Exponential Functions

This lesson introduces exponential functions — functions of the form y = aˣ where the variable is in the exponent. These functions model growth and decay in contexts ranging from population dynamics to radioactive decay, and they are central to the Edexcel 9MA0 specification.

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What Is an Exponential Function?

An exponential function has the form:

y = aˣ

where a is a positive constant called the base and x is the exponent (the variable). The key feature is that the variable appears as the power, not as the base.

Key Properties of y = aˣ (where a > 0, a ≠ 1)

Property	Detail
Domain	All real numbers — you can raise a to any power
Range	y > 0 — an exponential function is always positive
y-intercept	(0, 1) — because a⁰ = 1 for any valid base
Asymptote	The x-axis (y = 0) is a horizontal asymptote
Growth/Decay	If a > 1 the function increases (growth); if 0 < a < 1 the function decreases (decay)

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Graphs of y = aˣ

When a > 1 (Exponential Growth)

Consider y = 2ˣ:

x	-3	-2	-1	0	1	2	3
y	1/8	1/4	1/2	1	2	4	8

The graph rises steeply for positive x and approaches zero for large negative x. It always passes through (0, 1).

When 0 < a < 1 (Exponential Decay)

Consider y = (1/2)ˣ = 2⁻ˣ:

x	-3	-2	-1	0	1	2	3
y	8	4	2	1	1/2	1/4	1/8

The graph falls as x increases — this is a reflection of y = 2ˣ in the y-axis.

Exam Tip: The graph of y = a⁻ˣ is the reflection of y = aˣ in the y-axis. This means y = (1/a)ˣ and y = a⁻ˣ are identical functions.

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Transformations of Exponential Functions

Vertical Stretch: y = kaˣ

Multiplying by k stretches the graph vertically by scale factor k. The y-intercept becomes (0, k) instead of (0, 1). The asymptote remains y = 0.

Vertical Translation: y = aˣ + c

Adding c translates the graph up by c units. The asymptote moves to y = c.

Example: y = 3ˣ + 2 has asymptote y = 2 and y-intercept (0, 3) since 3⁰ + 2 = 3.

Horizontal Translation: y = a^(x + d)

Replacing x with (x + d) translates the graph left by d units. The y-intercept becomes (0, a^d).

Reflection: y = a⁻ˣ

As noted above, negating x reflects the graph in the y-axis, converting growth into decay and vice versa.

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Comparing Exponential Functions

For bases a > 1, a larger base means faster growth:

• y = 10ˣ grows much faster than y = 2ˣ
• Both pass through (0, 1)
• For x > 0, the larger base gives a larger y-value

For decay curves, a smaller base (closer to 0) means faster decay.

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Exponential Functions in Context

Exponential functions model situations where the rate of change is proportional to the current value:

• Population growth — each generation produces offspring proportional to the current population.
• Compound interest — interest earned is proportional to the current balance.
• Radioactive decay — the number of atoms decaying per second is proportional to the number present.
• Cooling — Newton's law of cooling says the rate of temperature loss is proportional to the temperature difference.

At A-Level you will encounter models such as:

• P = P₀ × aᵗ (population)
• A = A₀ × (1 + r/100)ⁿ (compound interest)
• N = N₀ × (1/2)^(t/h) (radioactive half-life)

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Sketching Tips for the Exam

1. Always mark the y-intercept — calculate the value when x = 0.
2. Draw the asymptote as a dashed line and label it.
3. Show the general shape — rising for growth, falling for decay.
4. Label the curve with its equation.
5. If asked about intersections, set the two functions equal and solve.

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Summary

• An exponential function has the form y = aˣ where the variable is the exponent.
• All exponential graphs pass through (0, 1), have the x-axis as an asymptote, and are always positive.
• If a > 1 the function grows; if 0 < a < 1 it decays.
• Transformations follow the standard rules: vertical/horizontal stretches and translations, reflections.
• Exponential functions model real-world growth and decay where the rate of change is proportional to the current value.

A-Level Deep Dive: Exponential Functions

Spec mapping

Edexcel 9MA0-01 specification section 6 — Exponentials and logarithms covers the function $y = a^x$y=ax and its graph, where $a$a is positive (refer to the official specification document for exact wording). Although this looks innocuous, exponential functions sit at the centre of a synoptic web that covers half of Pure Mathematics. The ideas connect directly to section 2 (Algebra and functions, transformations of graphs) because every exponential question on Paper 1 dresses $y = a^x$y=ax in some combination of vertical/horizontal stretches, translations and reflections; to section 9 (Differentiation) through the result $\frac{d}{dx}a^x = a^x \ln a$dxd​ax=axlna; and to section 14 (Modelling with differential equations) where exponential growth and decay model populations, cooling, radioactivity and compound interest. The inverse relationship with logarithms (also in section 6) means that the moment a candidate confidently sketches $y = a^x$y=ax they have implicitly understood $y = \log_a x$y=loga​x as its reflection in $y = x$y=x. Rates of change and the concept of proportional growth — "the derivative is proportional to the function value" — are also examined in section 9 and form the bridge between exponentials and the natural exponential $e^x$ex. The Edexcel formula booklet does not list the result $\frac{d}{dx}a^x = a^x \ln a$dxd​ax=axlna, but it does list the change-of-base relationship; index laws are assumed prior knowledge.

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Worked example with full mark scheme

Question (8 marks):

(a) Sketch the graph of $y = 2^{x-1} - 3$y=2x−1−3, showing clearly the coordinates of any intercepts with the axes and the equation of any asymptote. (4)

(b) Find the exact value of $x$x for which $2^{x-1} - 3 = 5$2x−1−3=5. (2)

(c) The function $f$f is defined by $f(x) = 2^{x-1} - 3$f(x)=2x−1−3 for $x \in \mathbb{R}$x∈R. Find $f^{-1}(x)$f−1(x) and state its domain. (2)

Solution with mark scheme:

(a) Step 1 — identify the parent function and transformations.

Start from $y = 2^x$y=2x (parent), apply a horizontal translation by $+1$+1 to get $y = 2^{x-1}$y=2x−1, then a vertical translation by $-3$−3 to get $y = 2^{x-1} - 3$y=2x−1−3.

M1 — recognising both transformations correctly. The horizontal translation is "right by 1" because the input is $x - 1$x−1. A common error is "left by 1" — the sign is inverted relative to vertical translations.

Step 2 — find the y-intercept.

Set $x = 0$x=0: $y = 2^{-1} - 3 = \tfrac{1}{2} - 3 = -\tfrac{5}{2}$y=2−1−3=21​−3=−25​.

A1 — y-intercept at $(0, -\tfrac{5}{2})$(0,−25​).

Step 3 — find the x-intercept.

Set $y = 0$y=0: $2^{x-1} = 3$2x−1=3, so $x - 1 = \log_2 3$x−1=log2​3, giving $x = 1 + \log_2 3$x=1+log2​3.

A1 — x-intercept at $(1 + \log_2 3, 0)$(1+log2​3,0). Decimal approximations are not required and may lose the exact-form mark; the question demands "coordinates" but exact form is the safer presentation.

Step 4 — identify the horizontal asymptote.

As $x \to -\infty$x→−∞, $2^{x-1} \to 0^{+}$2x−1→0+, so $y \to -3$y→−3. The asymptote is $y = -3$y=−3.

A1 — horizontal asymptote correctly identified and labelled on the sketch. Note that this is $y = -3$y=−3, not the parent's $y = 0$y=0 — the vertical translation has shifted the asymptote down by 3. Failing to update the asymptote after a vertical shift is the single most common error on this style of question.

(b) Step 1 — rearrange.

$2^{x-1} - 3 = 5 \implies 2^{x-1} = 8$2x−1−3=5⟹2x−1=8.

M1 — isolating the exponential.

Step 2 — recognise $8 = 2^3$8=23.

$2^{x-1} = 2^3 \implies x - 1 = 3 \implies x = 4$2x−1=23⟹x−1=3⟹x=4.

A1 — exact answer $x = 4$x=4. Using $\log_2$log2​ here would also work but is unnecessary when both sides reduce to the same base.

(c) Step 1 — set $y = 2^{x-1} - 3$y=2x−1−3 and swap.

Let $y = 2^{x-1} - 3$y=2x−1−3. Make $x$x the subject: $y + 3 = 2^{x-1}$y+3=2x−1, then $\log_2(y + 3) = x - 1$log2​(y+3)=x−1, so $x = 1 + \log_2(y + 3)$x=1+log2​(y+3). Swap $x$x and $y$y:

M1 — correct rearrangement and use of $\log_2$log2​ as the inverse of $2^{(\cdot)}$2(⋅).

$f^{-1}(x) = 1 + \log_2(x + 3)$f−1(x)=1+log2​(x+3), valid for $x + 3 > 0$x+3>0, i.e. $x > -3$x>−3.

A1 — both the formula and the domain $x > -3$x>−3 stated explicitly. The domain of $f^{-1}$f−1 equals the range of $f$f, which is $y > -3$y>−3 from part (a).

Total: 8 marks (M3 A5).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A culture of bacteria has population $P$P (millions) at time $t$t hours given by $P = 1.6 \cdot 3^{t/4}$P=1.6⋅3t/4.

(a) State the initial population. (1)

(b) Show that $\log_3 P = \dfrac{t}{4} + \log_3 1.6$log3​P=4t​+log3​1.6. (2)

(c) Find the time, in hours, at which the population first exceeds 10 million, giving your answer to 3 significant figures. (3)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.1a) — substituting $t = 0$t=0 to obtain $P = 1.6$P=1.6 million.

(b)

• M1 (AO1.1b) — applying $\log_3$log3​ to both sides of $P = 1.6 \cdot 3^{t/4}$P=1.6⋅3t/4 and using $\log_3(ab) = \log_3 a + \log_3 b$log3​(ab)=log3​a+log3​b.
• A1 (AO2.1) — correctly simplifying $\log_3(3^{t/4}) = t/4$log3​(3t/4)=t/4 to obtain the printed answer. Marks here reward reasoning: the candidate must show both log-law applications.

(c)

• M1 (AO3.1a) — setting $1.6 \cdot 3^{t/4} > 10$1.6⋅3t/4>10 and isolating $3^{t/4} > 6.25$3t/4>6.25.
• M1 (AO1.1b) — taking logs to obtain $t/4 > \log_3 6.25$t/4>log3​6.25 or equivalent using natural logs: $t/4 > \ln 6.25 / \ln 3$t/4>ln6.25/ln3.
• A1 (AO1.1b) — $t > 4 \log_3 6.25 \approx 6.67$t>4log3​6.25≈6.67 hours (3 s.f.).

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is a typical Paper 1 modelling question — AO1 dominates because the procedural log-manipulation steps must be solid, but the AO3 mark for translating "first exceeds 10 million" into a mathematical inequality is what separates routine answers from full marks.

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Synoptic links

Connects to:

1. Section 6 — Logarithms as inverse exponentials: the function $y = \log_a x$y=loga​x is exactly the reflection of $y = a^x$y=ax in the line $y = x$y=x. Every property of exponentials has a mirror property of logarithms — $a^x > 0$ax>0 for all $x$x becomes "the domain of $\log_a$loga​ is $x > 0$x>0"; the asymptote $y = 0$y=0 for $a^x$ax becomes the asymptote $x = 0$x=0 for $\log_a x$loga​x. Strong A* candidates exploit this duality to halve their memorisation.

2. Section 9 — Differentiation $\frac{d}{dx}a^x = a^x \ln a$dxd​ax=axlna: this beautiful result says the gradient of $y = a^x$y=ax at any point is proportional to $y$y itself, with constant of proportionality $\ln a$lna. When $a = e$a=e the constant is 1, which is the defining property of $e$e. Every exponential growth/decay question in section 14 ultimately rests on this differentiation rule.

3. Section 2 — Transformations of graphs: $y = a \cdot b^{cx + d} + k$y=a⋅bcx+d+k involves up to four transformations (vertical stretch by $a$a, horizontal stretch by $1/c$1/c, horizontal shift by $-d/c$−d/c, vertical shift by $k$k). Sketching such a graph forces complete fluency with both the order of transformations and their effect on intercepts and asymptotes.

4. Section 14 — Modelling exponential growth and decay: the differential equation $\frac{dN}{dt} = kN$dtdN​=kN has general solution $N = N_0 e^{kt}$N=N0​ekt, which can equivalently be written $N = N_0 a^t$N=N0​at where $a = e^k$a=ek. Every exponential model — radioactive decay, Newton's law of cooling, compound interest, populations, drug clearance — is an instance of this single differential equation, and recognising the structure across contexts is a hallmark of synoptic A* thinking.

5. Section 11 — Binomial expansion of $(1 + x)^n$(1+x)n: continuous compounding $(1 + r/n)^{nt}$(1+r/n)nt converges to $e^{rt}$ert as $n \to \infty$n→∞. Expanding $(1 + x)^n$(1+x)n for small $x$x via the binomial series produces $1 + nx + \binom{n}{2}x^2 + \ldots$1+nx+(2n​)x2+…, and letting $n \to \infty$n→∞ with $nx$nx fixed recovers the Taylor expansion of $e^{nx}$enx. This is one of the deepest synoptic links in the entire specification.

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Mark-scheme literacy

Exponential-function questions on 9MA0 split AO marks fairly evenly between AO1 and AO2, with AO3 appearing on modelling parts:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–65%	Sketching graphs, computing intercepts/asymptotes, applying transformations, manipulating exponential expressions, basic log-laws
AO2 (reasoning / interpretation)	25–35%	Justifying the position of an asymptote after transformation, identifying domain/range of inverse functions, recognising when to take logs of which base, interpreting the meaning of a derivative in context
AO3 (problem-solving)	5–20%	Setting up exponential equations from worded modelling stems, deciding which variable to isolate, formulating "first exceeds" or "halves" inequalities

Examiner-rewarded phrasing: "since $a^x > 0$ax>0 for all real $x$x, the curve never crosses the asymptote"; "the horizontal asymptote of $y = a^x + k$y=ax+k is $y = k$y=k, shifted from the parent's $y = 0$y=0"; "since the function is one-to-one on its domain, the inverse exists". Phrases that lose marks: "$2^x = 0$2x=0 when $x \to -\infty$x→−∞" (it tends to 0 but never equals 0 — A2 deduction); "the asymptote is $y = 0$y=0" written for a vertically-translated graph (failing to update the asymptote); using $\log$log without specifying base in a context that requires a particular base.

A specific Edexcel pattern to watch: when a question asks for an answer "to 3 significant figures" on a logarithmic answer, candidates often round prematurely mid-calculation. Always carry full calculator precision until the final line, then round once. Examiners can tell from the value whether you rounded $\log_3 6.25$log3​6.25 to 1.65 (loses precision) or kept the full 1.6505... internally.

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Grade-band model answers

3-mark question

Question: The graph of $y = 5^x - 4$y=5x−4 is drawn. State (i) the equation of the horizontal asymptote and (ii) the y-intercept.

Grade C response (~210 words):

(i) The asymptote of $y = 5^x$y=5x is $y = 0$y=0. Subtracting 4 shifts the graph down by 4 units, so the new asymptote is $y = -4$y=−4.

(ii) When $x = 0$x=0, $y = 5^0 - 4 = 1 - 4 = -3$y=50−4=1−4=−3. So the y-intercept is at $(0, -3)$(0,−3).