point equals the y-value at that point, so

\dfrac{d}{dx}(e^x) = e^x

. Using this property, find the gradient of

y = e^x

x = 0

Explanation

Since

\dfrac{d}{dx}(e^x) = e^x

, the gradient at

x = 0

e^0 = 1

. This also means the tangent to

y = e^x

at its y-intercept

(0, 1)

has gradient

1

Q10·Challenging

What is the equation of the horizontal asymptote of

y = e^x

$y = 1$$y = 0$$y = e$$x = 0$

Explanation

x \to -\infty

e^x \to 0

. The graph gets closer and closer to the x-axis but never touches it, so the horizontal asymptote is

y = 0

. Note that

e^x > 0

for all

x

, confirming the function never reaches the asymptote.

Q11·Challenging

Both

y = e^x

and

y = e^{-x}

are graphed on the same axes. They pass through a common point on the y-axis. What is the y-coordinate of this shared y-intercept?

Explanation

For

y = e^x

x = 0

y = e^0 = 1

. For

y = e^{-x}

x = 0

y = e^0 = 1

. Both curves pass through the point

(0, 1)

. This is because

e^{-x}

is the reflection of

e^x

in the y-axis, and reflections in the y-axis preserve the y-intercept.

Q12·Challenging

Which statement correctly describes the behaviour of

y = e^{-x}

x

increases?

It increases without bound.It decreases towards $y = 0$ but never reaches it.It reaches a minimum value of $-1$.It remains constant.

Explanation

Since

e^{-x} = \dfrac{1}{e^x}

, as

x

increases,

e^x

grows without bound, so

e^{-x}

decreases towards

0

. Because

e^{-x} > 0

for all

x

, the function never actually reaches

0

. The horizontal asymptote is

y = 0