What is 6.1: Exponential functions?

For a>0 with a1, the exponential function y=a^x has domain R, range y>0, horizontal asymptote y=0 and intercept (0,1).

How many marks are 6.1 questions worth?

6.1 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 6.1 appeared in?

We have not yet indexed a verified past-paper appearance for 6.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

6.1

Know and use the function aˣ and its graph, where a is positive; know and use the function eˣ and its graph.

Draft — not yet indexed

Exponential functions

Worked answers and methods for 6.1 on Edexcel A-level Maths 9MA0.

Explanation

For $a>0$ with $a\neq1$ , the exponential function $y=a^x$ has domain $\mathbb{R}$ , range $y>0$ , horizontal asymptote $y=0$ and intercept $(0,1)$ .
If $a>1$ the graph is increasing, while if $0<a<1$ it is decreasing; $a=1$ gives the constant function $y=1$ .
The natural exponential $e^x$ follows the same graph facts with base $e$ , and transformations such as $Ae^{kx}+c$ scale, reflect and translate the basic graph.
An exponential graph never reaches its horizontal asymptote; treating $a^0$ as $0$ instead of $1$ is a common error.

The basic exponential graphs approach the

x

-axis, their horizontal asymptote

y=0

; the base determines whether the curve rises or falls.

Worked example

For $y=2e^x-5$ , state the domain, range, horizontal asymptote and $y$ -intercept.

1.The function $e^x$ is defined and positive for every real $x$ .
2.Multiplying by $2$ preserves positivity, then subtracting $5$ gives $y>-5$ and moves the asymptote to $y=-5$ .
3.At $x=0$ , $y=2e^0-5=-3$ .

Answer: Domain $x\in\mathbb{R}$ Range $y>-5$ Horizontal asymptote $y=-5$ $y$ -intercept $(0,-3)$

Common mistakes

Don't write $a^0=0$ instead of $1$ , so the $y$ -intercept is wrong.
Don't draw an exponential curve crossing its horizontal asymptote.

Exam tip

State the asymptote, intercept, domain and range before sketching the transformed exponential graph.

Worked practice

Tier 1 · Easy

For the graph

y=3^x

, state the

y

-intercept and the horizontal asymptote.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$y$ -intercept $(0,1)$ Horizontal asymptote $y=0$	2
Notes
At $x=0$ , $y=3^0=1$ , giving the intercept $(0,1)$ . As $x\to-\infty$ , $3^x\to0$ without reaching zero, so the horizontal asymptote is $y=0$ .

(2 marks)

Tier 2 · Standard

Describe the geometrical relationship between the graphs

y=2^x

and

y=2^{-x}

, and find their point of intersection.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	The graphs are reflections of each other in the $y$ -axis. Intersection $(0,1)$	3
Notes
Replacing $x$ by $-x$ reflects a graph in the $y$ -axis, so $y=2^{-x}$ is the reflection of $y=2^x$ . At an intersection, $2^x=2^{-x}$ , hence $x=-x$ and $x=0$ . Both functions then have value $1$ , so the intersection is $(0,1)$ .

(3 marks)

Tier 3 · Hard

The curve

y=3^x+3^{2-x}

has a minimum point. Prove that its minimum value is

6

and find the corresponding value of

x

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Minimum value $6$ , attained when $x=1$	5
Notes
Let $u=3^x$ , so $u>0$ and $3^{2-x}=9/u$ . Then $y=u+9/u$ . Since $y-6=(u^2-6u+9)/u=(u-3)^2/u\geq0$ , it follows that $y\geq6$ . Equality occurs when $u=3$ , so $3^x=3$ and $x=1$ . Hence the minimum point is $(1,6)$ .

(5 marks)

Tier 1 · Easy

For the graph

y=\left(\dfrac13\right)^x

, state whether the function is increasing or decreasing, and state its range.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	The function is decreasing Range $y>0$	2
Notes
The base $1/3$ lies between $0$ and $1$ , so the exponential function decreases as $x$ increases. A positive-base exponential is always positive, so its range is $y>0$ .

(2 marks)

Tier 2 · Standard

The graph of

y=2^x

is transformed to the graph of

y=3-2^{x-1}

. Describe the transformations, and state the range and horizontal asymptote of the transformed graph.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Translate $1$ unit to the right, reflect in the $x$ -axis, then translate $3$ units upwards Range $y<3$ Horizontal asymptote $y=3$	4
Notes
Replacing $x$ by $x-1$ translates the graph $1$ unit right. Multiplication by $-1$ reflects it in the $x$ -axis, and adding $3$ translates it $3$ units upwards. Since $2^{x-1}>0$ , the transformed values satisfy $y<3$ and approach the horizontal asymptote $y=3$ .

(4 marks)

Tier 3 · Hard

Find the exact coordinates of both intersections of the curves

y=4^x

and

y=5-4^{-x}

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$\left(\log_4\!\left(\dfrac{5-\sqrt{21}}2\right),\dfrac{5-\sqrt{21}}2\right)$ and $\left(\log_4\!\left(\dfrac{5+\sqrt{21}}2\right),\dfrac{5+\sqrt{21}}2\right)$	5
Notes
Let $u=4^x$ , so $u>0$ and $4^{-x}=1/u$ . At an intersection, $u=5-1/u$ , hence $u^2-5u+1=0$ . Thus $u=(5\pm\sqrt{21})/2$ ; both roots are positive. Since $u=4^x=y$ , the two corresponding coordinates are the stated points.

(5 marks)

Tier 2 · Standard

The graph of

y=a^x

, where

a>0

, passes through

\left(3,\dfrac1{64}\right)

. Find

a

. State whether the function is increasing or decreasing, and state its

y

-intercept and horizontal asymptote.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$a=\dfrac14$ The function is decreasing $y$ -intercept $(0,1)$ Horizontal asymptote $y=0$	4
Notes
The point gives $a^3=1/64=(1/4)^3$ . Since $a>0$ , $a=1/4$ . As $0<a<1$ , the function is decreasing. Also $a^0=1$ , giving the $y$ -intercept $(0,1)$ , and $a^x\to0$ as $x\to\infty$ , so the horizontal asymptote is $y=0$ .

(4 marks)

Tier 3 · Hard

The curve

y=Aq^x+c

, where

A>0

and

0<q<1

, passes through

(0,11)

(1,7)

and

(2,5)

. Determine

A

q

and

c

. Hence state the range and horizontal asymptote of the curve.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$A=8$ , $q=\dfrac12$ and $c=3$ Range $y>3$ Horizontal asymptote $y=3$	6
Notes
The three points give $A+c=11$ , $Aq+c=7$ and $Aq^2+c=5$ . The quantities $11-c$ , $7-c$ and $5-c$ are consecutive terms of a geometric progression, and each is nonzero since $A>0$ and $q>0$ , so $\dfrac{7-c}{11-c}=\dfrac{5-c}{7-c}$ . Hence $(7-c)^2=(11-c)(5-c)$ , which gives $c=3$ . It follows that $A=8$ and $q=(7-3)/8=1/2$ . Since $8(1/2)^x>0$ for all real $x$ and tends to zero as $x\to\infty$ , the range is $y>3$ and the horizontal asymptote is $y=3$ .

(6 marks)

Tier 3 · Hard

Starting with

y=e^x

, consider the two transformations

A

: a stretch parallel to the

y

-axis with scale factor

3

, and

B

: a translation by the vector

\begin{pmatrix}\ln2\\-4\end{pmatrix}

. Find the equations of the images obtained by applying

A

then

B

, and by applying

B

then

A

. Show that the two images do not intersect. Hence find the exact values of

x

for which the images lie on opposite sides of the

x

-axis.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$A$ then $B$ : $y=\dfrac32e^x-4$ $B$ then $A$ : $y=\dfrac32e^x-12$ The images do not intersect $\ln\!\left(\dfrac83\right)<x<\ln8$	6
Notes
Applying $A$ first gives $y=3e^x$ ; translating right by $\ln2$ and down by $4$ then gives $y=3e^{x-\ln2}-4=\tfrac32e^x-4$ . Applying $B$ first gives $y=e^{x-\ln2}-4=\tfrac12e^x-4$ ; the vertical stretch then gives $y=\tfrac32e^x-12$ . The first image is exactly $8$ units above the second for every $x$ , so they cannot intersect. They are on opposite sides of the $x$ -axis exactly when the first is positive and the second is negative: $\tfrac32e^x>4$ and $\tfrac32e^x<12$ . Thus $8/3<e^x<8$ , and since $\ln$ is increasing, $\ln(8/3)<x<\ln8$ .

(6 marks)

Q10

Tier 3 · Hard

10.

Starting with the graph of

y=e^x

, apply, in order, a stretch parallel to the

x

-axis with scale factor

2

, a reflection in the

x

-axis, and a translation by the vector

\begin{pmatrix}-1\\4\end{pmatrix}

. Find the equation, range, horizontal asymptote and exact

x

-intercept of the resulting graph. Find also the image of the point

(0,1)

under these transformations.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$y=4-e^{(x+1)/2}$ Range $y<4$ and horizontal asymptote $y=4$ $x$ -intercept $\left(2\ln4-1,0\right)$ The image of $(0,1)$ is $(-1,3)$	7
Notes
The horizontal stretch gives $y=e^{x/2}$ , reflection gives $y=-e^{x/2}$ , and translating one unit left and four units up gives $y=4-e^{(x+1)/2}$ . The exponential term is positive, so $y<4$ and approaches the horizontal asymptote $y=4$ . Setting $y=0$ gives $e^{(x+1)/2}=4$ , hence $x=2\ln4-1$ . The point $(0,1)$ maps successively to $(0,1)$ , $(0,-1)$ and $(-1,3)$ .

(7 marks)

Verified exam appearances

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Other points in 6 Exponentials and logarithms

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