Chapter 25 · Higher tier

GCSE Maths Functions and Advanced Graphs (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Use function notation f(x), evaluate functions and solve equations of the form f(x) = k
Find inverse functions and composite functions such as fg(x), and solve equations involving them
Sketch the graphs of y = f(x) + a, y = f(x + a), y = −f(x), y = f(−x), and recognise stretches y = af(x) and y = f(ax)
Recognise, sketch and use the graphs of y = sin x, y = cos x and y = tan x (shape, period, key values), and solve trigonometric equations
Recognise and use exponential graphs y = k^x and solve growth and decay problems, including compound interest
Use the equation x² + y² = r² of a circle and find the equation of a tangent at a point
Estimate gradients of curves with tangents, and understand average and instantaneous rates of change
Estimate the area under a graph with trapezia and rectangles, and interpret it in context
Solve equations numerically by iteration, and work with general iterative processes

You should already know

Substituting into expressions and rearranging formulae (Chapter 8)
Gradients and equations of straight lines, including perpendicular lines (Chapter 12)
Quadratic and reciprocal graphs (Chapter 13) and solving quadratics (Chapter 14)
Compound interest and percentage multipliers (Chapter 5)
Trigonometry in right-angled triangles (Chapter 20)
Speed–time graphs (Chapters 13 and 16)

Where this leads

Circle theorems, including the tangent–radius theorem (Chapter 26)
A-level maths: calculus, which finds exact gradients and areas
Further work on sequences, series and growth models

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: A7, A12, A13, A15, A16, A20, R15, R16.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

25.1 Function notation: Worksheet (PDF) · Answers (PDF) · lesson page
25.2 Inverse functions: Worksheet (PDF) · Answers (PDF) · lesson page
25.3 Composite functions: Worksheet (PDF) · Answers (PDF) · lesson page
25.4 Graph transformations: Worksheet (PDF) · Answers (PDF) · lesson page
25.5 Trigonometric graphs: Worksheet (PDF) · Answers (PDF) · lesson page
25.6 Exponential graphs, growth and decay: Worksheet (PDF) · Answers (PDF) · lesson page
25.7 The equation of a circle and tangents: Worksheet (PDF) · Answers (PDF) · lesson page
25.8 Gradients of curves and rates of change: Worksheet (PDF) · Answers (PDF) · lesson page
25.9 Areas under graphs: Worksheet (PDF) · Answers (PDF) · lesson page
25.10 Iteration and iterative processes: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWork out the value of 3x² − 2x + 5 when x = −2.

Done

FMake x the subject of y = 3x − 5.

Done

FFind the gradient of the straight line through (1, 2) and (5, 10).

Done

FA value is increased by 15%. What single number do you multiply by?

Done

FSolve x² − 5x + 6 = 0.

Done

25.1

Function notation

Edexcel 1MA1: A7
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Function notationComing soon

A function is a rule that takes an input number, does something to it, and gives exactly one output number. You already know functions as function machines: put a number in, the machine follows its instructions, and a number comes out. The machine below multiplies the input by 3 and then subtracts 5.

The function machine for f(x) = 3x − 5. The input is x and the output is f(x).

Mathematicians give a function a name, usually a single letter such as f, g or h, and write the rule with function notation:

f(x) = 3x − 5

This is read “f of x equals 3x minus 5”. The letter x is a placeholder for whatever number goes in. The bracket is not a multiplication: f(x) does not mean f times x. It means “the output of f when the input is x”.

Finding an output. To find f(4) we put 4 in wherever the x was: f(4) = 3 × 4 − 5 = 7. So the function turns 4 into 7. Use brackets for negative inputs: f(−2) = 3 × (−2) − 5 = −11.

Finding an input. Sometimes we are told the output and must find the input. To solve f(x) = 19 we write the rule in place of f(x) and solve the equation: 3x − 5 = 19, so 3x = 24 and x = 8. For a quadratic function the equation may have two answers. For example, if g(x) = x², then g(x) = 9 gives x = 3 or x = −3, because two different inputs can give the same output.

Other inputs. The input does not have to be a number. If f(x) = 3x − 5, then f(a + 2) means put the whole expression a + 2 in place of x: f(a + 2) = 3(a + 2) − 5 = 3a + 1. Always put brackets round the input before you expand.

Which inputs are allowed? The set of allowed inputs is called the domain. For f(x) = 12 ÷ x we cannot divide by 0, so x = 0 is not allowed, and for g(x) = √x negative inputs are not allowed (at GCSE). Functions also appear in real life: a taxi fare C(d) = 2.5 + 1.8d gives the cost in pounds for a journey of d miles.

Key ideas

A function is a rule: each input gives exactly one output. f(x) is the output when the input is x.
f(a) means: replace every x in the rule by a (use brackets), then work it out.
To solve f(x) = k, write the rule in place of f(x) and solve the equation. A quadratic can have two solutions.
f(a + 2) or f(2a): put the whole expression in brackets in place of x, then expand and simplify.
Any letter can name a function (f, g, h, C, P …).

Worked example 25.1.1

f(x) = 5 − 3x. Work out f(−2).

Key words. A function is a rule that turns each input into one output. f(x) is read “f of x” and means the output when the input is x.

Method. To find f(a), substitute the number a for x everywhere in the rule, then work out the answer.

Write down the function. f(x) is a rule that says what to do to the input x. The letter x is a placeholder for whichever number we put in.
f(x) = −3x + 5
Replace every x with −2, using brackets. The number inside the bracket of f(−2) is the input, so it goes where the x was. Brackets matter with negative numbers.
f(−2) = −3 × (−2) + 5
Work out each part separately (powers first, then multiplying). BIDMAS: indices before multiplication, then add or subtract.
= 5 + 6
Add and subtract. Work from left to right.
= 11

Check. Work it out again on a calculator, typing the brackets exactly as shown: the result is 11. ✓

Watch out. Put brackets round negative inputs. For example (−3)² = 9 but −3² = −9.

Worked example 25.1.2

g(x) = x² − 4x + 1. Work out g(−3).

Key words. A function is a rule that turns each input into one output. f(x) is read “f of x” and means the output when the input is x.

Method. To find f(a), substitute the number a for x everywhere in the rule, then work out the answer.

Write down the function. g(x) is a rule that says what to do to the input x. The letter x is a placeholder for whichever number we put in.
g(x) = x² − 4x + 1
Replace every x with −3, using brackets. The number inside the bracket of g(−3) is the input, so it goes where the x was. Brackets matter with negative numbers.
g(−3) = (−3)² − 4 × (−3) + 1
Work out each part separately (powers first, then multiplying). BIDMAS: indices before multiplication, then add or subtract.
= 9 + 12 + 1
Add and subtract. Work from left to right.
= 22

Check. Use a different arrangement: x(x − 4) + 1 with x = −3 gives (−3) × ((−3) − 4) + 1 = 22. ✓

Watch out. Put brackets round negative inputs. For example (−3)² = 9 but −3² = −9.

Worked example 25.1.3

f(x) = 3x + 4. Solve f(x) = 19.

Key words. The input is the number that goes into the function and the output is the number that comes out.

Method. To solve f(x) = 19, put the rule in for f(x) and solve the equation for x.

Write the equation. We are told that the output is 19, so f(x) = 19. Put the rule in for f(x).
3x + 4 = 19
Subtract 4 from both sides. The last thing that happens to the x-part is that 4 is added. The opposite of adding 4 is subtracting 4. Do it to both sides to keep the equation balanced.
3x = 15
Divide both sides by 3. The unknown part is being multiplied by 3. The opposite of multiplying by 3 is dividing by 3.
x = 5
Write the answer. x is now on its own.
x = 5

Check. Put each solution back into the function: f(5) = 3 × 5 + 4 = 19. ✓

Watch out. Do not confuse the two jobs: finding f(3) means put 3 in and get an output; solving f(x) = 3 means the output is 3 and you must find the input.

Worked example 25.1.4

g(x) = x² − 2x. Solve g(x) = 8.

Key words. The input is the number that goes into the function and the output is the number that comes out.

Method. To solve g(x) = 8, put the rule in for g(x) and solve the equation for x.

Write the equation. We are told that the output is 8, so g(x) = 8. Put the rule in for g(x).
x² − 2x = 8
Move everything to one side so that the other side is 0. To factorise, a quadratic equation must equal 0. Subtract 8 from both sides.
x² − 2x − 8 = 0
Factorise the quadratic. If two things multiply to give 0, at least one of them must be 0. Factorising turns the equation into a product that equals 0.
(x − 4)(x + 2) = 0
Set each bracket equal to 0 and solve. One bracket must be 0.
x − 4 = 0 ⇒ x = 4
x + 2 = 0 ⇒ x = −2
Write the solutions. A quadratic equation can have two solutions, so give both.
x = −2 or x = 4

Check. Put each solution back into the function: g(−2) = (−2)² − 2 × (−2) = 8; g(4) = 4² − 2 × 4 = 8. ✓

Watch out. Do not confuse the two jobs: finding f(3) means put 3 in and get an output; solving f(x) = 3 means the output is 3 and you must find the input.

Worked example 25.1.5

f(x) = x² + 3x. Write f(a + 2) in the form a² + pa + q.

Method. To find f(a + 2), replace x in the rule by a + 2 (in brackets) and simplify.

Write the function. f(x) is a rule for the input x. If the input is something else, we put that in the same places.
f(x) = x² + 3x
Replace every x with a + 2. The whole expression in the brackets of f(a + 2) is the input. Use brackets round it.
f(a + 2) = (a + 2)² + 3(a + 2)
Expand any brackets. Multiply out each bracket, one term at a time. For a squared bracket, multiply the bracket by itself.
(a + 2)² = a² + 4a + 4
3(a + 2) = 3a + 6
Collect like terms. Add together the terms with the same power of the letter.
f(a + 2) = a² + 7a + 10

Check. Test with a = 2: the input is 2 + 2 = 4, and the original rule gives 28. The answer gives 2² + 7 × 2 + 10 = 28. They agree. ✓

Watch out. Replace every x, and expand squares properly: (x + 1)² is (x + 1)(x + 1), not x² + 1.

Watch out

f(x) is not f multiplied by x. Also keep the two jobs apart: finding f(3) means put 3 in and get an output; solving f(x) = 3 means the output is 3 and you must find the input. And with negative inputs always use brackets: if f(x) = x², then f(−3) = (−3)² = 9, not −9.

25.2

Inverse functions

Edexcel 1MA1: A7
Grade 6-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Inverse functionsComing soon

Some jobs can be undone. If you put your socks on and then your shoes, to undo it you take your shoes off first and then your socks. A function can be undone in the same way. The inverse function of f, written f⁻¹, takes the output of f and returns the original input. If f(4) = 7 then f⁻¹(7) = 4.

The function machine for f(x) = 3x − 5 and its inverse. Each step is reversed, and the order of the steps is reversed too.

Method 1: reverse the machine. Write down what f does to x in order: “multiply by 3, then subtract 5”. The inverse does the opposite operations in the opposite order: “add 5, then divide by 3”. So f⁻¹(x) = (x + 5) ÷ 3.

Method 2: algebra. This works for every function. (1) Write y = f(x). (2) Rearrange the equation to make x the subject, using the same steps you used to rearrange formulae in Chapter 8. (3) Replace y by x in the answer, because the input of a function is normally called x. For y = 3x − 5 we get y + 5 = 3x, so x = (y + 5) ÷ 3, and f⁻¹(x) = (x + 5) ÷ 3. When x appears twice, such as in f(x) = (2x + 1) ÷ (x − 3), multiply out the fraction first, collect all the x terms on one side and factorise.

Finding one value. To find f⁻¹(7) you do not need the whole inverse. It is the input that f turns into 7, so just solve f(x) = 7.

The graph of the inverse. If (a, b) is on the graph of y = f(x), then (b, a) is on the graph of y = f⁻¹(x), because the roles of input and output swap. So the two graphs are reflections of each other in the line y = x.

The blue line is y = f(x) = 2x + 1 and the orange line is y = f⁻¹(x) = (x − 1)/2. The grey dashed line is y = x. The points (2, 5) and (5, 2) are mirror images.

Not every function has an inverse. If two different inputs give the same output, the inverse would not know which one to return. For example g(x) = x² sends both 3 and −3 to 9. To make an inverse we restrict the inputs, for instance to x ≥ 0, and then g⁻¹(x) = √x. The questions in this chapter will say when this is needed.

Key ideas

The inverse function f⁻¹ undoes f: if f(a) = b then f⁻¹(b) = a.
To find it: write y = f(x), make x the subject, then swap y for x. Or reverse the machine: opposite operations, opposite order.
To find f⁻¹(b) for one value, solve f(x) = b.
The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.
f⁻¹(x) does not mean 1 ÷ f(x).

Worked example 25.2.1

f(x) = 3x − 5. Find f⁻¹(x).

Key words. The inverse function f⁻¹ reverses f: if f takes 3 to 11, then f⁻¹ takes 11 back to 3. Subject means the letter on its own on one side.

Method. To find the inverse function f⁻¹, write y = f(x), make x the subject, then swap the letters. The inverse undoes what the function did.

Write the function as y = … f(x) is the output, which we call y. We want to turn the rule round so that x is given in terms of y.
y = 3x − 5
Describe what the function does to x, in order. Think of a function machine. To undo it we must reverse every step, in reverse order.
Swap the sides so the x part is on the left. We now rearrange to make x the subject.
3x − 5 = y
Add 5 to both sides. The last thing that happens is that 5 is subtracted. The opposite of subtracting 5 is adding 5. Do it to both sides.
3x = y + 5
Divide both sides by 3. The unknown part is being multiplied by 3. The opposite of multiplying by 3 is dividing by 3.
x = y + 53
Write x in terms of y. Now x is the subject.
x = y + 53
Swap the letters: write x for the input of the inverse function. A function is written with x as its input, so we rename y as x.
f⁻¹(x) = x + 53

Check. Test with a number: f(2) = 1. Then f⁻¹(1) = 13 + 53 = 2, which takes us back to the input 2. ✓

Watch out. The notation f⁻¹(x) does not mean 1 divided by f(x). It means the inverse function.

Worked example 25.2.2

f(x) = 2x + 13. Find f⁻¹(x).

Key words. The inverse function f⁻¹ reverses f: if f takes 3 to 11, then f⁻¹ takes 11 back to 3. Subject means the letter on its own on one side.

Method. To find the inverse function f⁻¹, write y = f(x), make x the subject, then swap the letters. The inverse undoes what the function did.

Write the function as y = … f(x) is the output, which we call y. We want to turn the rule round so that x is given in terms of y.
y = 2x + 13
Describe what the function does to x, in order. Think of a function machine. To undo it we must reverse every step, in reverse order.
Swap the sides so the x part is on the left. We now rearrange to make x the subject.
2x + 13 = y
Multiply both sides by 3. The expression is being divided by 3. The opposite of dividing by 3 is multiplying by 3.
2x + 1 = 3y
Subtract 1 from both sides. The last thing that happens to the x-part is that 1 is added. The opposite of adding 1 is subtracting 1. Do it to both sides to keep the equation balanced.
2x = 3y − 1
Divide both sides by 2. The unknown part is being multiplied by 2. The opposite of multiplying by 2 is dividing by 2.
x = 3y − 12
Write x in terms of y. Now x is the subject.
x = 3y − 12
Swap the letters: write x for the input of the inverse function. A function is written with x as its input, so we rename y as x.
f⁻¹(x) = 3x − 12

Check. Test with a number: f(2) = 53. Then f⁻¹(53) = 3 × 532 − 12 = 2, which takes us back to the input 2. ✓

Watch out. The notation f⁻¹(x) does not mean 1 divided by f(x). It means the inverse function.

Worked example 25.2.3

f(x) = 8x − 1, x ≠ 1. Find f⁻¹(x).

Key words. The inverse function f⁻¹ reverses f: if f takes 3 to 11, then f⁻¹ takes 11 back to 3. Subject means the letter on its own on one side.

Method. To find the inverse function f⁻¹, write y = f(x), make x the subject, then swap the letters. The inverse undoes what the function did.

Write the function as y = … f(x) is the output, which we call y. We want to turn the rule round so that x is given in terms of y.
y = 8x − 1
Describe what the function does to x, in order. Think of a function machine. To undo it we must reverse every step, in reverse order.
Swap the sides so the x part is on the left. We now rearrange to make x the subject.
8x − 1 = y
The unknown is on the bottom of the fraction: multiply both sides by it, then divide both sides by the other side. Multiplying both sides by x − 1 removes the fraction. Then dividing both sides by y leaves x − 1 on its own.
8 = y(x − 1)
8y = x − 1
Add 1 to both sides. The last thing that happens is that 1 is subtracted. The opposite of subtracting 1 is adding 1. Do it to both sides.
x = y + 8y
Write x in terms of y. Now x is the subject.
x = y + 8y
Swap the letters: write x for the input of the inverse function. A function is written with x as its input, so we rename y as x.
f⁻¹(x) = x + 8x

Check. Test with a number: f(2) = 8. Then f⁻¹(8) = 88 + 1 = 2, which takes us back to the input 2. ✓

Watch out. The notation f⁻¹(x) does not mean 1 divided by f(x). It means the inverse function.

Worked example 25.2.4

f(x) = (x − 2)³ + 1. Find f⁻¹(x).

Key words. The inverse function f⁻¹ reverses f: if f takes 3 to 11, then f⁻¹ takes 11 back to 3. Subject means the letter on its own on one side.

Method. To find the inverse function f⁻¹, write y = f(x), make x the subject, then swap the letters. The inverse undoes what the function did.

Write the function as y = … f(x) is the output, which we call y. We want to turn the rule round so that x is given in terms of y.
y = (x − 2)³ + 1
Describe what the function does to x, in order. Think of a function machine. To undo it we must reverse every step, in reverse order.
Swap the sides so the x part is on the left. We now rearrange to make x the subject.
(x − 2)³ + 1 = y
Subtract 1 from both sides. The last thing that happens to the x-part is that 1 is added. The opposite of adding 1 is subtracting 1. Do it to both sides to keep the equation balanced.
(x − 2)³ = y − 1
Take the cube root of both sides. The opposite of cubing is the cube root.
x − 2 = ∛(y − 1)
Add 2 to both sides. The last thing that happens is that 2 is subtracted. The opposite of subtracting 2 is adding 2. Do it to both sides.
x = ∛(y − 1) + 2
Write x in terms of y. Now x is the subject.
x = ∛(y − 1) + 2
Swap the letters: write x for the input of the inverse function. A function is written with x as its input, so we rename y as x.
f⁻¹(x) = ∛(x − 1) + 2

Check. Test with a number: f(2) = 1. Then f⁻¹(1) = ∛(1 − 1) + 2 = 2, which takes us back to the input 2. ✓

Watch out. The notation f⁻¹(x) does not mean 1 divided by f(x). It means the inverse function.

Worked example 25.2.5

f(x) = 2x + 1x − 3, x ≠ 3. Find f⁻¹(x).

Method. Rearrange y = f(x) to make x the subject. When x appears twice, multiply out the fraction, collect the x terms and factorise.

Write the function as y = … This time x appears twice (top and bottom), so we cannot simply undo operations one by one. We will rearrange carefully.
y = 2x + 1x − 3
Multiply both sides by the bottom. This removes the fraction.
y(x − 3) = 2x + 1
Expand the bracket. Multiply y by each term inside.
xy − 3y = 2x + 1
Get all the x terms on one side and everything else on the other. We need x on its own eventually, so collect the terms containing x together.
xy − 2x = 3y + 1
Factorise the left side. Take x out as a common factor so that x appears only once.
x(y − 2) = 3y + 1
Divide both sides by the bracket. This leaves x on its own.
x = 3y + 1y − 2
Swap the letters. Write x for the input of the inverse function.
f⁻¹(x) = 3x + 1x − 2

Check. Test: f(4) = 9, and f⁻¹(9) = 4, the number we started with. ✓

Watch out. Do not forget that y multiplies both terms of the bracket when you expand.

Watch out

Do not confuse f⁻¹(x) with 1 ÷ f(x). They are different: for f(x) = 3x − 5, f⁻¹(x) = (x + 5) ÷ 3 but 1 ÷ f(x) = 1 ÷ (3x − 5). Also, when rearranging, divide or add to the whole of the other side: x = (y + 5) ÷ 3, not y + 5 ÷ 3.

25.3

Composite functions

Edexcel 1MA1: A7
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Composite functionsComing soon

A composite function is made by joining two functions: the output of the first becomes the input of the second. Suppose g(x) = x + 2 and f(x) = 3x. If we send a number through g and then straight into f, we have one new function.

The composite function fg(x). The function g acts first, then f acts on the result.

We write this composite function as fg(x), which means f(g(x)). Be careful about the order. The function nearest the x acts first. So fg(x) means “do g, then do f”. It is read from right to left, which feels backwards at first.

With a number. To find fg(4), first work out g(4) = 4 + 2 = 6, then put the 6 into f: f(6) = 18. So fg(4) = 18.

With algebra. To find fg(x) as one rule, replace the x in f by the whole of g(x), in brackets: f(x) = 3x becomes 3(x + 2) = 3x + 6. So fg(x) = 3x + 6. Check: 3 × 4 + 6 = 18. ✓

The order matters. If we do it the other way, gf(x) = g(f(x)) = 3x + 2, which is a different function. Usually fg and gf are not equal. A function composed with itself is written ff(x) or f²(x).

Solving. To solve an equation such as fg(x) = 20, first build the single function fg(x), then solve it like any other equation.

Inverse of a composite. To undo “do g, then f” you must undo f first and then g. So (fg)⁻¹ = g⁻¹f⁻¹. The order reverses, just like taking off your shoes and then your socks.

Key ideas

fg(x) means f(g(x)): do g first, then f.
For a number: work out the inside first, then put that answer into the outside function.
For algebra: replace the x in the outer function by the whole inner function (in brackets), expand and simplify.
fg(x) and gf(x) are usually different. ff(x) means f applied twice.
To solve fg(x) = k: find fg(x) first, then solve the equation.
(fg)⁻¹ = g⁻¹f⁻¹: the order reverses.

Worked example 25.3.1

f(x) = 3x − 2 and g(x) = x². Work out fg(3).

Method. Work from the inside out: put the number into the function nearest it, then put that answer into the other function.

Remember what fg(3) means. The function nearest the number, g, is applied first. Its output is then used as the input to f.
fg(3) = f(g(3))
Work out the inside: g(3). Put 3 into g.
g(3) = 3² = 9
Use that answer as the input to f: work out f(9). The output 9 goes into f.
f(9) = 3 × 9 − 2 = 25
Write the answer.
fg(3) = 25

Check. Another way: find fg(x) = 3x² − 2 first, then put in 3: 3 × 3² − 2 = 25. ✓

Watch out. fg(3) is not f(3) × g(3). It means f of (g of 3).

Worked example 25.3.2

f(x) = 2x + 3 and g(x) = x² − 1. Find fg(x).

Key words. A composite function is two functions joined: the output of the first becomes the input of the second.

Method. fg(x) means: do g first, then f. Replace the x in the outer function by the whole inner function.

Remember what fg(x) means. fg(x) is the same as f(g(x)). The function written closest to the x (g) is applied first, then f is applied to the result.
fg(x) = f(g(x))
Write down the two functions.
g(x) = x² − 1
f(x) = 2x + 3
Put g(x) into f: replace the x in f(x) with x² − 1. g(x) is the output of the first function, and it becomes the input of f. Use brackets.
fg(x) = 2(x² − 1) + 3
Expand the brackets. Multiply out each bracket, one term at a time.
2(x² − 1) = 2x² − 2
Simplify. Collect like terms.
fg(x) = 2x² + 1

Check. Test with x = 2: g(2) = 3, then f(3) = 9. The answer gives 2 × 2² + 1 = 9. They match. ✓

Watch out. Order matters: fg(x) and gf(x) are usually different. The function nearest the x goes first.

Worked example 25.3.3

f(x) = 2x + 3 and g(x) = x² − 1. Find gf(x). Is gf(x) the same as fg(x)?

Key words. A composite function is two functions joined: the output of the first becomes the input of the second.

Method. gf(x) means: do f first, then g. Replace the x in the outer function by the whole inner function.

Remember what gf(x) means. gf(x) is the same as g(f(x)). The function written closest to the x (f) is applied first, then g is applied to the result.
gf(x) = g(f(x))
Write down the two functions.
f(x) = 2x + 3
g(x) = x² − 1
Put f(x) into g: replace the x in g(x) with 2x + 3. f(x) is the output of the first function, and it becomes the input of g. Use brackets.
gf(x) = (2x + 3)² − 1
Expand the brackets. Multiply out each bracket, one term at a time.
(2x + 3)² = 4x² + 12x + 9
Simplify. Collect like terms.
gf(x) = 4x² + 12x + 8

Check. Test with x = 2: f(2) = 7, then g(7) = 48. The answer gives 4 × 2² + 12 × 2 + 8 = 48. They match. ✓

Watch out. Order matters: fg(x) and gf(x) are usually different. The function nearest the x goes first.

Worked example 25.3.4

f(x) = 3x − 1 and g(x) = x² + 2. Solve fg(x) = 32.

Method. First find fg(x) by putting the inner function into the outer one. Then solve fg(x) = 32 as an ordinary equation.

Remember what fg(x) means. fg(x) is the same as f(g(x)). The function written closest to the x (g) is applied first, then f is applied to the result.
fg(x) = f(g(x))
Write down the two functions.
g(x) = x² + 2
f(x) = 3x − 1
Put g(x) into f: replace the x in f(x) with x² + 2. g(x) is the output of the first function, and it becomes the input of f. Use brackets.
fg(x) = 3(x² + 2) − 1
Expand the brackets. Multiply out each bracket, one term at a time.
3(x² + 2) = 3x² + 6
Simplify. Collect like terms.
fg(x) = 3x² + 5
Write the equation. We are told that the output is 32, so fg(x) = 32. Put the rule in for fg(x).
3x² + 5 = 32
Subtract 5 from both sides. The last thing that happens to the x-part is that 5 is added. The opposite of adding 5 is subtracting 5. Do it to both sides to keep the equation balanced.
3x² = 27
Divide both sides by 3. The unknown part is being multiplied by 3. The opposite of multiplying by 3 is dividing by 3.
x² = 9
Take the square root of both sides. Remember the positive and negative roots. Both 3 and −3 square to give 9.
x = 3 or x = −3
Finish each equation. Solve each of the two equations with inverse operations.
x = 3
x = −3

Check. Test: fg(−3) = 3 × (−3)² + 5 = 32; fg(3) = 3 × 3² + 5 = 32. ✓

Watch out. Do not put the number k into the functions one at a time. Build the single function first.

Worked example 25.3.5

f(x) = x + 2 and g(x) = 3x. Find (fg)⁻¹(x), and show that it equals g⁻¹f⁻¹(x).

Method. Find fg(x), then invert it. Then find the two separate inverses and compose them in the reverse order.

Find fg(x). Replace the x in f by g(x) = 3x.
fg(x) = 3x + 2
Invert it: y = 3x + 2. Subtract 2 from both sides, then divide by 3.
y − 2 = 3x ⇒ x = y − 23
Write the inverse.
(fg)⁻¹(x) = x − 23
Find the inverses of f and g separately. Reverse each operation.
f⁻¹(x) = x − 2 and g⁻¹(x) = x ÷ 3
Compose in the reverse order: g⁻¹f⁻¹(x). First undo f (subtract 2), then undo g (divide by 3).
g⁻¹(x − 2) = (x − 2) ÷ 3 = x − 23
Compare. The two answers are the same.
(fg)⁻¹ = g⁻¹f⁻¹ ✓

Check. fg(5) = 3 × 5 + 2 = 17, and (fg)⁻¹(17) = (17 − 2) ÷ 3 = 5. ✓

Watch out. If you apply the inverses in the same order as the functions (f⁻¹ first) you get a different, wrong answer.

Watch out

The function nearest the x acts first. fg(2) is not f(2) × g(2) and it is not g(f(2)). It is f(g(2)): work out g(2) first. And when you replace x in the outer function, put brackets round the inner function: 3(x + 2) = 3x + 6, but 3x + 2 is wrong.

25.4

Graph transformations

Edexcel 1MA1: A12, A13
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Graph transformationsComing soon

If you know the graph of y = f(x), you can sketch the graphs of many related functions without calculating anything new. Each point on the old graph moves to a new position, and the new graph is the old graph moved, flipped or stretched. This is called a transformation of the graph.

The rules are in the table. In each case, a point (x, y) on y = f(x) goes to the point shown.

New function	What happens	Point (x, y) goes to
y = f(x) + a	translation up a units (down if a is negative)	(x, y + a)
y = f(x + a)	translation a units to the left (right if a is negative)	(x − a, y)
y = −f(x)	reflection in the x-axis	(x, −y)
y = f(−x)	reflection in the y-axis	(−x, y)
y = af(x) (extension)	stretch parallel to the y-axis, scale factor a	(x, ay)
y = f(ax) (extension)	stretch parallel to the x-axis, scale factor 1/a	(x/a, y)

The graph below shows y = f(x) = x² − 4x + 3 (dashed) and four transformed graphs (solid orange). Look at how the minimum point B(2, −1) moves each time.

The dashed curve is y = f(x). The orange curves are the transformed graphs. In f(x) + 2 the minimum B(2, −1) moves to (2, 1). In f(x + 2) it moves to (0, −1). In −f(x) it becomes a maximum at (2, 1). In f(−x) it moves to (−2, −1).

Why does f(x + 2) move the graph to the left? The new function gives the same output as f but when the input is 2 smaller: for example the new function at x = 0 gives f(0 + 2) = f(2). So whatever f does at x = 2, the new graph does at x = 0: everything has shifted 2 to the left.

Everything moves together. Turning points, intercepts and asymptotes move with the graph. The reciprocal graph y = 1/x has asymptotes x = 0 and y = 0. The graph of y = 1/x + 2 has asymptotes x = 0 and y = 2, and the graph of y = 1/(x − 3) has asymptotes x = 3 and y = 0.

The dashed curve is y = 1/x. The orange curve is y = 1/(x − 3) + 2: it has been translated 3 right and 2 up, so its asymptotes are x = 3 and y = 2.

Completing the square. A quadratic written as y = (x + p)² + q is the graph of y = x² translated by p to the left and q up, so its turning point is (−p, q). This links to the work on completing the square in Chapter 14.

Key ideas

f(x) + a: translate up a. f(x + a): translate a to the left. Together: the vector (−a, b) for f(x + a) + b.
−f(x): reflect in the x-axis. f(−x): reflect in the y-axis.
af(x): stretch in the y-direction by scale factor a. f(ax): stretch in the x-direction by scale factor 1/a.
Changes outside the bracket affect y. Changes inside the bracket affect x (and work the opposite way).
Turning points, roots and asymptotes move with the graph: transform the key points first.

Worked example 25.4.1

The graph of y = f(x) has a minimum at B(2, −1), crosses the x-axis at A(1, 0) and C(3, 0), and crosses the y-axis at D(0, 3). Find the coordinates of the images of A, B, C and D on the graph of y = f(x) + 2, and sketch it.

Method. y = f(x) + 2 is a translation by the vector 0
2 (2 units up). Apply the rule to each key point.

Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?
y = f(x) + 2 is a translation by the vector 0
2 (2 units up)
Use the rule for this type. The graph moves up (or down) by the number added to f(x). Only the y-coordinates change: add the number to y.
Apply the rule to each point. Work out the new coordinates.
A (1, 0) → (1, 2)
B (2, −1) → (2, 1)
C (3, 0) → (3, 2)
D (0, 3) → (0, 5)
Sketch the new graph. Join the new points with a smooth curve of the same shape (flipped or stretched if the transformation does that). The old graph is shown dashed for comparison.

Check. Check one point: B′ = (2, 1). The new equation y = x² − 4x + 5 gives y = 2² − 4 × 2 + 5 = 1 when x = 2. ✓

Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.

Worked example 25.4.2

The graph of y = f(x) is the same as in the previous example. Find the coordinates of the images of A, B, C and D on the graph of y = f(x − 3), and sketch it.

Method. y = f(x − 3) is a translation by the vector 3
0 (3 units to the right). Apply the rule to each key point.

Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?
y = f(x − 3) is a translation by the vector 3
0 (3 units to the right)
Use the rule for this type. Inside the bracket the change is the opposite way round to what you expect: y = f(x + a) moves the graph a units to the LEFT. Only the x-coordinates change: subtract a.
Apply the rule to each point. Work out the new coordinates.
A (1, 0) → (4, 0)
B (2, −1) → (5, −1)
C (3, 0) → (6, 0)
D (0, 3) → (3, 3)
Sketch the new graph. Join the new points with a smooth curve of the same shape (flipped or stretched if the transformation does that). The old graph is shown dashed for comparison.

Check. Check one point: B′ = (5, −1). The new equation y = x² − 10x + 24 gives y = 5² − 10 × 5 + 24 = −1 when x = 5. ✓

Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.

Worked example 25.4.3

The graph of y = f(x) = x³ − 3x² passes through A(0, 0), B(2, −4) and C(3, 0). Find the images of A, B and C on y = −f(x) and sketch it.

Method. y = −f(x) is a reflection in the x-axis. Apply the rule to each key point.

Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?
y = −f(x) is a reflection in the x-axis
Use the rule for this type. Minus f(x) turns the graph upside down (reflection in the x-axis). Only the y-coordinates change sign.
Apply the rule to each point. Work out the new coordinates.
A (0, 0) → (0, 0)
B (2, −4) → (2, 4)
C (3, 0) → (3, 0)
Sketch the new graph. Join the new points with a smooth curve of the same shape (flipped or stretched if the transformation does that). The old graph is shown dashed for comparison.

Check. Check one point: B′ = (2, 4). The new equation y = −x³ + 3x² gives y = −2³ + 3 × 2² = 4 when x = 2. ✓

Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.

Worked example 25.4.4

For the same graph y = f(x) = x³ − 3x², find the images of A, B and C on y = f(−x) and sketch it.

Method. y = f(−x) is a reflection in the y-axis. Apply the rule to each key point.

Identify the transformation. Look at what has been done to f(x): is it added outside the bracket, changed inside the bracket, or multiplied?
y = f(−x) is a reflection in the y-axis
Use the rule for this type. f(−x) flips the graph left to right (reflection in the y-axis). Only the x-coordinates change sign.
Apply the rule to each point. Work out the new coordinates.
A (0, 0) → (0, 0)
B (2, −4) → (−2, −4)
C (3, 0) → (−3, 0)
Sketch the new graph. Join the new points with a smooth curve of the same shape (flipped or stretched if the transformation does that). The old graph is shown dashed for comparison.

Check. Check one point: B′ = (−2, −4). The new equation y = −x³ − 3x² gives y = −(−2)³ − 3 × (−2)² = −4 when x = −2. ✓

Watch out. The effect of a change inside the bracket is the opposite of what you might expect: f(x + 2) moves the graph LEFT, and f(2x) makes it narrower.

Worked example 25.4.5

f(x) = x² − 4x. Find, in expanded form, the equation of the graph y = f(x + 2).

Method. Write the new function using f, replace f by its rule, then simplify.

Write the original function.
f(x) = x² − 4x
Say what the new function is in terms of f. The new graph is y = f(x + 2).
y = f(x + 2)
Put the rule in. Replace every x in the rule by (x + 2).
(x + 2)² − 4(x + 2)
Expand and simplify. Multiply out any brackets and collect like terms.
y = x² − 4

Check. Test with x = 3: the new rule gives 5; using the original rule with the changed input gives the same number. ✓

Watch out. Inside the bracket the effect is the opposite of what you expect: f(x + 2) moves the graph 2 to the left.

Worked example 25.4.6

Write y = x² + 6x + 11 in the form y = (x + p)² + q. Hence write down the coordinates of the turning point and describe the transformation from y = x².

Method. Complete the square. The completed-square form is a transformation of y = x²: a translation left or right, then up or down.

Halve the number in front of x. Completing the square: x² + bx = (x + b/2)² − (b/2)².
6 ÷ 2 = 3
Write the bracket and subtract the square of the half. (x + 3)² expands to x² + 6x + 9, which has 9 too many, so subtract it.
x² + 6x = (x + 3)² − 9
Add the constant term.
y = (x + 3)² + 2
Read the turning point. The graph of y = x² has its minimum at (0, 0). Replacing x by x + 3 moves it 3 left, and adding 2 moves it 2 up.
minimum at (−3, 2)

Check. Expand: (x + 3)² + 2 = x² + 6x + 11 = x² + 6x + 11. ✓

Watch out. The x-coordinate of the turning point is the opposite sign to the number inside the bracket.

Watch out

A change inside the bracket does the opposite of what you expect: y = f(x + 2) moves 2 units to the left, not the right, and y = f(2x) is narrower, not wider. Also, −f(x) flips the graph over (y changes sign) but f(−x) flips it sideways (x changes sign). Only the coordinate that is affected changes.

25.5

Trigonometric graphs

Edexcel 1MA1: A12, A13
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Trigonometric graphsComing soon

In Chapter 20 you used sin, cos and tan for angles in right-angled triangles, so the angles were between 0° and 90°. A calculator will also give sin, cos and tan of any angle, even 210° or −40°. When we plot these values against the angle we get three famous graphs. Always make sure your calculator is in degree mode.

The sine graph, y = sin x. It starts at 0, rises to a maximum of 1 at 90°, falls back to 0 at 180°, reaches a minimum of −1 at 270° and returns to 0 at 360°. After that the pattern repeats. The length of one complete pattern is the period, which is 360° for sine. All the values lie between −1 and 1.

The graph of y = sin x for 0° ≤ x ≤ 360°. Period 360°, maximum 1, minimum −1.

The cosine graph, y = cos x. It has the same wave shape but starts at its maximum: cos 0° = 1, then 0 at 90°, a minimum of −1 at 180°, 0 at 270° and back to 1 at 360°. The period is again 360°. The cosine graph is the sine graph moved 90° to the left: cos x = sin(x + 90°). This is exactly the transformation f(x + 90) from the last section.

The graph of y = cos x. Period 360°, maximum 1, minimum −1.

The tangent graph, y = tan x. This one looks very different. tan x = sin x ÷ cos x, so it is not defined where cos x = 0, that is at 90° and 270°. There the graph has vertical asymptotes (lines it gets closer and closer to but never touches). Between asymptotes the graph rises from very negative to very positive. It is 0 at 0°, 180° and 360°, and it equals 1 at 45°. Its period is 180°, and it has no maximum or minimum value.

The graph of y = tan x. The dashed lines are the asymptotes x = 90° and x = 270°. Period 180°.

It helps to know the key values. They are given to 2 decimal places here; some are exact (0, ±1, ±0.5).

x	0°	30°	60°	90°	120°	150°	180°	210°	240°	270°	300°	330°	360°
sin x	0.00	0.50	0.87	1.00	0.87	0.50	0.00	−0.50	−0.87	−1.00	−0.87	−0.50	0.00
cos x	1.00	0.87	0.50	0.00	−0.50	−0.87	−1.00	−0.87	−0.50	0.00	0.50	0.87	1.00
tan x	0.00	0.58	1.73	–	−1.73	−0.58	0.00	0.58	1.73	–	−1.73	−0.58	0.00

Using symmetry. Look at the sine graph: the hump between 0° and 180° is symmetrical about 90°, so sin 150° = sin 30° = 0.5. The part between 180° and 360° is the same hump turned upside down, so sin 210° = −sin 30° = −0.5. For any angle, use the graph to decide whether the value is positive or negative, and use the related acute angle for the size.

Solving equations such as sin x = 0.3. The solutions are where the horizontal line y = 0.3 crosses the graph. The calculator button gives only the first solution, 17.5°. Use the symmetry of the graph for the others: for sin, the second solution is 180° − the first; for cos it is 360° − the first; for tan, add 180°. Then add or subtract 360° (180° for tan) to find solutions in the required range.

Transformations. The ideas of the last section work on these graphs too. y = 2 sin x has the same period but amplitude 2 (maximum 2, minimum −2). y = sin 2x completes a wave in 180° instead of 360°, so its period is 360° ÷ 2 = 180°. y = sin x + 1 is moved up 1, and y = −sin x is turned upside down.

Left: y = 2 sin x (blue) is twice as tall as y = sin x (grey). Right: y = sin 2x (blue) has a period of 180°, so it fits two waves into 360°.

Key ideas

sin x and cos x: period 360°, maximum 1, minimum −1. cos x starts at 1; sin x starts at 0.
tan x: period 180°, asymptotes at 90° and 270°, passes through (45°, 1), no maximum or minimum.
Use symmetry: sin 150° = sin 30°, cos 120° = −cos 60°, tan 135° = −tan 45°. Decide the sign from the graph.
To solve sin x = k: first solution from the calculator, second solution 180° − first (cos: 360° − first; tan: first + 180°), then add or subtract full periods to reach the range.
y = a sin x: amplitude a. y = sin bx: period 360°/b. y = sin x + d: shift up d.

Worked example 25.5.1

Given that sin 30° = 0.5, write down the value of sin 210°. Do not use a calculator.

Method. Use the symmetry of the sin graph: the size of the value is the same as for the related acute angle, and the sign comes from the quarter.

Find how 210° is related to 30°. Use the cycle of 360°: angles in the first quarter are 0° to 90°, second 90° to 180°, third 180° to 270°, fourth 270° to 360°.
210° = 180° + 30° (a third-quarter angle)
Decide whether the value is positive or negative. Look at the graph of sin x: in the third quarter the curve is below the x-axis.
sin 210° is negative.
Use the same size with the correct sign. The graph is a mirror image (or turned image) of the first quarter, so the size of the value is the same.
sin 210° = −0.5 = −0.5

Check. On a calculator, sin 210° = −0.500, which matches −0.5. ✓

Watch out. Do not forget the sign. Check whether the curve is above or below the axis at that angle.

Worked example 25.5.2

Solve sin x = 0.3 for 0° ≤ x ≤ 360°. Give your answers to 1 decimal place.

Key words. The period of a graph is the length of one complete cycle before it repeats. sin and cos repeat every 360°; tan repeats every 180°.

Method. Find the first solution with the inverse function, then use the symmetry and the repeat pattern of the sin graph to find the rest in the range.

Write the equation and picture the graph. The solutions of sin x = 0.3 are the x-values where the horizontal line y = 0.3 crosses the graph of y = sin x. Between 0° and 360° it crosses 2 times.
Use the inverse button on the calculator for the first (principal) solution. The calculator gives one answer, which may not be in the range we want.
sin⁻¹(0.3) = 17.5°
Use the symmetry of the graph to find the other solutions. The sine graph is symmetrical about x = 90° (and x = 270°). So the second solution in the same cycle is 180° − the first.
180 − 17.5 = 162.5
Write the solutions. There are 2 solutions in the range 0° ≤ x ≤ 360°.
17.5, 162.5

Check. Put each answer into the calculator: sin 17.5° = 0.30; sin 162.5° = 0.30 (rounding to 2 d.p. should give about 0.30). ✓

Watch out. The calculator only gives one answer. Always ask yourself where the graph crosses the line again.

Worked example 25.5.3

Solve cos x = −0.4 for 0° ≤ x ≤ 360°. Give your answers to 1 decimal place.

Key words. The period of a graph is the length of one complete cycle before it repeats. sin and cos repeat every 360°; tan repeats every 180°.

Method. Find the first solution with the inverse function, then use the symmetry and the repeat pattern of the cos graph to find the rest in the range.

Write the equation and picture the graph. The solutions of cos x = −0.4 are the x-values where the horizontal line y = −0.4 crosses the graph of y = cos x. Between 0° and 360° it crosses 2 times.
Use the inverse button on the calculator for the first (principal) solution. The calculator gives one answer, which may not be in the range we want.
cos⁻¹(−0.4) = 113.6°
Use the symmetry of the graph to find the other solutions. The cosine graph is symmetrical about x = 0° and x = 180°. So the second solution is the negative of the first, then add 360° to bring it into the range.
−113.6 + 360 = 246.4
Add or subtract whole cycles until every answer is in the range. The graph repeats every 360°. Keep only answers from 0° to 360°.
113.6, 246.4
Write the solutions. There are 2 solutions in the range 0° ≤ x ≤ 360°.
113.6, 246.4

Check. Put each answer into the calculator: cos 113.6° = −0.40; cos 246.4° = −0.40 (rounding to 2 d.p. should give about −0.40). ✓

Watch out. The calculator only gives one answer. Always ask yourself where the graph crosses the line again.

Worked example 25.5.4

Solve tan x = 2 for 0° ≤ x ≤ 360°. Give your answers to 1 decimal place.

Key words. The period of a graph is the length of one complete cycle before it repeats. sin and cos repeat every 360°; tan repeats every 180°.

Method. Find the first solution with the inverse function, then use the symmetry and the repeat pattern of the tan graph to find the rest in the range.

Write the equation and picture the graph. The solutions of tan x = 2 are the x-values where the horizontal line y = 2 crosses the graph of y = tan x. Between 0° and 360° it crosses 2 times.
Use the inverse button on the calculator for the first (principal) solution. The calculator gives one answer, which may not be in the range we want.
tan⁻¹(2) = 63.4°
Use the symmetry of the graph to find the other solutions. The tangent graph repeats every 180°, so the other solutions are found by adding (or subtracting) 180°.
63.4 + 180 = 243.4
Add or subtract whole cycles until every answer is in the range. The graph repeats every 180°. Keep only answers from 0° to 360°.
63.4, 243.4
Write the solutions. There are 2 solutions in the range 0° ≤ x ≤ 360°.
63.4, 243.4

Check. Put each answer into the calculator: tan 63.4° = 2.00; tan 243.4° = 2.00 (rounding to 2 d.p. should give about 2.00). ✓

Watch out. The calculator only gives one answer. Always ask yourself where the graph crosses the line again.

Worked example 25.5.5

The graph shows y = a sin(bx). Find the values of a and b and write down the equation.

Method. Read the amplitude, the period and the middle line from the graph, then match them to y = a sin(bx) + d (or cos).

Read the maximum and minimum values. The graph oscillates between a highest and lowest value.
maximum = 2, minimum = −2
Find the amplitude and the middle line. The amplitude is half the distance between the maximum and the minimum. The middle line is halfway between them.
amplitude = (2 − (−2)) ÷ 2 = 2
middle line: y = (2 + (−2)) ÷ 2 = 0
Read the period (the length of one full cycle). The standard sin graph has period 360°. If one cycle takes 120° then the angle is multiplied by 3.
period = 120° ⇒ b = 360 ÷ 120 = 3
Write the equation. Use y = a sin (bx) + d with a = 2, b = 3, d = 0.
y = 2 sin 3x

Check. Test one point: when x = 0, the equation gives 0.00, and the graph passes through that value. ✓

Watch out. The period is the length of one whole cycle, not half of it.

Watch out

The calculator gives only one answer to sin x = 0.3, but there are two in each 360°. Use the graph (or symmetry) to find the others. Also check that the calculator is in degree mode, and for equations like 2 sin x = 1, divide by the 2 first to get sin x = 0.5.

Exercise 25E Trigonometric graphs

Calculator allowed

FUse your calculator (in degree mode) to complete the table of values for y = cos x. Give each value to 1 decimal place.

x	0°	60°	90°	120°	180°	270°	360°
cos x

Method. Type cos of each angle into the calculator, with the calculator set to degrees, and round to 1 decimal place.

Check the calculator mode. It must show D or DEG. In radian mode all the answers would be wrong.
Work out each value. Press cos, then the angle, then equals.
cos 0° = 1.000 ≈ 1.0
cos 60° = 0.500 ≈ 0.5
cos 90° = 0.000 ≈ 0.0
cos 120° = −0.500 ≈ −0.5
cos 180° = −1.000 ≈ −1.0
cos 270° = 0.000 ≈ 0.0
cos 360° = 1.000 ≈ 1.0
Write them in the table.
x 0° 60° 90° 120° 180° 270° 360°
cos x 1.0 0.5 0.0 −0.5 −1.0 0.0 1.0

Check. The values match the shape of the cosine graph: 1, then down to −1 at 180°, then back up to 1. ✓

Watch out. cos 90° and cos 270° are exactly 0. A calculator may show a tiny number like 6 × 10⁻¹⁷; that is a rounding error.

Done

FFor each graph, write down the maximum value, the minimum value and the period.

(a)y = sin x

(b)y = cos x

(c)y = tan x

Done

FGiven that sin 30° = 0.5, write down the value of

(a)sin 150°

(b)sin 210°

(c)sin 330°

Done

FGiven that cos 60° = 0.5, write down the value of

(a)cos 120°

(b)cos 240°

(c)cos 300°

Done

FGiven that tan 45° = 1, write down the value of

(a)tan 135°

(b)tan 225°

(c)tan 315°

Done

FSolve each equation for 0° ≤ x ≤ 360°. Give your answers to 1 decimal place.

(a)sin x = 0.5

(b)sin x = 0.8

(c)sin x = −0.5

(d)sin x = −0.9

Done

FSolve each equation for 0° ≤ x ≤ 360°. Give your answers to 1 decimal place.

(a)cos x = 0.5

(b)cos x = −0.5

(c)cos x = 0.2

(d)cos x = −0.7

Done

FSolve each equation for 0° ≤ x ≤ 360°. Give your answers to 1 decimal place.

(a)tan x = 1

(b)tan x = −1

(c)tan x = 3

(d)tan x = −0.5

Done

PSSolve each equation for −180° ≤ x ≤ 180°. Give your answers to 1 decimal place.

(a)sin x = 0.5

(b)cos x = 0.5

(c)tan x = 1

(d)sin x = −0.4

Done

PSSolve sin x = 0.3 for 0° ≤ x ≤ 720°. Give your answers to 1 decimal place. How many solutions are there?

Done

PSEach graph is of the form y = a sin(bx) + d or y = a cos(bx) + d. Write down the equation of each graph.

(a)

(b)

(c)

Done

PSFor each graph, write down the maximum, the minimum, the period and the y-intercept.

(a)y = 2 cos x

(b)y = sin 3x + 1

(c)y = 4 cos 2x − 1

Done

PSThe depth of water in a harbour, in metres, is modelled by h = 5 + 3 sin(30t)°, where t is the time in hours after midnight, for 0 ≤ t ≤ 12.

(a)Work out the depth at 3 a.m. and at 9 a.m.

(b)Write down the greatest and the least depth.

(c)At what times is the depth 6.5 m?

Done

RThe graph of y = cos(x − 90°) is a translation of the graph of y = cos x. Describe the translation and show that cos(x − 90°) = sin x by completing a table for x = 0°, 30°, 90° and 150°.

Method. Use the translation rule f(x − 90) and compare values from a table.

Describe the translation. Subtracting 90 inside the bracket moves the graph 90 to the right.
translation by the vector (90°, 0)
Work out the values. For each x, find x − 90 then take the cosine.
x x − 90° cos(x − 90°) sin x
0° -90° 0.00 0.00
30° -60° 0.50 0.50
90° 0° 1.00 1.00
150° 60° 0.50 0.50
Compare. The two columns are equal in every row.
So cos(x − 90°) = sin x

x	x − 90°	cos(x − 90°)	sin x
0°	-90°	0.00	0.00
30°	-60°	0.50	0.50
90°	0°	1.00	1.00
150°	60°	0.50	0.50

Check. The cosine graph starts at its maximum at x = 0. Moving it 90° right puts the maximum at 90°, which is where sin x has its maximum. ✓

Watch out. A translation to the right means subtracting inside the bracket.

Done

RExplain why the equation sin x = 2 has no solutions, but the equation tan x = 2000 does have solutions.

Done

PSSolve 2 sin x + 1 = 0 for 0° ≤ x ≤ 360°.

Done

PSSolve 2 cos x + 1 = 0.2 for 0° ≤ x ≤ 360°. Give your answers to 1 decimal place.

Done

RThe graphs of y = sin x (blue) and y = cos x (orange) are shown. They cross at two points between 0° and 360°. Show that at these points tan x = 1, and find the two values of x.

Done

25.6

Exponential graphs, growth and decay

Edexcel 1MA1: A12, R16
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Exponential graphs, growth and decayComing soon

Imagine a rumour. On day 0 one person knows it. Each day every person who knows it tells one more person, so the number doubles: 1, 2, 4, 8, 16, … The number after x days is 2^x. A function with the unknown x in the power is called an exponential function. The general form is y = k^x, where k is a positive number called the base.

To draw the graph, make a table of values. For y = 2^x: 2⁰ = 1 (any number to the power 0 is 1), 2¹ = 2, 2² = 4, 2³ = 8, and for negative powers 2⁻¹ = 1/2, 2⁻² = 1/4. Every graph of y = k^x has the same key features:

it passes through (0, 1), because k⁰ = 1;
it is always above the x-axis (y is never zero or negative), and the x-axis is an asymptote;
if k > 1 the graph rises faster and faster: this is exponential growth;
if 0 < k < 1 the graph falls, steeply at first and then more slowly towards the axis: this is exponential decay.

Growth (k > 1) on the left: blue is 2^x and orange is 3^x. Decay (0 < k < 1) on the right: blue is (1/2)^x and orange is (1/3)^x. All pass through (0, 1) and never touch the x-axis.

Growth and decay problems. Exponential functions model real situations where something changes by the same percentage in every period: compound interest, a growing population, a falling car value, a medicine leaving the body. You met the idea in Chapter 5. The method is:

Find the multiplier for one period: for an increase of r% it is 1 + r/100; for a decrease it is 1 − r/100.
The amount after n periods is starting amount × (multiplier)ⁿ.
Use a calculator for the power. Keep all the digits until the final rounding.

To find when an amount reaches a target, build a table of values period by period (or use trial and improvement) until you pass the target. For a quantity that halves every T units of time (a half-life) the amount is start × (1/2)^{time ÷ T}.

If a graph passes through (0, a), then y = a × k^x: the a is the starting value. A second point gives k.

Key ideas

y = k^x passes through (0, 1), is always positive, and has the x-axis as an asymptote. k > 1: growth. 0 < k < 1: decay.
k⁰ = 1 and k⁻ⁿ = 1/kⁿ.
Multiplier: increase r% = 1 + r/100; decrease r% = 1 − r/100. After n periods: start × multiplierⁿ.
To find a time, use a table or trial and improvement.
Through (0, a): y = a × k^x. Use a second point to find k.

Worked example 25.6.1

Complete the table of values for y = 2^x for x from −2 to 4, and describe the shape of the graph.

Key words. In 2^x the number 2 is the base and x is the power (or index). A function with x in the power is an exponential function.

Method. Substitute each value of x into y = 2^x and work out the power.

Remember what a power means. 2^x means 2 multiplied by itself x times. A power of 0 gives 1. A negative power means 1 divided by the positive power.
Work out each value. Take the powers one at a time.
x = −2: 2⁻² = 1 ÷ 2² = 1 ÷ 4 = 0.25
x = −1: 2⁻¹ = 1 ÷ 2¹ = 1 ÷ 2 = 0.5
x = 0: 2⁰ = 1 (any number to the power 0 is 1)
x = 1: 2¹ = 2
x = 2: 2² = 2 × 2 = 4
x = 3: 2³ = 2 × 2 × 2 = 8
x = 4: 2⁴ = 2 × 2 × 2 × 2 = 16
Write the values in the table.
x −2 −1 0 1 2 3 4
2^x 0.25 0.5 1 2 4 8 16

x	−2	−1	0	1	2	3	4
2^x	0.25	0.5	1	2	4	8	16

Check. Each value in the table is 2 times the one before it (reading left to right). ✓

Watch out. A negative power does not make the answer negative: 2⁻³ is 1/8, not −8.

Worked example 25.6.2

£3000 is invested at 4% per year compound interest. How much is it worth after 5 years?

Key words. The multiplier is the number you multiply by to apply a percentage change in one step. Compound means each change is applied to the new amount, not the original.

Method. Growth and decay problems use a multiplier. Find the multiplier, raise it to the power of the number of years, and multiply by the starting amount.

Find the growth multiplier. An increase of 4% means we add 4% of the amount. So we are left with 100% + 4% = 104% of the amount, and as a decimal that is the multiplier.
104% = 104 ÷ 100 = 1.04
Write the formula. The amount is multiplied by 1.04 once for each year. After n years we multiply n times, which is the multiplier to the power n.
V = (starting amount) × (multiplier)ⁿ
Put in the numbers: start = £3000, multiplier = 1.04, n = 5. Replace each letter by its value.
V = £3000 × 1.04⁵
Work it out on a calculator. Do the power 1.04⁵ first, then multiply by 3000. Keep all the digits until the final rounding.
1.04⁵ = 1.21665
V = £3000 × 1.21665 = 3649.96
Round sensibly. Money is rounded to the nearest penny (2 decimal places).
V = £3649.96

Check. Check the direction: the amount increases, so the answer should be bigger than £3000. It is £3649.96. ✓

Watch out. Do not work out the percentage once and multiply by the number of years. That gives simple growth. Compound growth multiplies by the multiplier every period, so use a power.

Worked example 25.6.3

A car is worth £14 000 when new. Its value falls by 15% each year. Work out its value after 3 years.

Key words. The multiplier is the number you multiply by to apply a percentage change in one step. Compound means each change is applied to the new amount, not the original.

Method. Growth and decay problems use a multiplier. Find the multiplier, raise it to the power of the number of years, and multiply by the starting amount.

Find the decay multiplier. A decrease of 15% means we take away 15% of the amount. So we are left with 100% − 15% = 85% of the amount, and as a decimal that is the multiplier.
85% = 85 ÷ 100 = 0.85
Write the formula. The amount is multiplied by 0.85 once for each year. After n years we multiply n times, which is the multiplier to the power n.
V = (starting amount) × (multiplier)ⁿ
Put in the numbers: start = £14000, multiplier = 0.85, n = 3. Replace each letter by its value.
V = £14000 × 0.85³
Work it out on a calculator. Do the power 0.85³ first, then multiply by 14000. Keep all the digits until the final rounding.
0.85³ = 0.614125
V = £14000 × 0.614125 = 8597.75
Round sensibly. Money is rounded to the nearest penny (2 decimal places).
V = £8597.75

Check. Check the direction: the amount decreases, so the answer should be smaller than £14000. It is £8597.75. ✓

Watch out. Do not work out the percentage once and multiply by the number of years. That gives simple growth. Compound growth multiplies by the multiplier every period, so use a power.

Worked example 25.6.4

£800 is invested at 3% compound interest per year. After how many whole years will it first be worth at least £1000?

Method. Use the multiplier repeatedly, building a table, until the amount reaches the target.

Find the multiplier. Each year the amount is multiplied by the same number.
103% = 1.03
Work out the amount at the end of each year. Multiply by 1.03 each time. Stop as soon as the amount reaches or passes £1000.
After n years Amount
0 £800.00
1 £824.00
2 £848.72
3 £874.18
4 £900.41
5 £927.42
6 £955.24
7 £983.90
8 £1013.42
Read off the answer. The first row that reaches or passes £1000 is after 8 years.
n = 8

After n years	Amount
0	£800.00
1	£824.00
2	£848.72
3	£874.18
4	£900.41
5	£927.42
6	£955.24
7	£983.90
8	£1013.42

Check. Use the formula: £800 × 1.03⁸ = £1013.42 (above £1000) but after 7 years it is £983.90 (not yet there). ✓

Watch out. Count the years carefully. The row marked 0 is the start, before any year has passed.

Worked example 25.6.5

The graph of y = a × k^x passes through (0, 3) and (4, 48). Find a and k.

Method. Use the point on the y-axis to find a, then use the second point to find k.

Write the general form. An exponential graph through the given points has the form y = a × k^x, where a is the starting value and k is the multiplier.
y = a × k^x
Use the point with x = 0 to find a. Any number to the power 0 is 1, so when x = 0 we get y = a × 1 = a.
(0, 3) ⇒ a = 3
Use the point (4, 48) to find k. Substitute x and y and then divide by a.
48 = 3 × k⁴
k⁴ = 48 ÷ 3 = 16
Find the number that gives this value when raised to the power. Ask: which number multiplied by itself 4 times gives 16?
2⁴ = 2 × 2 × 2 × 2 = 16, so k = 2
Write the equation.
y = 3 × 2^x

Check. Test the second point: x = 4 gives 3 × 2⁴ = 3 × 16 = 48. ✓

Watch out. Divide by a before looking for the power. Do not square-root or cube-root until you have kⁿ alone.

Watch out

A 20% increase followed by a 20% decrease does not return you to the start. Do not add the percentage on every year: multiply by the multiplier every year. And remember a negative power does not make a negative answer: 2⁻³ = 1/8.

Exercise 25F Exponential graphs, growth and decay

Calculator allowed

FComplete a table of values for each function and describe what happens to y as x increases.

(a)y = 2^x, for x = -2, 0, 2, 4

Key words. In 2^x the number 2 is the base and x is the power (or index). A function with x in the power is an exponential function.

Method. Substitute each value of x into y = 2^x and work out the power.

Remember what a power means. 2^x means 2 multiplied by itself x times. A power of 0 gives 1. A negative power means 1 divided by the positive power.
Work out each value. Take the powers one at a time.
x = −2: 2⁻² = 1 ÷ 2² = 1 ÷ 4 = 0.25
x = 0: 2⁰ = 1 (any number to the power 0 is 1)
x = 2: 2² = 2 × 2 = 4
x = 4: 2⁴ = 2 × 2 × 2 × 2 = 16
Write the values in the table.
x −2 0 2 4
2^x 0.25 1 4 16

x	−2	0	2	4
2^x	0.25	1	4	16

Check. Each value in the table is 2 times the one before it (reading left to right). ✓

Watch out. A negative power does not make the answer negative: 2⁻³ is 1/8, not −8.

(b)y = 3^x, for x = -2, -1, 0, 1, 2

Key words. In 2^x the number 2 is the base and x is the power (or index). A function with x in the power is an exponential function.

Method. Substitute each value of x into y = 3^x and work out the power.

Remember what a power means. 3^x means 3 multiplied by itself x times. A power of 0 gives 1. A negative power means 1 divided by the positive power.
Work out each value. Take the powers one at a time.
x = −2: 3⁻² = 1 ÷ 3² = 1 ÷ 9 = 19
x = −1: 3⁻¹ = 1 ÷ 3¹ = 1 ÷ 3 = 13
x = 0: 3⁰ = 1 (any number to the power 0 is 1)
x = 1: 3¹ = 3
x = 2: 3² = 3 × 3 = 9
Write the values in the table.
x −2 −1 0 1 2
3^x 19 13 1 3 9

x	−2	−1	0	1	2
3^x	19	13	1	3	9

Check. Each value in the table is 3 times the one before it (reading left to right). ✓

Watch out. A negative power does not make the answer negative: 2⁻³ is 1/8, not −8.

(c)y = 10^x, for x = -2, -1, 0, 1, 2

Key words. In 2^x the number 2 is the base and x is the power (or index). A function with x in the power is an exponential function.

Method. Substitute each value of x into y = 10^x and work out the power.

Remember what a power means. 10^x means 10 multiplied by itself x times. A power of 0 gives 1. A negative power means 1 divided by the positive power.
Work out each value. Take the powers one at a time.
x = −2: 10⁻² = 1 ÷ 10² = 1 ÷ 100 = 0.01
x = −1: 10⁻¹ = 1 ÷ 10¹ = 1 ÷ 10 = 0.1
x = 0: 10⁰ = 1 (any number to the power 0 is 1)
x = 1: 10¹ = 10
x = 2: 10² = 10 × 10 = 100
Write the values in the table.
x −2 −1 0 1 2
10^x 0.01 0.1 1 10 100

x	−2	−1	0	1	2
10^x	0.01	0.1	1	10	100

Check. Each value in the table is 10 times the one before it (reading left to right). ✓

Watch out. A negative power does not make the answer negative: 2⁻³ is 1/8, not −8.

(d)y = 0.5^x, for x = -3, -1, 0, 2, 4

Key words. In 2^x the number 2 is the base and x is the power (or index). A function with x in the power is an exponential function.

Method. Substitute each value of x into y = 0.5^x and work out the power.

Remember what a power means. 0.5^x means 0.5 multiplied by itself x times. A power of 0 gives 1. A negative power means 1 divided by the positive power.
Work out each value. Take the powers one at a time.
x = −3: 0.5⁻³ = 1 ÷ 0.5³ = 1 ÷ 0.125 = 8
x = −1: 0.5⁻¹ = 1 ÷ 0.5¹ = 1 ÷ 0.5 = 2
x = 0: 0.5⁰ = 1 (any number to the power 0 is 1)
x = 2: 0.5² = 0.5 × 0.5 = 0.25
x = 4: 0.5⁴ = 0.5 × 0.5 × 0.5 × 0.5 = 0.0625
Write the values in the table.
x −3 −1 0 2 4
0.5^x 8 2 1 0.25 0.0625

x	−3	−1	0	2	4
0.5^x	8	2	1	0.25	0.0625

Check. Moving one step to the right multiplies the value by 0.5. ✓

Watch out. A negative power does not make the answer negative: 2⁻³ is 1/8, not −8.

Done

FThe graph of y = 2^x is shown.

(a)Use the graph to estimate the value of 2^2.5.

(b)Use the graph to estimate the value of x when 2^x = 10.

Done

FThe diagram shows the graphs of y = 2^x, y = 3^x and y = (1/2)^x. Match each equation to a graph A, B or C.

(a)y = 2^x

(b)y = 3^x

(c)y = (1/2)^x

Done

FIs each function an example of exponential growth or exponential decay?

(a)y = 1.05^x

(b)y = 0.9^x

(c)y = 7^x

(d)y = 0.75^x

(e)y = 0.99^x

Done

PSSolve each equation by trying whole-number powers.

(a)2^x = 32

(b)3^x = 81

(d)0.5^x = 64

Done

PS£3000 is invested at 4% compound interest per year. How much is it worth after 6 years? Give your answer to the nearest penny.

Done

PSThe population of a town is 24 000 and grows by 2.5% each year. Estimate the population after 8 years. Give your answer to the nearest hundred.

Done

PSA motorbike costs £9500 new and loses 12% of its value each year. Work out its value after 4 years to the nearest £10.

Done

PSA sample of a radioactive substance has a mass of 500 g. Its mass falls by 6% each day. Work out the mass after 10 days. Give your answer to 3 significant figures.

Done

PS£800 is invested at 3% compound interest per year. How many whole years will it take to reach at least £1000?

Method. Use the multiplier repeatedly, building a table, until the amount reaches the target.

Find the multiplier. Each year the amount is multiplied by the same number.
103% = 1.03
Work out the amount at the end of each year. Multiply by 1.03 each time. Stop as soon as the amount reaches or passes £1000.
After n years Amount
0 £800.00
1 £824.00
2 £848.72
3 £874.18
4 £900.41
5 £927.42
6 £955.24
7 £983.90
8 £1013.42
Read off the answer. The first row that reaches or passes £1000 is after 8 years.
n = 8

After n years	Amount
0	£800.00
1	£824.00
2	£848.72
3	£874.18
4	£900.41
5	£927.42
6	£955.24
7	£983.90
8	£1013.42

Check. Use the formula: £800 × 1.03⁸ = £1013.42 (above £1000) but after 7 years it is £983.90 (not yet there). ✓

Watch out. Count the years carefully. The row marked 0 is the start, before any year has passed.

Done

PSA patient takes 200 mg of a drug. Each hour, 20% of the drug in the body is removed. After how many hours is there first less than 50 mg left in the body?

Method. Use the multiplier repeatedly, building a table, until the amount falls to the target.

Find the multiplier. Each hour the amount is multiplied by the same number.
80% = 0.8
Work out the amount at the end of each hour. Multiply by 0.8 each time. Stop as soon as the amount falls to or below 50 mg.
After n hours Amount
0 200.0 mg
1 160.0 mg
2 128.0 mg
3 102.4 mg
4 81.9 mg
5 65.5 mg
6 52.4 mg
7 41.9 mg
Read off the answer. The first row that is at or below 50 mg is after 7 hours.
n = 7

After n hours	Amount
0	200.0 mg
1	160.0 mg
2	128.0 mg
3	102.4 mg
4	81.9 mg
5	65.5 mg
6	52.4 mg
7	41.9 mg

Check. Use the formula: 200 × 0.8⁷ = 41.9 mg (below 50) but after 6 hours it is 52.4 mg (not yet there). ✓

Watch out. Count the hours carefully. The row marked 0 is the start, before any hour has passed.

Done

PSThe mass of a sample is M = 80 × (1/2)^t/5 grams after t days, so the mass halves every 5 days. Work out the mass after

(a)5 days

(b)15 days

Done

PSEach graph has the form y = a × k^x and passes through the two points given. Find a and k.

(a)(0, 3) and (4, 48)

(b)(0, 5) and (3, 135)

(c)(0, 2) and (2, 50)

Done

PS£1000 is invested at a fixed annual compound interest rate. After 3 years it is worth £1331. Find the annual percentage rate.

Done

RJo is offered two savings plans for £100. Plan A adds £10 at the end of each year. Plan B adds 10% of the current balance at the end of each year. Which plan is worth more after 20 years, and by how much? Explain why the difference grows.

Method. Work out both balances after 20 years using a formula, and compare.

Plan A adds the same amount each year. This is a straight-line pattern: balance = 100 + 10 × years.
After 20 years: 100 + 10 × 20 = £300
Plan B multiplies by 1.1 each year. This is exponential growth: balance = 100 × 1.1^years.
After 20 years: 100 × 1.1²⁰ = £672.75
Compare.
£672.75 − £300 = £372.75
Look at a table to see why the gap grows. In plan B the interest is paid on the interest as well, so each year’s increase is bigger than the last.
years Plan A Plan B
0 £100 £100.00
1 £110 £110.00
2 £120 £121.00
5 £150 £161.05
10 £200 £259.37
20 £300 £672.75

years	Plan A	Plan B
0	£100	£100.00
1	£110	£110.00
2	£120	£121.00
5	£150	£161.05
10	£200	£259.37
20	£300	£672.75

Check. Plan B’s increase in year 20 is about £61.16, much more than Plan A’s £10. ✓

Watch out. Linear growth adds the same amount each time; exponential growth multiplies, so its increases keep getting bigger.

Done

RFor the graph of y = 2^x, work out the average rate of change (gradient of the chord) from x = 0 to x = 1 and from x = 3 to x = 4. What does this tell you about the graph?

Done

RMia says: “If I keep going far enough left, the graph of y = 2^x will eventually touch the x-axis.” Is she right? Use values of 2^x to explain.

Method. Work out the value of 2^x for more and more negative x.

Work out some values. A negative power is a reciprocal.
x 2^x
−1 1/2 = 0.5
−2 1/4 = 0.25
−5 1/32 = 0.03125
−10 1/1024 = 0.00097656
−20 1/1048576 = 0.00000095
Think about 1 divided by a huge number. It gets very close to 0 but is never exactly 0, and never negative.
So y > 0 for every x.
Conclude. The x-axis is an asymptote: the curve gets as close as you like but never touches it.
Mia is wrong.

x	2^x
−1	1/2 = 0.5
−2	1/4 = 0.25
−5	1/32 = 0.03125
−10	1/1024 = 0.00097656
−20	1/1048576 = 0.00000095

Check. If 2^x = 0 then no power of 2 could ever give 0, because 2 × 2 × … is never 0. ✓

Watch out. “Asymptote” means very close but never touching.

Done

RA graph with equation y = a × b^x (where b > 0) passes through (1, 6) and (3, 24). Find the values of a and b.

Done

25.7

The equation of a circle and tangents

Edexcel 1MA1: A16
Grade 8-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The equation of a circle and tangentsComing soon

Draw a circle with its centre at the origin (0, 0) and radius r. Pick any point (x, y) on the circle. The radius to that point, the horizontal distance x and the vertical distance y form a right-angled triangle. By Pythagoras’ theorem (Chapter 20), x² + y² = r². This is true for every point on the circle, and false for every other point. So:

The circle with centre (0, 0) and radius r has equation x² + y² = r².

The circle x² + y² = 25 has radius 5. The point P(3, 4) is on it because 3² + 4² = 9 + 16 = 25. The 3-4-5 triangle shows Pythagoras at work.

Reading the equation. The number on the right is r², so you must take the square root to get the radius. x² + y² = 49 is a circle of radius 7. If the equation is x² + y² = 20 the radius is √20 = 2√5 (Chapter 6).

Is a point on the circle? Work out x² + y² for the point and compare with r². If they are equal the point is on the circle; if x² + y² is smaller the point is inside; if bigger it is outside. If you know one coordinate you can find the other: put it in the equation and solve. There are usually two answers, one above and one below the axis (or left and right).

Tangents. A tangent to a circle is a straight line that touches the circle at exactly one point. A key fact (proved in Chapter 26) is that the tangent is perpendicular to the radius at the point of contact. So to find the equation of the tangent at a point (a, b) on the circle:

find the gradient of the radius from (0, 0) to (a, b): b ÷ a;
the tangent gradient is the negative reciprocal: −a ÷ b (Chapter 12);
use y − b = m(x − a) and rearrange to y = mx + c.

The tangent (red) touches the circle at (3, 4) and meets the radius (purple) at a right angle. The radius has gradient 4/3, so the tangent has gradient −3/4.

Key ideas

Circle, centre (0, 0), radius r: x² + y² = r². The right-hand side is r², not r.
A point is on the circle if x² + y² = r², inside if it is less, outside if it is more.
A tangent touches the circle at one point and is perpendicular to the radius there.
Tangent at (a, b): gradient of radius = b/a; gradient of tangent = −a/b; then y − b = m(x − a).
The tangent at (a, b) can also be written ax + by = r².

Worked example 25.7.1

Write down the equation of the circle with centre (0, 0) and radius 7.

Key words. The radius is the distance from the centre to the edge. A point is on a circle if its coordinates make the equation true.

Method. A circle with centre (0, 0) and radius r has equation x² + y² = r².

Remember the equation of a circle with its centre at the origin. Pick any point (x, y) on the circle. The radius to that point, the horizontal distance x and the vertical distance y make a right-angled triangle. Pythagoras gives x² + y² = (radius)².
x² + y² = r²
Put in the radius. Here the radius is r = 7.
r² = 7² = 49
Write the equation. Keep x² + y² on the left.
x² + y² = 49

Check. Test the point (7, 0), which is on the circle: 7² + 0² = 49. ✓

Watch out. The right-hand side is r², not r. A circle of radius 5 has equation x² + y² = 25, not = 5.

Worked example 25.7.2

The point (3, k) is on the circle x² + y² = 25. Find the possible values of k.

Method. Substitute the known coordinate into x² + y² = r², then solve for the other. There are usually two answers.

Substitute the known coordinate into the circle equation. The point is on the circle, so its coordinates satisfy the equation.
x = 3: (3)² + y² = 25
Work out the square and rearrange for y². Subtract the known square from both sides.
9 + y² = 25
y² = 25 − 9 = 16
Square root both sides. Remember there are two roots: positive and negative. This is because there are two points on the circle with this x-coordinate, one on each side.
y = ±√16 = ±4

Check. Check one: (3)² + (4)² = 9 + 16 = 25. ✓

Watch out. A square root has two answers, positive and negative.

Worked example 25.7.3

Is the point (2, 5) inside, outside or on the circle x² + y² = 25?

Method. Work out x² + y² for the point and compare it with r².

Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
(2)² + (5)² = 4 + 25 = 29
Compare with r². The circle has r² = 25. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
29 > 25
Write the conclusion.
The point (2, 5) is outside the circle.

Check. The distance from the origin is √29 ≈ 5.39 and the radius is 5.00: bigger, so outside. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Worked example 25.7.4

Find the equation of the tangent to the circle x² + y² = 25 at the point (3, 4).

Key words. A tangent is a straight line that touches a circle at exactly one point. It meets the radius at that point at a right angle.

Method. The tangent is perpendicular to the radius. Find the gradient of the radius, flip it and change the sign, then use the point to find the equation.

Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
(3)² + (4)² = 9 + 16 = 25 ✓
Find the gradient of the radius. The radius joins the centre (0, 0) to the point. Gradient = change in y ÷ change in x.
gradient of radius = (4 − 0) ÷ (3 − 0) = 43
Find the gradient of the tangent. A tangent is perpendicular to the radius at the point of contact (Chapter 26). For perpendicular lines the gradients multiply to give −1: flip the fraction and change the sign.
gradient of tangent = −1 ÷ 43 = −34
Use y − y₁ = m(x − x₁) with the point. The tangent has gradient m and passes through (a, b).
y − 4 = −34(x − 3)
Expand and rearrange to y = mx + c. Multiply out the bracket: −34 × 3 = −94. Then add 4 to both sides.
y = −34x + 94 + 4
y = −3x4 + 254
(Optional) A neater form. Multiplying through by the bottom of the fraction removes the fractions. The result always has this pattern: ax + by = r².
3x + 4y = 25

Check. Put the point back in: y = −34 × 3 + 254 = 4, and the point has y = 4. ✓

Watch out. The tangent gradient is the negative reciprocal of the radius gradient. A common slip is to flip the fraction but forget the sign change.

Worked example 25.7.5

Find the equation of the tangent to the circle x² + y² = 100 at the point (−6, 8).

Key words. A tangent is a straight line that touches a circle at exactly one point. It meets the radius at that point at a right angle.

Method. The tangent is perpendicular to the radius. Find the gradient of the radius, flip it and change the sign, then use the point to find the equation.

Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
((−6))² + (8)² = 36 + 64 = 100 ✓
Find the gradient of the radius. The radius joins the centre (0, 0) to the point. Gradient = change in y ÷ change in x.
gradient of radius = (8 − 0) ÷ ((−6) − 0) = −43
Find the gradient of the tangent. A tangent is perpendicular to the radius at the point of contact (Chapter 26). For perpendicular lines the gradients multiply to give −1: flip the fraction and change the sign.
gradient of tangent = −1 ÷ −43 = 34
Use y − y₁ = m(x − x₁) with the point. The tangent has gradient m and passes through (a, b).
y − 8 = 34(x − (−6))
Expand and rearrange to y = mx + c. Multiply out the bracket: 34 × (−6) = −92. Then add 8 to both sides.
y = 34x + 92 + 8
y = 3x4 + 252
(Optional) A neater form. Multiplying through by the bottom of the fraction removes the fractions. The result always has this pattern: ax + by = r².
−6x + 8y = 100

Check. Put the point back in: y = 34 × (−6) + 252 = 8, and the point has y = 8. ✓

Watch out. The tangent gradient is the negative reciprocal of the radius gradient. A common slip is to flip the fraction but forget the sign change.

Worked example 25.7.6

Show that the line 3x + 4y = 25 is a tangent to the circle x² + y² = 25.

Method. Substitute the equation of the line into the circle equation. A tangent touches at one point only, so the resulting quadratic has exactly one solution (a repeated root).

Rearrange the line to make y the subject. We will substitute this into the circle equation.
3x + 4y = 25 ⇒ y = −3x + 254
Substitute into the circle equation. Where the line meets the circle, both equations are true at once. Replace y in x² + y² = 25.
x² + (−3x + 254)² = 25
Expand, clear the fractions and simplify to a quadratic equal to 0. Multiply every term by 16 to clear the fractions, then move everything to one side.
x² − 6x + 9 = 0
Factorise. Look for a perfect square.
(x − 3)² = 0
Count the solutions. A perfect square gives only one value of x, so the line meets the circle at exactly one point. That is what a tangent does.
x = 3 only. (Discriminant b² − 4ac = 0.)

Check. The point of contact is (3, 4); it satisfies both equations. ✓

Watch out. If the discriminant were positive the line would cut the circle twice (a secant); if negative it would miss the circle.

Watch out

The equation is x² + y² = r². A circle of radius 5 has equation x² + y² = 25, not = 5. And for the tangent, do not use the gradient of the radius: the tangent is perpendicular so its gradient is the negative reciprocal.

25.8

Gradients of curves and rates of change

Edexcel 1MA1: A15, R15
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Gradients of curves and rates of changeComing soon

For a straight line the gradient is the same everywhere, so it tells us a steady rate of change: for a distance–time graph the gradient is the speed (Chapter 12). For a curve the steepness changes from place to place, so the gradient depends on where you look. Think of a car accelerating: its speed is different at different moments.

The gradient at a point on a curve. To find how steep a curve is at one point, draw a tangent: a straight line that just touches the curve at that point and has the same direction as the curve there. The gradient of the tangent is the gradient of the curve at that point. It is called the instantaneous rate of change. To find it:

Use a ruler to draw the tangent so that the curve is balanced equally on either side of the point.
Choose two points that are far apart on the tangent, and read their coordinates.
Work out rise ÷ run.

The tangent (red) to the curve y = x²/2 at the point (2, 2). Two points on the tangent are (0, −2) and (4, 6). The rise is 8 and the run is 4, so the gradient is 2.

A hand-drawn tangent is never perfect, so an answer is an estimate. In an exam, answers within a small tolerance are accepted.

Average rate of change. Often we only know the graph at two points. The average rate of change between them is the gradient of the straight line joining them, called a chord: (change in y) ÷ (change in x). For a distance–time graph this is the average speed over the interval.

As the second point slides towards (2, 2) the chord (green) turns and gets closer to the tangent (red). The gradient of the tangent is the limit of the gradients of the chords.

Closing in on the tangent. To estimate the gradient at x = 3 for y = x², work out the gradient of the chord from x = 3 to x = 3 + h for smaller and smaller h. The answers 7, 6.5, 6.1, 6.01 get closer and closer to 6, which is the gradient of the tangent. In algebra the chord gradient is (y-change) ÷ h; for y = x² between x = 3 and 3 + h it simplifies to 6 + h, and when h is tiny this is almost 6. (At A level you will find such gradients exactly using calculus; here we estimate.)

Units. The gradient has units of (y-unit) per (x-unit). On a distance–time graph in metres and seconds the gradient is in m/s; on a speed–time graph it is m/s² (acceleration); on a graph of cost in pounds against items bought it is £ per item.

Key ideas

The gradient of a curve at a point = the gradient of the tangent there = the instantaneous rate of change. It is found by rise ÷ run using two points on the tangent.
Average rate of change between two points = gradient of the chord = (change in y) ÷ (change in x).
As the second point moves closer, the chord gradient gets closer to the tangent gradient.
Gradient units: (y-unit) per (x-unit), e.g. m/s on a distance–time graph.
Hand-drawn tangents give estimates, not exact answers.

Worked example 25.8.1

The diagram shows the graph of y = x² ÷ 2. Draw a tangent at the point where x = 2 and use it to estimate the gradient of the curve at that point.

Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.

Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.

Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at x = 2 and has the curve balanced evenly on both sides of that point; draw along the ruler.
Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.
(0, −2) and (4, 6)
Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.
run = 4 − 0 = 4
rise = 6 − (−2) = 8
Work out the gradient. Gradient = rise ÷ run.
gradient = 8 ÷ 4 = 2

Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 1.8 and 2.2). The exact value for this curve is 2. ✓

Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.

Worked example 25.8.2

The distance, s metres, travelled by a car in t seconds is shown by the graph s = t² ÷ 4. Estimate the speed of the car after 4 seconds.

Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.

Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at t = 4 and has the curve balanced evenly on both sides of that point; draw along the ruler.
Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.
(1, −2) and (7, 10)
Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.
run = 7 − 1 = 6
rise = 10 − (−2) = 12
Work out the gradient. Gradient = rise ÷ run.
gradient = 12 ÷ 6 = 2
Interpret the answer. The gradient of the tangent is the rate of change at that instant.
The gradient of a distance–time graph is speed. So at 4 seconds the car is travelling at 2 m/s.

Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 1.8 and 2.2). The exact value for this curve is 2. ✓

Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.

Worked example 25.8.3

A ball rolls down a ramp. After t seconds it has travelled s = t² metres. Work out its average speed between t = 2 and t = 5.

Key words. A chord is a straight line joining two points on a curve. The average rate of change is the gradient of the chord.

Method. The average rate of change over an interval = (change in the output) ÷ (change in the input). It is the gradient of the straight line (chord) joining the two points on the graph.

Find the value at each end of the interval. To use the rule, substitute each value into the formula.
at t = 2: s = 2² = 4
at t = 5: s = 5² = 25
Find the change in the output and the change in the input. Subtract the first from the second.
change in s = 25 − 4 = 21m
change in t = 5 − 2 = 3s
Divide. The average rate of change is the gradient of the chord joining the two points: change in output ÷ change in input.
average rate = 21 ÷ 3 = 7 m/s

Check. Picture the chord from (2, 4) to (5, 25): rise 21 over run 3 gives 7. ✓

Watch out. An average rate is only an average over the whole interval. The rate at a single instant can be higher or lower.

Worked example 25.8.4

Estimate the gradient of the curve y = x² at the point (3, 9) by finding the gradient of chords from (3, 9) to points with x = 4, 3.5, 3.1 and 3.01.

Method. Use a chord from the fixed point to a second point very close to it. As the second point gets closer, the gradient of the chord approaches the gradient of the tangent.

Find the value at x = 3. This is the fixed point where we want the gradient.
y = 3² = 9
For each step size h, find the second point and the gradient of the chord. Chord gradient = (change in y) ÷ (change in x) = (y₂ − 9) ÷ h. Use smaller and smaller h so the second point slides towards the first.
h x-value y-value gradient of chord
1 4 16 7
0.5 3.5 12.25 6.5
0.1 3.1 9.61 6.1
0.01 3.01 9.0601 6.01
Look at the pattern. As h gets smaller the chord gets closer to the tangent, so the chord gradients settle down towards one number.
The gradients are getting closer to 6.

h	x-value	y-value	gradient of chord
1	4	16	7
0.5	3.5	12.25	6.5
0.1	3.1	9.61	6.1
0.01	3.01	9.0601	6.01

Check. The last chord gradient, 6.01, is very close to 6. ✓

Watch out. Use the same fixed point every time, and divide by h, the change in x.

Worked example 25.8.5

Show that the gradient of the chord of y = x² from x = 3 to x = 3 + h is 6 + h. What does this suggest for the gradient at x = 3?

Method. Find the gradient of the chord from x = 3 to x = 3 + h, simplify it, then see what happens when h gets tiny.

Write the two points on the curve. Take x = 3 and a nearby value x = 3 + h. Substitute into y = x².
(3, 9) and (3 + h, (3 + h)²)
Find the change in y. Expand (3 + h)² and take away 9.
(3 + h)² − 9 = h² + 6h
Divide by the change in x, which is h. Gradient of the chord = change in y ÷ change in x. Every term has a factor h, so divide each term by h.
gradient = h + 6
Let h become very small (almost 0). Any term containing h becomes almost 0, so only the number is left. This is the gradient of the tangent.
gradient of the tangent at x = 3 is 6

Check. Try h = 0.01 numerically: ((3 + 0.01)² − 9) ÷ 0.01 = 6.0100, close to 6. ✓

Watch out. Do not put h = 0 before dividing; that would give 0 ÷ 0. Simplify first, then let h become small.

Watch out

A tangent touches the curve at one point; it does not cut through it there. Do not use two points on the curve to find the gradient of a tangent: that gives a chord (an average rate). Take two points on the tangent line itself, far apart, and give the units of the gradient.

Exercise 25H Gradients of curves and rates of change

Non-calculator

FIn each diagram a tangent has been drawn to the curve at the marked point. Use the two points shown on the tangent to estimate the gradient of the curve at that point.

(a)y = x² ÷ 2 at x = 2

(b)y = x² ÷ 2 at x = −2

(c)y = x³ ÷ 4 at x = 2

Done

PSThe distance, s metres, travelled by a cyclist in t seconds is given by s = t² ÷ 4. A tangent has been drawn on the graph. Use it to estimate the cyclist’s speed after

(a)2 seconds

(b)4 seconds

Done

FA rocket’s height is s = t² metres after t seconds. Work out its average speed between

(a)t = 0 and t = 2

(b)t = 2 and t = 4

(c)t = 4 and t = 8

Done

PSThe value of an investment is V = 100 × 1.1^t pounds after t years. Work out the average rate of change of its value between

(a)t = 0 and t = 2

(b)t = 2 and t = 4

Done

PSA cyclist’s distance is s = 0.5t² + 2t metres after t seconds. Estimate her speed at t = 4 by working out the average speeds over the intervals from t = 4 to 5, 4.5, 4.1 and 4.01.

Method. Use a chord from the fixed point to a second point very close to it. As the second point gets closer, the gradient of the chord approaches the gradient of the tangent.

Find the value at t = 4. This is the fixed point where we want the gradient.
s = 4²2 + 2 × 4 = 16
For each step size h, find the second point and the gradient of the chord. Chord gradient = (change in s) ÷ (change in t) = (s₂ − 16) ÷ h. Use smaller and smaller h so the second point slides towards the first.
h t-value s-value gradient of chord
1 5 22.5 6.5
0.5 4.5 19.125 6.25
0.1 4.1 16.605 6.05
0.01 4.01 16.06005 6.005
Look at the pattern. As h gets smaller the chord gets closer to the tangent, so the chord gradients settle down towards one number.
The gradients are getting closer to 6.

h	t-value	s-value	gradient of chord
1	5	22.5	6.5
0.5	4.5	19.125	6.25
0.1	4.1	16.605	6.05
0.01	4.01	16.06005	6.005

Check. The last chord gradient, 6.005, is very close to 6. ✓

Watch out. Use the same fixed point every time, and divide by h, the change in x.

Done

RFor each curve, find the gradient of the chord from x = a to x = a + h, simplified, and use it to write down the gradient of the curve at x = a.

(a)y = x², a = 3

(b)y = x³, a = 1

(c)y = x², a = 5

(d)y = x³, a = −2

Done

PSA car’s distance is s = t² metres after t seconds. (a) Work out its average speed over the first 4 seconds. (b) Work out the gradient of the chord from t = 4 to t = 4.01 and use it to estimate the speed at t = 4. (c) Explain why the two answers differ.

Done

PSThe population of a colony of bacteria is P = 1000 × 1.1^t after t hours. (a) Work out the average rate of increase between t = 2 and t = 5. (b) Estimate the rate of increase at t = 3 using the interval from t = 3 to t = 3.01. Give answers to the nearest whole number of bacteria per hour.

Done

PSFor each graph, give the units of the gradient and say what the gradient represents.

(a)A graph of cost (£) against number of items bought has gradient 3.5.

(b)A graph of speed (m/s) against time (s) has gradient 2.

(c)A graph of distance (km) against time (hours) has gradient 60.

(d)A graph of litres of fuel left against distance (km) has gradient −0.08.

Done

FThe curve is y = x²(x − 3). Five points are marked. Write down the points where the gradient of the curve is

(a)zero

(b)positive

(c)negative

Done

RA student says: “The average speed of the train over the whole journey is 80 km/h, so the speed at the half-way point must be 80 km/h.” Explain why the student may be wrong.

Done

RThe curve y = x² − 4x has its lowest point at (2, −4). Explain why the gradient of the curve is 0 at that point.

Done

RFor y = x², compare the gradient of the chord from x = 3 to x = 5 with the gradient of the tangent at x = 3 (which is 6). Is the chord gradient bigger or smaller, and why does that make sense on the graph?

Done

RUse the chord gradient 2a + h for the curve y = x² between x = a and x = a + h to find the value of x at which the gradient of the curve is 10.

Done

25.9

Areas under graphs

Edexcel 1MA1: A15, R15
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Areas under graphsComing soon

In Chapter 13 you used distance–time graphs, where the gradient is the speed. Now look at a speed–time graph. If a car travels at a steady 20 m/s for 10 seconds, it goes 20 × 10 = 200 m. On the graph, 20 is the height and 10 is the width of a rectangle, so the area of the rectangle is the distance travelled. This works for every speed–time graph:

area under a speed–time graph = distance travelled.

In general, the area under any graph is (vertical quantity) × (horizontal quantity): the units multiply. m/s × s = m; litres/min × min = litres; kW × hours = kWh; pounds per hour × hours = pounds.

A speed–time graph: the car accelerates for 10 s, travels at 20 m/s for 30 s and then slows to a stop in 10 s. The shaded area is the distance travelled.

Straight-line graphs. When the graph is made of straight lines, split the area into rectangles, triangles and trapezia, and add up the areas. For a trapezium, area = ½ × (sum of the parallel sides) × width, where the parallel sides are the two heights.

Curved graphs. When the graph is a curve we cannot find the area exactly with these shapes, but we can estimate it. Cut the area into thin vertical strips of equal width. Treat each strip as a trapezium (the top is the straight line joining two points on the curve) and add the areas. With more strips the estimate gets better. You can also use rectangles, using the height at the left (or right) edge of each strip.

Estimating the area under y = t²/2 + 2 between t = 0 and t = 6. The trapezia (left) lie above this upward-bending curve, so they give an over-estimate; the left-hand rectangles lie below the rising curve, so they give an under-estimate.

Over or under? If the curve bends upwards (convex), the straight tops of the trapezia lie above the curve, so the estimate is too big (an over-estimate). If the curve bends downwards, the trapezia lie below it and the estimate is too small. For rectangles on an increasing graph, left-hand heights give an under-estimate and right-hand heights give an over-estimate; on a decreasing graph it is the other way round.

Key ideas

Area under a speed–time graph = distance. In general area = (vertical value) × (horizontal value) and the units multiply.
Split straight-line graphs into rectangles, triangles and trapezia. Trapezium area = ½ × (sum of parallel sides) × width.
For a curve, use equal-width strips and trapezia (or rectangles) to estimate the area. More strips give a better estimate.
Average speed = total distance ÷ total time (area ÷ time).
A convex (upward-bending) curve: trapezium estimate is too big. A concave curve: too small.

Worked example 25.9.1

The speed–time graph shows a car’s journey (shown above). Work out the total distance travelled.

Method. The area under a speed–time graph is the distance travelled. Split it into rectangles, triangles and trapezia, find each area and add.

Remember what the area represents. On a graph with time on the horizontal axis and speed on the vertical axis, area = speed × time = distance travelled. So the area under the graph is the distance travelled.
Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.
Part 1 (from 0 to 10 s): triangle: ½ × 10 × 20 = 100
Part 2 (from 10 to 40 s): rectangle: 20 × 30 = 600
Part 3 (from 40 to 50 s): triangle: ½ × 10 × 20 = 100
Add the areas.
100 + 600 + 100 = 800

Check. Total time = 50 s, so the average speed is 800 ÷ 50 = 16 m/s. This lies between the lowest and highest values on the graph. ✓

Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.

Worked example 25.9.2

The speed of a cyclist, v m/s, after t seconds is given by v = t²/2 + 2. Use 3 strips of equal width to estimate the distance travelled in the first 6 seconds. Is your answer an under-estimate or an over-estimate?

Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).

Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.

Split the area into strips of equal width. The region from time = 0 to 6 is 6 wide. With 3 strips, each strip has width 6 ÷ 3 = 2.
Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 2 4 6
speed 2 4 10 20
Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.
Strip 1: ½ × (2 + 4) × 2 = 6
Strip 2: ½ × (4 + 10) × 2 = 14
Strip 3: ½ × (10 + 20) × 2 = 30
Add the strip areas to get the estimate.
6 + 14 + 30 = 50
Comment on the accuracy. The curve bends upwards (convex), so each trapezium top lies above the curve. So the estimate is an over-estimate. More strips would give a better estimate.
(The exact area is 48.00 m, so the estimate is bigger.)

time	0	2	4	6
speed	2	4	10	20

Check. Use the trapezium rule formula: ½ × 2 × [2 + 20 + 2 × (4 + 10)] = 50.00. ✓

Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.

Worked example 25.9.3

Use 3 rectangles, with heights taken at the left-hand side of each strip, to estimate the same area for v = t²/2 + 2 between t = 0 and t = 6. Is it an under-estimate or an over-estimate?

Method. Split the area into strips, use the height at the left (or right) edge of each as a rectangle, and add the rectangle areas.

Split into strips of equal width. Width of each strip = (6 − 0) ÷ 3 = 2.
Use the height at the left-hand edge of each strip. Each strip is treated as a rectangle whose height is the value of the graph at its left edge.
time 0 2 4 6
speed 2 4 10 20
Area of each rectangle = height × width.
Strip 1: 2 × 2 = 4
Strip 2: 4 × 2 = 8
Strip 3: 10 × 2 = 20
Add.
4 + 8 + 20 = 32
Say whether it is an over- or under-estimate. The graph is increasing. Using the left-hand heights means each rectangle is below the curve for most of the strip. So the estimate is an under-estimate.
estimate 32.00 m < true area 48.00 m

time	0	2	4	6
speed	2	4	10	20

Check. The more strips we use, the closer the estimate gets to the true value (48.00). ✓

Watch out. For an increasing graph, left-hand rectangles give an under-estimate and right-hand rectangles give an over-estimate. For a decreasing graph it is the other way round.

Worked example 25.9.4

Water flows into a tank. The graph shows the flow rate in litres per minute against time in minutes (rate rises from 0 to 12 in 4 minutes, stays at 12 until 10 minutes, then falls to 0 at 14 minutes). Work out the total volume of water that flows into the tank.

Method. The area under a flow rate–time graph is the volume of water that flows in. Split it into rectangles, triangles and trapezia, find each area and add.

Remember what the area represents. On a graph with time on the horizontal axis and flow rate on the vertical axis, area = flow rate × time = volume of water that flows in. So the area under the graph is the volume of water that flows in.
Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.
Part 1 (from 0 to 4 minutes): triangle: ½ × 4 × 12 = 24
Part 2 (from 4 to 10 minutes): rectangle: 12 × 6 = 72
Part 3 (from 10 to 14 minutes): triangle: ½ × 4 × 12 = 24
Add the areas.
24 + 72 + 24 = 120

Check. Total time = 14 minutes, so the average flow rate is 120 ÷ 14 = 607 litres/min. This lies between the lowest and highest values on the graph. ✓

Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.

Worked example 25.9.5

A train accelerates uniformly from rest to a speed of V m/s in 10 s, travels at V m/s for 30 s, then slows uniformly to rest in 20 s. The total distance travelled is 450 m. Find V.

Method. Write the area under the graph in terms of V (a rectangle and two triangles) and set it equal to 450.

Sketch the graph. The shape is a trapezium: it rises to height V over 10 s, is flat for 30 s and falls over 20 s.
Area of the first triangle. ½ × base × height.
½ × 10 × V = 5V
Area of the rectangle. Base 30, height V.
30 × V = 30V
Area of the second triangle. Base 20, height V.
½ × 20 × V = 10V
Add them and set equal to the total distance. The area is the distance.
5V + 30V + 10V = 45V = 450
Solve. Divide both sides by 45.
V = 450 ÷ 45 = 10

Check. Total area = ½ × (60 + 30) × 10 = 450 using the trapezium formula with parallel sides 60 and 30. ✓

Watch out. Keep V as a letter until the end. Do not guess a number first.

Watch out

The area under a distance–time graph does not mean anything useful (m × s). Only use the area under a speed–time graph (or other rate graph). Also, when estimating with trapezia, remember the ½ and use the two heights as the parallel sides.

Exercise 25I Areas under graphs

Calculator allowed

FEach graph shows speed (m/s) against time (s). Work out the distance travelled.

(a)

(b)

(c)

Done

PSWork out the average speed for the journeys shown on the graphs above (the same three graphs).

(a)

(b)

(c)

Done

PSA car accelerates uniformly from rest to 24 m/s in 8 s, then travels at a constant 24 m/s for 20 s, and finally slows uniformly to rest in 6 s.

(a)Work out the acceleration during the first 8 seconds.

(b)Work out the total distance travelled.

(c)Work out the average speed for the whole journey.

Done

FUse strips of equal width and the trapezium method to estimate the area under each curve.

(a)y = x²2 + 2 from x = 0 to x = 6, using 3 strips

Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).

Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.

Split the area into strips of equal width. The region from time = 0 to 6 is 6 wide. With 3 strips, each strip has width 6 ÷ 3 = 2.
Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 2 4 6
speed 2 4 10 20
Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.
Strip 1: ½ × (2 + 4) × 2 = 6
Strip 2: ½ × (4 + 10) × 2 = 14
Strip 3: ½ × (10 + 20) × 2 = 30
Add the strip areas to get the estimate.
6 + 14 + 30 = 50
Comment on the accuracy. The curve bends upwards (convex), so each trapezium top lies above the curve. So the estimate is an over-estimate. More strips would give a better estimate.
(The exact area is 48.00 m, so the estimate is bigger.)

time	0	2	4	6
speed	2	4	10	20

Check. Use the trapezium rule formula: ½ × 2 × [2 + 20 + 2 × (4 + 10)] = 50.00. ✓

Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.

(b)y = x² + 1 from x = 0 to x = 4, using 4 strips

Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).

Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.

Split the area into strips of equal width. The region from time = 0 to 4 is 4 wide. With 4 strips, each strip has width 4 ÷ 4 = 1.
Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 1 2 3 4
speed 1 2 5 10 17
Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.
Strip 1: ½ × (1 + 2) × 1 = 1.5
Strip 2: ½ × (2 + 5) × 1 = 3.5
Strip 3: ½ × (5 + 10) × 1 = 7.5
Strip 4: ½ × (10 + 17) × 1 = 13.5
Add the strip areas to get the estimate.
1.5 + 3.5 + 7.5 + 13.5 = 26
Comment on the accuracy. The curve bends upwards (convex), so each trapezium top lies above the curve. So the estimate is an over-estimate. More strips would give a better estimate.
(The exact area is 25.33 m, so the estimate is bigger.)

time	0	1	2	3	4
speed	1	2	5	10	17

Check. Use the trapezium rule formula: ½ × 1 × [1 + 17 + 2 × (2 + 5 + 10)] = 26.00. ✓

Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.

(c)y = −x² + 16 from x = 0 to x = 4, using 4 strips

Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).

Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.

Split the area into strips of equal width. The region from time = 0 to 4 is 4 wide. With 4 strips, each strip has width 4 ÷ 4 = 1.
Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 1 2 3 4
speed 16 15 12 7 0
Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.
Strip 1: ½ × (16 + 15) × 1 = 15.5
Strip 2: ½ × (15 + 12) × 1 = 13.5
Strip 3: ½ × (12 + 7) × 1 = 9.5
Strip 4: ½ × (7 + 0) × 1 = 3.5
Add the strip areas to get the estimate.
15.5 + 13.5 + 9.5 + 3.5 = 42
Comment on the accuracy. The curve bends downwards (concave), so each trapezium top lies below the curve. So the estimate is an under-estimate. More strips would give a better estimate.
(The exact area is 42.67 m, so the estimate is smaller.)

time	0	1	2	3	4
speed	16	15	12	7	0

Check. Use the trapezium rule formula: ½ × 1 × [16 + 0 + 2 × (15 + 12 + 7)] = 42.00. ✓

Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.

Done

FUse rectangles to estimate the area under y = x²/2 + 2 between x = 0 and x = 6 with 3 strips, and decide if it is an over- or under-estimate.

(a)Left-hand rectangles

Method. Split the area into strips, use the height at the left (or right) edge of each as a rectangle, and add the rectangle areas.

Split into strips of equal width. Width of each strip = (6 − 0) ÷ 3 = 2.
Use the height at the left-hand edge of each strip. Each strip is treated as a rectangle whose height is the value of the graph at its left edge.
time 0 2 4 6
speed 2 4 10 20
Area of each rectangle = height × width.
Strip 1: 2 × 2 = 4
Strip 2: 4 × 2 = 8
Strip 3: 10 × 2 = 20
Add.
4 + 8 + 20 = 32
Say whether it is an over- or under-estimate. The graph is increasing. Using the left-hand heights means each rectangle is below the curve for most of the strip. So the estimate is an under-estimate.
estimate 32.00 m < true area 48.00 m

time	0	2	4	6
speed	2	4	10	20

Check. The more strips we use, the closer the estimate gets to the true value (48.00). ✓

Watch out. For an increasing graph, left-hand rectangles give an under-estimate and right-hand rectangles give an over-estimate. For a decreasing graph it is the other way round.

(b)Right-hand rectangles

Method. Split the area into strips, use the height at the left (or right) edge of each as a rectangle, and add the rectangle areas.

Split into strips of equal width. Width of each strip = (6 − 0) ÷ 3 = 2.
Use the height at the right-hand edge of each strip. Each strip is treated as a rectangle whose height is the value of the graph at its right edge.
time 0 2 4 6
speed 2 4 10 20
Area of each rectangle = height × width.
Strip 1: 4 × 2 = 8
Strip 2: 10 × 2 = 20
Strip 3: 20 × 2 = 40
Add.
8 + 20 + 40 = 68
Say whether it is an over- or under-estimate. The graph is increasing. Using the right-hand heights means each rectangle is above the curve for most of the strip. So the estimate is an over-estimate.
estimate 68.00 m > true area 48.00 m

time	0	2	4	6
speed	2	4	10	20

Check. The more strips we use, the closer the estimate gets to the true value (48.00). ✓

Watch out. For an increasing graph, left-hand rectangles give an under-estimate and right-hand rectangles give an over-estimate. For a decreasing graph it is the other way round.

(c)Left-hand rectangles for y = 16 − x² from 0 to 4 with 4 strips

Method. Split the area into strips, use the height at the left (or right) edge of each as a rectangle, and add the rectangle areas.

Split into strips of equal width. Width of each strip = (4 − 0) ÷ 4 = 1.
Use the height at the left-hand edge of each strip. Each strip is treated as a rectangle whose height is the value of the graph at its left edge.
time 0 1 2 3 4
speed 16 15 12 7 0
Area of each rectangle = height × width.
Strip 1: 16 × 1 = 16
Strip 2: 15 × 1 = 15
Strip 3: 12 × 1 = 12
Strip 4: 7 × 1 = 7
Add.
16 + 15 + 12 + 7 = 50
Say whether it is an over- or under-estimate. The graph is decreasing. Using the left-hand heights means each rectangle is above the curve for most of the strip. So the estimate is an over-estimate.
estimate 50.00 m > true area 42.67 m

time	0	1	2	3	4
speed	16	15	12	7	0

Check. The more strips we use, the closer the estimate gets to the true value (42.67). ✓

Watch out. For an increasing graph, left-hand rectangles give an under-estimate and right-hand rectangles give an over-estimate. For a decreasing graph it is the other way round.

Done

PSPetrol is pumped into a tank. The flow rate rises steadily from 0 to 20 litres per minute in 5 minutes, stays at 20 litres per minute for 10 minutes and then falls steadily to 10 litres per minute over the next 5 minutes. How many litres are pumped in the 20 minutes?

Done

PSA coach starts from rest, speeds up to 60 km/h in half an hour, travels at 60 km/h for 1.5 hours and stops in the next half hour. Work out the distance travelled in kilometres.

Done

PSA graph shows a constant rate. Work out the area under the graph for the time given and say what the area represents.

(a)Power 3 kW for 4 hours

(b)Flow 8 litres/s for 15 s

(c)Pay rate £12.50 per hour for 6 hours

(d)Speed 15 m/s for 40 s

Done

PSThe speed of a runner (m/s) at 2-second intervals is given by v = 0.35t². Use the trapezium method with 5 strips to estimate how far she runs in the first 10 seconds.

Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).

Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.

Split the area into strips of equal width. The region from time = 0 to 10 is 10 wide. With 5 strips, each strip has width 10 ÷ 5 = 2.
Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 2 4 6 8 10
speed 0 1.4 5.6 12.6 22.4 35
Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.
Strip 1: ½ × (0 + 1.4) × 2 = 1.4
Strip 2: ½ × (1.4 + 5.6) × 2 = 7
Strip 3: ½ × (5.6 + 12.6) × 2 = 18.2
Strip 4: ½ × (12.6 + 22.4) × 2 = 35
Strip 5: ½ × (22.4 + 35) × 2 = 57.4
Add the strip areas to get the estimate.
1.4 + 7 + 18.2 + 35 + 57.4 = 119
Comment on the accuracy. The curve bends upwards (convex), so each trapezium top lies above the curve. So the estimate is an over-estimate. More strips would give a better estimate.
(The exact area is 116.67 m, so the estimate is bigger.)

time	0	2	4	6	8	10
speed	0	1.4	5.6	12.6	22.4	35

Check. Use the trapezium rule formula: ½ × 2 × [0 + 35 + 2 × (1.4 + 5.6 + 12.6 + 22.4)] = 119.00. ✓

Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.

Done

RThe graph of y = 16 − x² bends downwards between x = 0 and x = 4. Explain whether the trapezium method gives an over-estimate or an under-estimate of the area, and show that using more strips changes the answer. (The exact area is 42.67 to 2 d.p.)

Done

RBen says: “The area under a distance–time graph is the distance travelled.” Explain why Ben is wrong, and say what the area under a speed–time graph represents.

Done

RThe straight line y = kx is drawn for x from 0 to 6. The area between the line and the x-axis is 54. Find k.

Done

25.10

Iteration and iterative processes

Edexcel 1MA1: A20, R16
Grade 8-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Iteration and iterative processesComing soon

Some equations cannot be solved with the methods you know. For example x³ − x − 1 = 0 cannot be factorised and there is no simple formula. But we can still find an answer as accurately as we like using iteration: repeating a calculation, using each answer as the input for the next, so that the numbers get closer and closer to the solution.

An iteration formula looks like x_n+1 = g(x_n). It says: to get the next term, put the current term into g. You are given a starting value x₀. For example, with x_n+1 = ∛(x_n + 1) and x₀ = 1: x₁ = ∛(1 + 1) = 1.2599…, then x₂ = ∛(1.2599 + 1) = 1.3122…, and so on. On a calculator type 1 =, then type ∛(Ans + 1) and press = repeatedly. The Ans button stores the full unrounded value, which keeps the working accurate.

Why it finds a solution. If the terms settle down to a value L, then x_n and x_n+1 are both about L, so L = g(L). So L solves the equation x = g(x). The iteration formula comes from rearranging the original equation into the form x = g(x). For x³ − x − 1 = 0 we can write x³ = x + 1 and so x = ∛(x + 1).

Where is the solution? To show that a solution lies between two numbers, put both into the expression. If one gives a negative answer and the other a positive answer (a change of sign), the graph crosses the x-axis between them. For f(x) = x³ − x − 1: f(1) = −1 and f(2) = 5, so there is a solution between 1 and 2.

A staircase (cobweb) diagram for xₙ₊₁ = ∛(xₙ + 1), starting at x₀ = 1. The blue curve is y = ∛(x + 1) and the grey line is y = x. Each step goes up to the curve and across to the line. The staircase closes in on where they cross: the solution of x = ∛(x + 1).

Giving the answer to a number of decimal places. Keep iterating until two (or three) successive terms agree to the required accuracy. To prove that the answer is correct to, say, 3 decimal places, substitute the two ends of the rounding interval (for 1.325 these are 1.3245 and 1.3255) into f(x): a change of sign proves the solution is between them.

Not every rearrangement works. The same equation can be rearranged in several ways. Rearranging x³ − x − 1 = 0 as x = x³ − 1 gives terms that fly away: starting at 1.5 we get 2.375, 12.4, 1905… (it diverges). In an exam you will always be told which formula to use.

The rearrangement xₙ₊₁ = xₙ³ − 1 starting from 1.5: the steps move away from the crossing point (the staircase spirals outwards), so the iteration diverges.

General iterative processes. Iteration also describes situations that repeat in steps, such as a population measured every year or the amount of a drug in the body measured every day. Write the rule as u_n+1 = au_n + b: the next amount is the last amount multiplied by a factor, plus (or minus) a fixed amount added or removed. Compound interest is the simplest case (b = 0). If the terms settle on a limiting value L, then L = aL + b, so L = b ÷ (1 − a).

Key ideas

Iteration: x_n+1 = g(x_n). Start at x₀ and repeat. Use the Ans button and keep all the digits.
The limit L of the terms satisfies L = g(L), which is a rearrangement of the original equation.
A change of sign of f(x) between two values shows a solution of f(x) = 0 lies between them.
For a solution to n d.p.: iterate until the terms agree to n d.p., then show a change of sign at the ends of the rounding interval.
Some rearrangements diverge. For u_n+1 = au_n + b, the limit (if there is one) is b ÷ (1 − a).

Worked example 25.10.1

A sequence is given by x_n+1 = ∛(x_n + 1) with x₀ = 1. Work out x₁, x₂, … x₅, giving your answers to 6 decimal places.

Key words. An iteration repeats the same calculation, using each answer as the next input. x₁, x₂ ... are the terms.

Method. Substitute each value into the formula to get the next one, starting from x₀.

Write the iteration formula and the starting value. An iteration is a recipe that gives each term from the one before. The starting value is x₀.
x_n+1 = ∛(x_n + 1)
x₀ = 1
Work out x₁: put x₀ into the right-hand side. Replace x_n by the value you have, using brackets for negatives.
x₁ = ∛(1 + 1) = 1.259921
Work out x₂: use the answer for x₁. Now x₁ is the input. On a calculator type the value, press =, then type the formula using the Ans key in place of x_n; each press of = gives the next term.
x₂ = ∛(1.259921 + 1) = 1.312294
Keep pressing = to find the rest. Do not round during the calculation: keep all the digits in the calculator memory and only round the values you write down.
n x_n
0 1.000000
1 1.259921
2 1.312294
3 1.322354
4 1.324269
5 1.324633

n	x_n
0	1.000000
1	1.259921
2	1.312294
3	1.322354
4	1.324269
5	1.324633

Check. The terms are settling down towards 1.3246: each new term is closer to the last. ✓

Watch out. Use the calculator’s Ans button so you keep every digit. Rounding too early can give a wrong final digit.

Worked example 25.10.2

Show that the equation x³ − x − 1 = 0 has a solution between x = 1 and x = 2.

Method. Work out f(x) at each end. A change of sign shows the graph crosses the x-axis, so there is a solution in between.

Substitute the lower value. Put the first value of x into the expression.
f(1) = 1³ − 1 − 1 = −1
Substitute the upper value. Put the second value in.
f(2) = 2³ − 2 − 1 = 5
Compare the signs. If the graph of a continuous function is below the x-axis at one value and above it at another, it must cross the axis between them. A crossing means f(x) = 0.
f(1) is negative and f(2) is positive: there is a change of sign.
State the conclusion.
So the equation x³ − x − 1 = 0 has a solution between x = 1 and x = 2.

Check. The root is about 1.325, which lies between 1 and 2. ✓

Watch out. You must show both values and say there is a change of sign. Saying only “one is negative” earns no marks.

Worked example 25.10.3

Show that the equation x³ + 4x − 3 = 0 can be rearranged to give x = 3 ÷ (x² + 4).

Method. To rearrange an equation into the form x = g(x), get an x (or x², x³) on its own on one side using inverse operations.

Start with the equation.
x³ + 4x − 3 = 0
Move the plain number to the right. Subtract it from both sides.
x³ + 4x = 3
Factorise the left side. Take out x as a common factor.
x(x² + 4) = 3
Divide both sides by the bracket. This leaves x on its own.
x = 3x² + 4

Check. At the solution (x ≈ 0.674) both sides agree: 0.674 and 0.674. ✓

Watch out. The same equation can be rearranged in several ways. Some rearrangements work for iteration and some do not (the numbers fly off), so always use the one the question gives.

Worked example 25.10.4

Use the iteration formula x_n+1 = ∛(x_n + 1) with x₀ = 1.5 to find a solution of x³ − x − 1 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.

Key words. The values converge when they get closer and closer to one number. A root (solution) of f(x) = 0 is a value of x that makes f(x) equal 0.

Method. Use the formula repeatedly until the terms agree to 3 d.p., then prove the answer by a change of sign.

Write the iteration formula and the starting value. We repeat the same calculation, each time putting the last answer back in.
x_n+1 = ∛(x_n + 1)
x₀ = 1.5
Calculate the terms using the calculator’s Ans button. Type the starting value and press =. Then type the formula with Ans in place of x_n and keep pressing = for each new term. Keep every digit; only round what you write down.
n x_n
0 1.500000
1 1.357209
2 1.330861
3 1.325884
4 1.324939
5 1.324760
6 1.324726
7 1.324719
Stop when the terms agree to 3 decimal places. The last few terms all round to 1.325, so the solution is close to this value.
x₅, x₆, x₇ ≈ 1.3248, 1.3247, 1.3247: all round to 1.325
Show the answer is correct to 3 d.p.: test the two ends of the rounding interval. Any number that rounds to 1.325 lies between 1.3245 and 1.3255. If f changes sign between these, the solution must lie in the interval.
f(1.3245) = 1.3245³ − 1.3245 − 1 = −0.00093
f(1.3255) = 1.3255³ − 1.3255 − 1 = 0.00334
Conclude. There is a change of sign, so the solution is between the two values, and both round to the same number.
x = 1.325 to 3 decimal places

n	x_n
0	1.500000
1	1.357209
2	1.330861
3	1.325884
4	1.324939
5	1.324760
6	1.324726
7	1.324719

Check. Put x = 1.325 into the original equation: 1.325³ − 1.325 − 1 = 0.0012, which is very close to 0. ✓

Watch out. The change-of-sign test must use the ends of the rounding interval (for example 1.3245 and 1.3255), not just the answer itself.

Worked example 25.10.5

A patient takes 20 mg of a drug at the start of each day. During each day 20% of the drug in the body is removed. The amount in the body just after each dose is u_n mg, with u₀ = 0, and u_n+1 = 0.8u_n + 20. Work out u₁, u₂, u₃ and u₄.

Method. An iterative process gives each term from the previous one. Apply the rule repeatedly.

Write the rule and the starting value. Each term is made from the one before.
u_n+1 = 0.8u_n + 20, u₀ = 0
Work out the terms one at a time. Put each term into the rule to get the next.
u₁ = 0.8 × 0 + 20 = 20
u₂ = 0.8 × 20 + 20 = 36
u₃ = 0.8 × 36 + 20 = 48.8
u₄ = 0.8 × 48.8 + 20 = 59.04
Write the answers. List the terms asked for.
20, 36, 48.8, 59.04

Check. Check the first step again: 0.8 × 0 + 20 = 20. ✓

Watch out. Use the previous term each time, not the starting value.

Watch out

Do not round during the iteration. If you round every term to 2 or 3 decimal places the errors build up and the final digit may be wrong. Use the Ans key. Also, do not show a root only with the answer: you must show a change of sign for the two ends of the rounding interval.

Exercise 25J Iteration and iterative processes

Calculator allowed

FUse each iteration formula and starting value to work out x₁, x₂ and x₃. Give your answers to 6 decimal places.

(a)x_n+1 = ∛(x_n + 1), x₀ = 1

Key words. An iteration repeats the same calculation, using each answer as the next input. x₁, x₂ ... are the terms.

Method. Substitute each value into the formula to get the next one, starting from x₀.

Write the iteration formula and the starting value. An iteration is a recipe that gives each term from the one before. The starting value is x₀.
x_n+1 = ∛(x_n + 1)
x₀ = 1
Work out x₁: put x₀ into the right-hand side. Replace x_n by the value you have, using brackets for negatives.
x₁ = ∛(1 + 1) = 1.259921
Work out x₂: use the answer for x₁. Now x₁ is the input. On a calculator type the value, press =, then type the formula using the Ans key in place of x_n; each press of = gives the next term.
x₂ = ∛(1.259921 + 1) = 1.312294
Keep pressing = to find the rest. Do not round during the calculation: keep all the digits in the calculator memory and only round the values you write down.
n x_n
0 1.000000
1 1.259921
2 1.312294
3 1.322354

n	x_n
0	1.000000
1	1.259921
2	1.312294
3	1.322354

Watch out. Use the calculator’s Ans button so you keep every digit. Rounding too early can give a wrong final digit.

(b)x_n+1 = 3x_n² + 4, x₀ = 1

Key words. An iteration repeats the same calculation, using each answer as the next input. x₁, x₂ ... are the terms.

Method. Substitute each value into the formula to get the next one, starting from x₀.

Write the iteration formula and the starting value. An iteration is a recipe that gives each term from the one before. The starting value is x₀.
x_n+1 = 3x_n² + 4
x₀ = 1
Work out x₁: put x₀ into the right-hand side. Replace x_n by the value you have, using brackets for negatives.
x₁ = 31² + 4 = 0.600000
Work out x₂: use the answer for x₁. Now x₁ is the input. On a calculator type the value, press =, then type the formula using the Ans key in place of x_n; each press of = gives the next term.
x₂ = 30.600000² + 4 = 0.688073
Keep pressing = to find the rest. Do not round during the calculation: keep all the digits in the calculator memory and only round the values you write down.
n x_n
0 1.000000
1 0.600000
2 0.688073
3 0.670624

n	x_n
0	1.000000
1	0.600000
2	0.688073
3	0.670624

Watch out. Use the calculator’s Ans button so you keep every digit. Rounding too early can give a wrong final digit.

(c)x_n+1 = x_n³5 + 15, x₀ = 0

Key words. An iteration repeats the same calculation, using each answer as the next input. x₁, x₂ ... are the terms.

Method. Substitute each value into the formula to get the next one, starting from x₀.

Write the iteration formula and the starting value. An iteration is a recipe that gives each term from the one before. The starting value is x₀.
x_n+1 = x_n³5 + 15
x₀ = 0
Work out x₁: put x₀ into the right-hand side. Replace x_n by the value you have, using brackets for negatives.
x₁ = 0³5 + 15 = 0.200000
Work out x₂: use the answer for x₁. Now x₁ is the input. On a calculator type the value, press =, then type the formula using the Ans key in place of x_n; each press of = gives the next term.
x₂ = 0.200000³5 + 15 = 0.201600
Keep pressing = to find the rest. Do not round during the calculation: keep all the digits in the calculator memory and only round the values you write down.
n x_n
0 0.000000
1 0.200000
2 0.201600
3 0.201639

n	x_n
0	0.000000
1	0.200000
2	0.201600
3	0.201639

Check. The terms are settling down towards 0.2016: each new term is closer to the last. ✓

Watch out. Use the calculator’s Ans button so you keep every digit. Rounding too early can give a wrong final digit.

(d)x_n+1 = √(3x_n + 2), x₀ = 3

Key words. An iteration repeats the same calculation, using each answer as the next input. x₁, x₂ ... are the terms.

Method. Substitute each value into the formula to get the next one, starting from x₀.

Write the iteration formula and the starting value. An iteration is a recipe that gives each term from the one before. The starting value is x₀.
x_n+1 = √(3x_n + 2)
x₀ = 3
Work out x₁: put x₀ into the right-hand side. Replace x_n by the value you have, using brackets for negatives.
x₁ = √(3 × 3 + 2) = 3.316625
Work out x₂: use the answer for x₁. Now x₁ is the input. On a calculator type the value, press =, then type the formula using the Ans key in place of x_n; each press of = gives the next term.
x₂ = √(3 × 3.316625 + 2) = 3.456859
Keep pressing = to find the rest. Do not round during the calculation: keep all the digits in the calculator memory and only round the values you write down.
n x_n
0 3.000000
1 3.316625
2 3.456859
3 3.517183

n	x_n
0	3.000000
1	3.316625
2	3.456859
3	3.517183

Watch out. Use the calculator’s Ans button so you keep every digit. Rounding too early can give a wrong final digit.

Done

FShow that each equation has a solution between the two given values.

(a)x³ − x − 1 = 0 between x = 1 and x = 2

(b)x³ + 4x − 3 = 0 between x = 0 and x = 1

(c)x³ − 5x + 1 = 0 between x = 0 and x = 1

(d)x² − 3x − 2 = 0 between x = 3 and x = 4

Done

RShow that each equation can be rearranged to give the iteration formula x = g(x) given.

(a)x³ + 2x − 7 = 0 can be rearranged to x = ∛(−2x + 7)

(b)x² − 3x − 2 = 0 can be rearranged to x = √(3x + 2)

(c)x² − 3x − 2 = 0 can be rearranged to x = 3x + 2x

(d)x³ − 5x + 1 = 0 can be rearranged to x = x³ + 15

Done

PSUse the iteration formula x_n+1 = ∛(x_n + 1) with x₀ = 1.5 to find a solution of x³ − x − 1 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.

Key words. The values converge when they get closer and closer to one number. A root (solution) of f(x) = 0 is a value of x that makes f(x) equal 0.

Method. Use the formula repeatedly until the terms agree to 3 d.p., then prove the answer by a change of sign.

Write the iteration formula and the starting value. We repeat the same calculation, each time putting the last answer back in.
x_n+1 = ∛(x_n + 1)
x₀ = 1.5
Calculate the terms using the calculator’s Ans button. Type the starting value and press =. Then type the formula with Ans in place of x_n and keep pressing = for each new term. Keep every digit; only round what you write down.
n x_n
0 1.500000
1 1.357209
2 1.330861
3 1.325884
4 1.324939
5 1.324760
6 1.324726
7 1.324719
Stop when the terms agree to 3 decimal places. The last few terms all round to 1.325, so the solution is close to this value.
x₅, x₆, x₇ ≈ 1.3248, 1.3247, 1.3247: all round to 1.325
Show the answer is correct to 3 d.p.: test the two ends of the rounding interval. Any number that rounds to 1.325 lies between 1.3245 and 1.3255. If f changes sign between these, the solution must lie in the interval.
f(1.3245) = 1.3245³ − 1.3245 − 1 = −0.00093
f(1.3255) = 1.3255³ − 1.3255 − 1 = 0.00334
Conclude. There is a change of sign, so the solution is between the two values, and both round to the same number.
x = 1.325 to 3 decimal places

n	x_n
0	1.500000
1	1.357209
2	1.330861
3	1.325884
4	1.324939
5	1.324760
6	1.324726
7	1.324719

Check. Put x = 1.325 into the original equation: 1.325³ − 1.325 − 1 = 0.0012, which is very close to 0. ✓

Watch out. The change-of-sign test must use the ends of the rounding interval (for example 1.3245 and 1.3255), not just the answer itself.

Done

PSUse the iteration formula x_n+1 = 3x_n² + 4 with x₀ = 1 to find a solution of x³ + 4x − 3 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.

Key words. The values converge when they get closer and closer to one number. A root (solution) of f(x) = 0 is a value of x that makes f(x) equal 0.

Method. Use the formula repeatedly until the terms agree to 3 d.p., then prove the answer by a change of sign.

Write the iteration formula and the starting value. We repeat the same calculation, each time putting the last answer back in.
x_n+1 = 3x_n² + 4
x₀ = 1
Calculate the terms using the calculator’s Ans button. Type the starting value and press =. Then type the formula with Ans in place of x_n and keep pressing = for each new term. Keep every digit; only round what you write down.
n x_n
0 1.000000
1 0.600000
2 0.688073
3 0.670624
4 0.674197
5 0.673470
6 0.673618
7 0.673588
8 0.673594
Stop when the terms agree to 3 decimal places. The last few terms all round to 0.674, so the solution is close to this value.
x₆, x₇, x₈ ≈ 0.6736, 0.6736, 0.6736: all round to 0.674
Show the answer is correct to 3 d.p.: test the two ends of the rounding interval. Any number that rounds to 0.674 lies between 0.6735 and 0.6745. If f changes sign between these, the solution must lie in the interval.
f(0.6735) = 0.6735³ + 4 × 0.6735 − 3 = −0.00050
f(0.6745) = 0.6745³ + 4 × 0.6745 − 3 = 0.00486
Conclude. There is a change of sign, so the solution is between the two values, and both round to the same number.
x = 0.674 to 3 decimal places

n	x_n
0	1.000000
1	0.600000
2	0.688073
3	0.670624
4	0.674197
5	0.673470
6	0.673618
7	0.673588
8	0.673594

Check. Put x = 0.674 into the original equation: 0.674³ + 4 × 0.674 − 3 = 0.0022, which is very close to 0. ✓

Watch out. The change-of-sign test must use the ends of the rounding interval (for example 1.3245 and 1.3255), not just the answer itself.

Done

PSUse the iteration formula x_n+1 = √(3x_n + 2) with x₀ = 3 to find a solution of x² − 3x − 2 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.

Key words. The values converge when they get closer and closer to one number. A root (solution) of f(x) = 0 is a value of x that makes f(x) equal 0.

Method. Use the formula repeatedly until the terms agree to 3 d.p., then prove the answer by a change of sign.

Write the iteration formula and the starting value. We repeat the same calculation, each time putting the last answer back in.
x_n+1 = √(3x_n + 2)
x₀ = 3
Calculate the terms using the calculator’s Ans button. Type the starting value and press =. Then type the formula with Ans in place of x_n and keep pressing = for each new term. Keep every digit; only round what you write down.
n x_n
0 3.000000
1 3.316625
2 3.456859
3 3.517183
4 3.542817
5 3.553653
6 3.558224
7 3.560151
… …
12 3.561534
13 3.561545
14 3.561550
Stop when the terms agree to 3 decimal places. The last few terms all round to 3.562, so the solution is close to this value.
x₁₂, x₁₃, x₁₄ ≈ 3.5615, 3.5615, 3.5615: all round to 3.562
Show the answer is correct to 3 d.p.: test the two ends of the rounding interval. Any number that rounds to 3.562 lies between 3.5615 and 3.5625. If f changes sign between these, the solution must lie in the interval.
f(3.5615) = 3.5615² − 3 × 3.5615 − 2 = −0.00022
f(3.5625) = 3.5625² − 3 × 3.5625 − 2 = 0.00391
Conclude. There is a change of sign, so the solution is between the two values, and both round to the same number.
x = 3.562 to 3 decimal places

n	x_n
0	3.000000
1	3.316625
2	3.456859
3	3.517183
4	3.542817
5	3.553653
6	3.558224
7	3.560151
…	…
12	3.561534
13	3.561545
14	3.561550

Check. Put x = 3.562 into the original equation: 3.562² − 3 × 3.562 − 2 = 0.0018, which is very close to 0. ✓

Watch out. The change-of-sign test must use the ends of the rounding interval (for example 1.3245 and 1.3255), not just the answer itself.

Done

PSUse the iteration formula x_n+1 = x_n³5 + 15 with x₀ = 0 to find a solution of x³ − 5x + 1 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.

Key words. The values converge when they get closer and closer to one number. A root (solution) of f(x) = 0 is a value of x that makes f(x) equal 0.

Method. Use the formula repeatedly until the terms agree to 3 d.p., then prove the answer by a change of sign.

Write the iteration formula and the starting value. We repeat the same calculation, each time putting the last answer back in.
x_n+1 = x_n³5 + 15
x₀ = 0
Calculate the terms using the calculator’s Ans button. Type the starting value and press =. Then type the formula with Ans in place of x_n and keep pressing = for each new term. Keep every digit; only round what you write down.
n x_n
0 0.000000
1 0.200000
2 0.201600
3 0.201639
4 0.201640
Stop when the terms agree to 3 decimal places. The last few terms all round to 0.202, so the solution is close to this value.
x₂, x₃, x₄ ≈ 0.2016, 0.2016, 0.2016: all round to 0.202
Show the answer is correct to 3 d.p.: test the two ends of the rounding interval. Any number that rounds to 0.202 lies between 0.2015 and 0.2025. If f changes sign between these, the solution must lie in the interval.
f(0.2015) = 0.2015³ − 5 × 0.2015 + 1 = 0.00068
f(0.2025) = 0.2025³ − 5 × 0.2025 + 1 = −0.00420
Conclude. There is a change of sign, so the solution is between the two values, and both round to the same number.
x = 0.202 to 3 decimal places

n	x_n
0	0.000000
1	0.200000
2	0.201600
3	0.201639
4	0.201640

Check. Put x = 0.202 into the original equation: 0.202³ − 5 × 0.202 + 1 = −0.0018, which is very close to 0. ✓

Watch out. The change-of-sign test must use the ends of the rounding interval (for example 1.3245 and 1.3255), not just the answer itself.

Done

PSUse the iteration formula x_n+1 = ∛(−2x_n + 7) with x₀ = 1.5 to find a solution of x³ + 2x − 7 = 0 correct to 2 decimal places. Show that your answer is correct to 2 decimal places.

Key words. The values converge when they get closer and closer to one number. A root (solution) of f(x) = 0 is a value of x that makes f(x) equal 0.

Method. Use the formula repeatedly until the terms agree to 2 d.p., then prove the answer by a change of sign.

Write the iteration formula and the starting value. We repeat the same calculation, each time putting the last answer back in.
x_n+1 = ∛(−2x_n + 7)
x₀ = 1.5
Calculate the terms using the calculator’s Ans button. Type the starting value and press =. Then type the formula with Ans in place of x_n and keep pressing = for each new term. Keep every digit; only round what you write down.
n x_n
0 1.500000
1 1.587401
2 1.563932
3 1.570303
4 1.568579
5 1.569046
6 1.568919
7 1.568954
Stop when the terms agree to 2 decimal places. The last few terms all round to 1.57, so the solution is close to this value.
x₅, x₆, x₇ ≈ 1.569, 1.569, 1.569: all round to 1.57
Show the answer is correct to 2 d.p.: test the two ends of the rounding interval. Any number that rounds to 1.57 lies between 1.565 and 1.575. If f changes sign between these, the solution must lie in the interval.
f(1.565) = 1.565³ + 2 × 1.565 − 7 = −0.03696
f(1.575) = 1.575³ + 2 × 1.575 − 7 = 0.05698
Conclude. There is a change of sign, so the solution is between the two values, and both round to the same number.
x = 1.57 to 2 decimal places

n	x_n
0	1.500000
1	1.587401
2	1.563932
3	1.570303
4	1.568579
5	1.569046
6	1.568919
7	1.568954

Check. Put x = 1.57 into the original equation: 1.57³ + 2 × 1.57 − 7 = 0.0099, which is very close to 0. ✓

Watch out. The change-of-sign test must use the ends of the rounding interval (for example 1.3245 and 1.3255), not just the answer itself.

Done

RUsing x_n+1 = x_n³ − 1 with x₀ = 1.5, work out x₁, x₂ and x₃. Does this iteration find the solution of x³ − x − 1 = 0 near 1.32? Explain.

Done

PSThe diagram shows the graph of y = √(3x + 2) (blue) and y = x (grey). A staircase has been drawn for the iteration x_n+1 = √(3x_n + 2) with x₀ = 1. Does the iteration converge? Estimate the value it converges to.

Done

PSA drug is given in doses of 50 mg. Between doses 30% of the drug in the body is lost. The amount just after each dose is u_n mg, where u_n+1 = 0.7u_n + 50 and u₀ = 0. Work out u₁ to u₄, and use L = 0.7L + 50 to find the limiting amount.

Done

PSA loan of £5000 has interest of 1% added each month and a repayment of £200 is made at the end of each month. The balance after n months is B_n, where B_n+1 = 1.01B_n − 200 and B₀ = 5000. Work out B₁, B₂ and B₃. After how many months does the balance first fall below £4000?

Method. Apply the rule repeatedly: multiply by 1.01 for the interest, then subtract 200.

Work out the first three balances. Use each balance to find the next.
B₁ = 1.01 × 5000 − 200 = 4850.00
B₂ = 1.01 × 4850.00 − 200 = 4698.50
B₃ = 1.01 × 4698.50 − 200 = 4545.48
Carry on until the balance is below 4000. Keep a table.
n B_n (£)
0 5000.00
1 4850.00
2 4698.50
3 4545.48
4 4390.94
5 4234.85
6 4077.20
7 3917.97
Read the answer. The first balance below £4000 is B₇ = £3917.97.
7 months

n	B_n (£)
0	5000.00
1	4850.00
2	4698.50
3	4545.48
4	4390.94
5	4234.85
6	4077.20
7	3917.97

Check. B₆ = £4077.20 is still above £4000. ✓

Watch out. Apply the interest before subtracting the repayment, as the rule says.

Done

PSA fish farm has P_n fish at the start of year n. Each year the population grows by 10% and then 50 fish are sold, so P_n+1 = 1.1P_n − 50. Work out P₁, P₂ and P₃ when

(a)P₀ = 600

(b)P₀ = 400

Done

RFor the fish farm in the last question, the amount of fish stays the same if P_n+1 = P_n = L. Find L and explain what happens if the farm starts with more than L fish and with fewer than L fish.

Done

PSRavi puts £100 into an account. At the end of each year 3% interest is added and then he adds another £100. Write an iteration formula for the amount u_n after n years with u₀ = 100, and find the amount after 3 years.

Done

RThe iteration x_n+1 = (x_n + 7 ÷ x_n) ÷ 2 with x₀ = 3 is used to find √7. Work out x₁, x₂ and x₃, and show why the limit L is √7.

Done

Rx_n+1 = 1 + 1 ÷ x_n, x₀ = 1. Work out x₁ to x₅. The terms settle on a limit L: show that L² − L − 1 = 0 and find L using the quadratic formula.

Done

Chapter summary

Skill	Method
Function notation	f(a): replace every x by a (brackets). Solve f(x) = k by putting the rule in and solving the equation.
Inverse function	y = f(x); make x the subject; swap y and x. f⁻¹(b) = a when f(a) = b. Graph: reflect in y = x.
Composite function	fg(x) = f(g(x)): g first. Replace x in f by g(x). (fg)⁻¹ = g⁻¹f⁻¹.
Graph transformations	f(x) + a: up; f(x + a): a left; −f(x): reflect in x-axis; f(−x): reflect in y-axis; af(x), f(ax): stretches.
Trigonometric graphs	sin and cos: period 360°, between −1 and 1; tan: period 180°, asymptotes at 90°, 270°. Solve using the inverse key, symmetry and the period.
Exponential graphs	y = k^x passes through (0, 1), is above the x-axis. k > 1 growth; 0 < k < 1 decay.
Growth and decay	Amount = start × (multiplier)ⁿ. Multiplier 1 + r/100 or 1 − r/100. Use a table to find a time.
Circle and tangent	x² + y² = r². Tangent at (a, b): gradient −a/b (perpendicular to the radius); y − b = m(x − a).
Gradient of a curve	Draw a tangent; rise ÷ run using two points on it (instantaneous rate of change). Chord gradient = average rate of change.
Area under a graph	Area = vertical × horizontal (units multiply). Speed–time: area = distance. Estimate curves with equal strips (trapezia or rectangles).
Iteration	x_n+1 = g(x_n) with the Ans key. Show a root with a change of sign at the ends of the rounding interval.
Iterative processes	u_n+1 = au_n + b; the limit L satisfies L = aL + b.

Can you…?

I can use function notation to evaluate f(a) and solve f(x) = k, including with algebraic inputs.
I can find an inverse function using algebra or a function machine, and I know its graph is a reflection in y = x.
I can find composite functions such as fg(x) and solve equations involving them.
I can sketch y = f(x) + a, f(x + a), −f(x) and f(−x) from the graph of y = f(x), and identify these transformations.
I can sketch and describe the graphs of sin, cos and tan, and solve equations such as sin x = 0.3 for a range of angles.
I can sketch exponential graphs and solve growth and decay problems with multipliers.
I can write down the equation of a circle with centre the origin and find the equation of a tangent at a point.
I can estimate the gradient of a curve using a tangent and explain average and instantaneous rates of change.
I can find and estimate areas under graphs and say what the area represents.
I can use an iteration formula, show that a solution lies in an interval, and give a root to a stated accuracy.

Key terms

Function: A rule that gives exactly one output for each input.
Inverse function: The function that undoes f, written f⁻¹.
Composite function: Two functions joined so the output of one is the input of the other, e.g. fg(x) = f(g(x)).
Domain: The set of allowed inputs of a function.
Transformation: A change to a graph such as a translation, reflection or stretch.
Period: The length of one complete cycle of a repeating graph.
Amplitude: Half the distance between the maximum and minimum of a wave.
Asymptote: A line that a graph gets closer and closer to but never touches.
Exponential function: A function with the unknown in the power, such as y = 2^x.
Multiplier: The number you multiply by to apply a percentage change in one step.
Tangent: A straight line that touches a curve (or circle) at exactly one point.
Chord: A straight line joining two points on a curve.
Instantaneous rate of change: The rate of change at one point: the gradient of the tangent.
Average rate of change: The change in the output divided by the change in the input: the gradient of a chord.
Iteration: Repeating a calculation, using each answer as the next input, to approach a solution.
Converge: To get closer and closer to one value.

Total 62 marks. Allow about 62 minutes.

Exercise 25X Exam-style questions

Non-calculator

Ff(x) = 5 − 2x. (a) Work out f(−3). (b) Solve f(x) = 9.[4 marks]

(a)Work out f(−3).

(b)Solve f(x) = 9.

Done

Ff(x) = 3x + 12. Find f⁻¹(x).[3 marks]

Done

PSf(x) = 2x + 3 and g(x) = x² − 4. (a) Find fg(x). (b) Solve gf(x) = 21.[6 marks]

(a)Find fg(x).

(b)Solve gf(x) = 21.

Done

FThe graph of y = f(x) has a minimum point at (3, −2). Write down the coordinates of the minimum point of (a) y = f(x − 1) + 4, (b) y = f(2x), and say what happens to the graph of y = −f(x).[5 marks]

(a)y = f(x − 1) + 4

(b)y = f(2x)

(c)y = −f(x)

Done

FSolve sin x = 0.7 for 0° ≤ x ≤ 360°. Give your answers to 1 decimal place.[3 marks]

Done

FThe graph shows y = a sin(bx) for 0° ≤ x ≤ 360°. Find the values of a and b.[3 marks]

Done

F£6000 is invested for 4 years at 2.5% per year compound interest. Work out the value of the investment at the end of 4 years. Give your answer to the nearest penny.[3 marks]

Done

PSA van costs £18 000 new. Its value falls by 14% each year. (a) Work out its value after 5 years, to the nearest £. (b) After how many whole years is it first worth less than £8000?[5 marks]

(a)Value after 5 years

(b)First year below £8000

Method. Use the multiplier repeatedly, building a table, until the amount falls to the target.

Find the multiplier. Each year the amount is multiplied by the same number.
86% = 0.86
Work out the amount at the end of each year. Multiply by 0.86 each time. Stop as soon as the amount falls to or below £8000.
After n years Amount
0 £18000
1 £15480
2 £13313
3 £11449
4 £9846
5 £8468
6 £7282
Read off the answer. The first row that is at or below £8000 is after 6 years.
n = 6

After n years	Amount
0	£18000
1	£15480
2	£13313
3	£11449
4	£9846
5	£8468
6	£7282

Check. Use the formula: £18000 × 0.86⁶ = £7282 (below £8000) but after 5 years it is £8468 (not yet there). ✓

Watch out. Count the years carefully. The row marked 0 is the start, before any year has passed.

Marks. 1 mark for a method of repeated multiplication, 1 mark for the correct answer.

Done

FThe circle C has equation x² + y² = 169. The point P(−5, 12) is on C. Find the equation of the tangent to C at P.[4 marks]

Done

RShow that the line 3x + 4y = 25 is a tangent to the circle x² + y² = 25.[4 marks]

Done

FThe graph shows the distance travelled by a cyclist, s metres, after t seconds. Draw a tangent at t = 4 and use it to estimate the speed of the cyclist after 4 seconds. Give the units of your answer.[3 marks]

Done

PSThe speed of a particle is v = 16 − t² m/s for 0 ≤ t ≤ 4. Use 4 strips of equal width to estimate the distance travelled in the 4 seconds. Is your answer an under-estimate or an over-estimate? Give a reason.[4 marks]

Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).

Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.

Split the area into strips of equal width. The region from time = 0 to 4 is 4 wide. With 4 strips, each strip has width 4 ÷ 4 = 1.
Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 1 2 3 4
speed 16 15 12 7 0
Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.
Strip 1: ½ × (16 + 15) × 1 = 15.5
Strip 2: ½ × (15 + 12) × 1 = 13.5
Strip 3: ½ × (12 + 7) × 1 = 9.5
Strip 4: ½ × (7 + 0) × 1 = 3.5
Add the strip areas to get the estimate.
15.5 + 13.5 + 9.5 + 3.5 = 42
Comment on the accuracy. The curve bends downwards (concave), so each trapezium top lies below the curve. So the estimate is an under-estimate. More strips would give a better estimate.
(The exact area is 42.67 m, so the estimate is smaller.)

time	0	1	2	3	4
speed	16	15	12	7	0

Check. Use the trapezium rule formula: ½ × 1 × [16 + 0 + 2 × (15 + 12 + 7)] = 42.00. ✓

Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.

Marks. 1 mark for the speeds 16, 15, 12, 7, 0, 1 mark for the trapezium areas, 1 mark for 42 m, 1 mark for under-estimate with the reason (the curve bends downwards).

Done

PS(a) Show that x³ + 4x − 3 = 0 has a solution between 0 and 1. (b) Use the iteration formula x_n+1 = 3 ÷ (x_n² + 4) with x₀ = 1 to find the solution correct to 3 decimal places. Show that your answer is correct to 3 decimal places.[6 marks]

(a)Show there is a solution between 0 and 1.

(b)Find the solution to 3 d.p.

Key words. The values converge when they get closer and closer to one number. A root (solution) of f(x) = 0 is a value of x that makes f(x) equal 0.

Method. Use the formula repeatedly until the terms agree to 3 d.p., then prove the answer by a change of sign.

Write the iteration formula and the starting value. We repeat the same calculation, each time putting the last answer back in.
x_n+1 = 3x_n² + 4
x₀ = 1
Calculate the terms using the calculator’s Ans button. Type the starting value and press =. Then type the formula with Ans in place of x_n and keep pressing = for each new term. Keep every digit; only round what you write down.
n x_n
0 1.000000
1 0.600000
2 0.688073
3 0.670624
4 0.674197
5 0.673470
6 0.673618
7 0.673588
8 0.673594
Stop when the terms agree to 3 decimal places. The last few terms all round to 0.674, so the solution is close to this value.
x₆, x₇, x₈ ≈ 0.6736, 0.6736, 0.6736: all round to 0.674
Show the answer is correct to 3 d.p.: test the two ends of the rounding interval. Any number that rounds to 0.674 lies between 0.6735 and 0.6745. If f changes sign between these, the solution must lie in the interval.
f(0.6735) = 0.6735³ + 4 × 0.6735 − 3 = −0.00050
f(0.6745) = 0.6745³ + 4 × 0.6745 − 3 = 0.00486
Conclude. There is a change of sign, so the solution is between the two values, and both round to the same number.
x = 0.674 to 3 decimal places

n	x_n
0	1.000000
1	0.600000
2	0.688073
3	0.670624
4	0.674197
5	0.673470
6	0.673618
7	0.673588
8	0.673594

Check. Put x = 0.674 into the original equation: 0.674³ + 4 × 0.674 − 3 = 0.0022, which is very close to 0. ✓

Watch out. The change-of-sign test must use the ends of the rounding interval (for example 1.3245 and 1.3255), not just the answer itself.

Marks. 1 mark for the terms, 1 mark for 0.674, 2 marks for the change-of-sign test at 0.6735 and 0.6745.

Done

RA population of 2000 insects grows by 5% each day. Show that it takes 15 days for the population to double (reach at least 4000).[4 marks]

Method. Use the multiplier repeatedly, building a table, until the amount reaches the target.

Find the multiplier. Each year the amount is multiplied by the same number.
105% = 1.05
Work out the amount at the end of each year. Multiply by 1.05 each time. Stop as soon as the amount reaches or passes 4000.
After n years Amount
0 2000
1 2100
2 2205
12 3592
13 3771
14 3960
15 4158
Read off the answer. The first row that reaches or passes 4000 is after 15 years.
n = 15

After n years	Amount
0	2000
1	2100
2	2205
12	3592
13	3771
14	3960
15	4158

Check. Use the formula: 2000 × 1.05¹⁵ = 4158 (above 4000) but after 14 years it is 3960 (not yet there). ✓

Watch out. Count the years carefully. The row marked 0 is the start, before any year has passed.

Marks. 1 mark for the multiplier 1.05, 1 mark for 2000 × 1.05¹⁴ < 4000, 1 mark for 2000 × 1.05¹⁵ > 4000, 1 mark for the conclusion.

Done

Rf(x) = x + 5 and g(x) = 3x − 2. Find (fg)⁻¹(x) and show that it is the same as g⁻¹f⁻¹(x).[5 marks]

Done

← Chapter 24: Statistics Chapter 26: Circle theorems →

GCSE Maths Functions and Advanced Graphs (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Function notation

Key ideas

Watch out

Exercise 25A Function notation

Inverse functions

Key ideas

Watch out

Exercise 25B Inverse functions

Composite functions

Key ideas

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Exercise 25C Composite functions

Graph transformations

Key ideas

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Exercise 25D Graph transformations

Trigonometric graphs

Key ideas

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Exercise 25E Trigonometric graphs

Exponential graphs, growth and decay

Key ideas

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Exercise 25F Exponential graphs, growth and decay

The equation of a circle and tangents

Key ideas

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Exercise 25G Circles and tangents

Gradients of curves and rates of change

Key ideas

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Exercise 25H Gradients of curves and rates of change

Areas under graphs

Key ideas

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Exercise 25I Areas under graphs

Iteration and iterative processes

Key ideas

Watch out

Exercise 25J Iteration and iterative processes

Chapter summary

Can you…?

Key terms

Exercise 25Z Check-up (non-calculator)

Exercise 25X Exam-style questions