How to use this menu
Six original tasks: two for criterion A (a short check and an end-of-unit test), one investigation for B, one communication task for C, and two real-life tasks for D that also assess C. Set at least one task for each criterion across the unit; nobody needs all six. The timings spread assessment through the unit rather than leaving it all to the end.
Descriptors: Year 10 Extended is assessed with the MYP Year 5 descriptors, paraphrased below each task. Check them against your school's current guide.
The menu at a glance
| Task | Criteria | When | Time | Format |
|---|---|---|---|---|
| A1: Quick check: functions, factorising and the vertex | A | After lesson 3 | 25 minutes | Individual, written, in class. 19 marks. |
| A2: End-of-unit test | A | End of unit, after lesson 7 | 50 minutes | Individual, written, in class, one lesson. 32 marks in three sections. |
| B1: Straight lines through a parabola | B | Mid-unit, after lesson 4 | 50 minutes | In class. Students may collect results in pairs, but each writes up their own findings. |
| C1: Explain it to someone who missed it | C | After lesson 6 | 35 minutes | Individual, written, in class or as homework. One page of words, equations and a sketch. |
| D1: A vegetable plot against a wall | D and C | End of unit, after lesson 7 | 50 minutes | Individual, in class. A written solution with explanations and a short note to the grower. |
| D2: Pricing a river tour | D and C | End of unit, after lesson 7 | 50 minutes | Individual, in class, one lesson. A written solution with explanations and a short reply to the operator. |
D1 and D2 are alternatives. Both assess criteria D and C, in two different real-life contexts: one is set, not both.
The unit's inquiry
- Key concept. Relationships: how an input, a rule and an output are tied together, and how a change to the rule changes the graph.
- Related concepts. Model, Equivalence, Change
- Global context. Scientific and technical innovation: modelling paths, areas and prices with functions so that a decision can be tested before it is made.
- Statement of inquiry. A function links an input to an output; one quadratic can be written in several forms, and each form answers a different question about the same graph.
MYP Year 5 descriptors, paraphrased
Each task's clarification, on its student sheet below, says what these levels look like in that task.
| Level | Criterion A: Knowing and understanding |
|---|---|
| 1–2 | Selects suitable mathematics for simple problems in familiar situations and applies it with some success. |
| 3–4 | Selects suitable mathematics for more complex problems in familiar situations and applies it successfully. |
| 5–6 | Selects suitable mathematics for challenging problems in familiar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| 7–8 | Selects suitable mathematics for challenging problems in both familiar and unfamiliar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| Level | Criterion B: Investigating patterns |
|---|---|
| 1–2 | With support, applies problem-solving techniques to find simple patterns and states predictions that fit them. |
| 3–4 | Applies problem-solving techniques to find simple patterns and suggests general rules that fit the findings. |
| 5–6 | Applies problem-solving techniques to find complex patterns, describes them as general rules that fit the findings and verifies that the rules hold. |
| 7–8 | Selects and applies problem-solving techniques to find complex patterns, describes them as general rules that fit correct findings, and proves, or verifies and justifies, those rules. |
| Level | Criterion C: Communicating |
|---|---|
| 1–2 | Uses limited mathematical language and limited forms of representation; the reasoning is hard to follow. |
| 3–4 | Uses some appropriate mathematical language and forms of representation; the reasoning can be followed and the work is adequately organised. |
| 5–6 | Usually uses appropriate language, moves between forms of representation with some success, and the reasoning is complete and coherent in a mostly logical structure. |
| 7–8 | Consistently uses appropriate language, moves effectively between forms of representation, and the reasoning is complete, coherent and concise in a logical structure. |
| Level | Criterion D: Applying mathematics in real-life contexts |
|---|---|
| 1–2 | Identifies some relevant elements of the real-life situation and selects, with some success, strategies to model it. |
| 3–4 | Identifies the relevant elements, selects adequate strategies to model the situation, applies them to reach a solution and discusses whether it makes sense. |
| 5–6 | Identifies the relevant elements, selects adequate modelling strategies and applies them to reach a valid solution, explains the degree of accuracy and explains whether the solution makes sense. |
| 7–8 | Identifies the relevant elements, selects appropriate modelling strategies and applies them successfully to reach a correct solution, justifies the degree of accuracy and justifies whether the solution makes sense. |
Task A1: Quick check: functions, factorising and the vertex
Why set it. A short check of lessons 1 to 3 while there is still time to reteach: domain and range, function notation with an expression as input, factorising with a leading coefficient above 1, and completing the square. Questions 7 and 8 are unfamiliar, so the top band is open to everyone.
When: After lesson 3. Time: 25 minutes. Format: Individual, written, in class. 19 marks.
The student sheet, with its task-specific clarification
When: After lesson 3. Time: 25 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
Answer every question you can and show your working: a correct method earns credit even if the final answer slips. Questions 7 and 8 are meant to make you think, so try something rather than leaving them blank.
Part 1. Functions
- g(x) = √(x + 3) − 2. State the largest set of inputs that gives a real output, and state the range of g. (2 marks)
- f(x) = x2 + 4x. Write f(t − 1) as an expression in t, simplified. (2 marks)
Part 2. Factorising
- Factorise 6x2 + x − 15. (3 marks)
- Factorise 12x2 − 27 completely. (2 marks)
- Explain why 3x2 + 5x + 1 cannot be written as a product of two brackets with whole-number coefficients. (2 marks)
Part 3. Forms of a quadratic
- Write 3x2 + 12x + 2 in the form a(x − h)2 + k and state the vertex. (3 marks)
Part 4. Think it through
- A ball is thrown from a platform. Its height after t seconds is h = −5t2 + 20t + 1, in metres. When is the ball highest, and how high is it then? Show how you know it is the greatest value. (3 marks)
- Show that x2 − 6x + 11 is never zero or negative, whatever the value of x. (2 marks)
How your work is judged
| Level | Criterion A in task A1: what it looks like |
|---|---|
| 1–2 | You answer some of questions 1 to 4 correctly, for example the domain in question 1 or the common factor in question 4. |
| 3–4 | You answer most of questions 1 to 6 correctly, using a method you can state, such as the pair method for question 3 or halving the x coefficient for question 6. |
| 5–6 | You answer questions 1 to 6 correctly, with working that shows your method, and make a good start on question 7 or 8. |
| 7–8 | You answer questions 1 to 6 correctly and solve questions 7 and 8 with reasoning that shows your answer is the greatest (or the least) value. |
Markscheme
| Task A1 question | Answer and marking |
|---|---|
| 1 | Domain x ≥ −3; range g(x) ≥ −2. 1 mark each. The root needs x + 3 ≥ 0; the root itself is at least 0, so g is at least −2, and −2 is reached at x = −3. x > −3 or g > −2 loses the end point. |
| 2 | t2 + 2t − 3. 1 mark for (t − 1)2 + 4(t − 1), 1 mark for the simplified result: t2 − 2t + 1 + 4t − 4. A common slip is (t − 1)2 = t2 − 1. |
| 3 | (2x − 3)(3x + 5). ac = −90, and the pair with sum 1 is 10 and −9: 6x2 + 10x − 9x − 15. 1 mark for the pair, 1 for the grouping, 1 for the brackets. Expanding to check is good practice. |
| 4 | 3(2x − 3)(2x + 3). 1 mark for the common factor 3, 1 for the difference of two squares. 3(4x2 − 9) alone earns 1. |
| 5 | ac = 3, so the pair must multiply to 3 and add to 5. The pairs of whole numbers with product 3 are 1 and 3 (sum 4) and −1 and −3 (sum −4): neither adds to 5. 1 mark for the target, 1 for checking the pairs. The discriminant 25 − 12 = 13 is not a square: also accept it, as a full answer. |
| 6 | 3(x + 2)2 − 10; vertex (−2, −10). 3(x2 + 4x) + 2 = 3((x + 2)2 − 4) + 2. 1 mark for a = 3 and h = −2, 1 for k = −10, 1 for the vertex. −2 comes from forgetting to multiply the 4 by 3. |
| 7 | 2 s; 21 m. h = −5(t − 2)2 + 21. 1 mark for the vertex form (or the vertex from t = −b ÷ 2a = 2), 1 for t = 2, 1 for 21 with the reason: the square term is never positive, so h is at most 21. |
| 8 | x2 − 6x + 11 = (x − 3)2 + 2. 1 mark for the completed square. 1 mark for the reason: a square is at least 0, so the whole is at least 2. Trying a few values is not a reason; the discriminant 36 − 44 < 0 is also accepted. |
| Marks to a level | Total 19. 1 to 5 marks: 1–2. 6 to 11: 3–4. 12 to 16: 5–6. 17 to 19, with at least 4 of the 5 marks on questions 7 and 8: 7–8. Use the marks as a guide, then check the student's profile against the clarification before settling the level. |
Exemplar answers
| Task A1 response | What it looks like |
|---|---|
| Level 3–4 profile | Question 1 right (2 marks); question 2 written as (t − 1)2 + 4(t − 1) then expanded to t2 + 4t − 5 (1 mark); question 3 finds the pair 10 and −9 and stops (1 mark); question 4 takes out 3 and stops at 3(4x2 − 9) (1 mark); question 5 blank; question 6 gives 3(x + 2)2 − 2 (2 marks); nothing on 7 or 8. 7 marks, matching 3–4: methods started, little carried to the end, nothing unfamiliar. |
| Level 7–8 answer to question 8 | “x2 − 6x + 11 = (x − 3)2 − 9 + 11 = (x − 3)2 + 2. A square can never be below 0, so the expression is at least 2, and its smallest value 2 comes at x = 3. It can never reach 0.” |
| Level 7–8 answer to question 7 | “h = −5(t2 − 4t) + 1 = −5((t − 2)2 − 4) + 1 = −5(t − 2)2 + 21. The bracket squared is never negative, so the most it can take away from 21 is 0, which happens at t = 2. The ball is highest after 2 s at 21 m.” |
Differentiation
A way in for anyone stuck
- Give the three-column pair table from lesson 2 (factors of ac, their sum) for questions 3 and 5.
- Offer a blank vertex-form frame a(x − …)2 + … for questions 6 and 7.
- Read questions 7 and 8 aloud on request and suggest writing the expression in vertex form first. Reading is not what is being assessed.
Stretch
- Question 7 again: when is the ball back at its starting height of 1 m? (4 s, by symmetry about t = 2.)
- For which values of c is x2 − 6x + c never zero or negative? (c > 9.)
Students learning in English as an additional language
| Word in task A1 | Plain meaning |
|---|---|
| domain | all the inputs a function can accept |
| range | all the outputs a function can give |
| vertex | the turning point of a parabola, its highest or lowest point |
| factorise | write as a product of brackets |
| leading coefficient | the number multiplying the x2 term |
Sentence starters for task A1:
- The inputs must be … because …
- I need two numbers with product … and sum …
- The greatest value is … because the square is never …
Task A2: End-of-unit test
Why set it. The whole unit at three levels of demand. Section 3 is unfamiliar on purpose: it is where the evidence for 7–8 comes from.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, written, in class, one lesson. 32 marks in three sections.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
Three sections, from familiar questions to ones you have not met in this form. Show your working everywhere: it earns credit even when an answer slips. Give exact answers (a root or a fraction) unless the question asks for a decimal.
Section 1. Familiar questions
- h(x) = 1 ÷ (x + 5). State the domain and the range of h. (2 marks)
- f(x) = x2 + 1. Write f(2a − 1) as an expression in a, simplified. (2 marks)
- Factorise 4x2 − 4x − 15. (3 marks)
- Solve 2x2 − 3x − 20 = 0 by factorising. (2 marks)
- Write x2 + 8x + 11 in vertex form and give the vertex. (2 marks)
Section 2. More complex questions
- Solve 3x2 − 4x − 2 = 0. Give both solutions exactly, in their simplest form. (3 marks)
- A quadratic graph crosses the x-axis at −1 and at 5, and also goes through the point (0, −10). Write its equation as y = ax2 + bx + c. (3 marks)
- (a) For which values of k does x2 + kx + 49 = 0 have exactly one solution? (b) For which values of k does it have no real solution? (3 marks)
- The point (2, 5) is on the graph of y = f(x). Give the image of this point on (a) y = f(x − 3) + 1, (b) y = −f(x), (c) y = f(2x). (3 marks)
Section 3. Unfamiliar questions
- The graph of y = x2 goes through two steps: a stretch by scale factor 4 parallel to the y-axis, and a translation 1 up. Write the equation that results when the stretch comes first, and when the translation comes first. Are they the same? Name one pair of transformations where the order can never matter. (3 marks)
- A water jet from a park fountain leaves the nozzle at ground level, lands 8 m away, and reaches its highest point, 3 m above the ground, halfway between. Find an equation for its path, with x the distance from the nozzle in metres. How high is the jet 2 m from the nozzle? (4 marks)
- Aroon says: “If a and c have opposite signs, then ax2 + bx + c = 0 always has two solutions.” Is he right? Explain. (2 marks)
How your work is judged
| Level | Criterion A in task A2: what it looks like |
|---|---|
| 1–2 | You answer some Section 1 questions correctly, for example the domain of h or a factor pair. |
| 3–4 | You answer most of Section 1 correctly and make progress in Section 2, for example the discriminant in question 8 or the zeros in question 7. |
| 5–6 | You answer Section 1 and most of Section 2 correctly, with clear methods, and make a start in Section 3. |
| 7–8 | You answer Sections 1 and 2 correctly and solve most of Section 3, with reasoning that explains why the result is true for every case. |
Markscheme
| Task A2 question | Answer and marking |
|---|---|
| 1 | Domain x ≠ −5; range y ≠ 0. 1 mark each. The range excludes 0 because 1 ÷ (x + 5) is never 0. |
| 2 | 4a2 − 4a + 2. 1 mark for (2a − 1)2 + 1, 1 mark for the expansion. 4a2 + 1 is the usual slip. |
| 3 | (2x − 5)(2x + 3). ac = −60, the pair with sum −4 is 6 and −10: 4x2 + 6x − 10x − 15. 1 mark for the pair, 1 for the grouping, 1 for the brackets. |
| 4 | x = 4 or x = −5 ÷ 2. (2x + 5)(x − 4) = 0. 1 mark for the factorisation, 1 for both solutions. |
| 5 | (x + 4)2 − 5; vertex (−4, −5). 1 mark each. (x + 4)2 − 16 + 11 is the working. |
| 6 | x = (2 + √10) ÷ 3 or x = (2 − √10) ÷ 3. Discriminant 16 + 24 = 40, √40 = 2√10, and (4 plus or minus 2√10) ÷ 6 simplifies by 2. 1 mark for the discriminant, 1 for the formula, 1 for the simplified exact form. About 1.72 and −0.39. |
| 7 | y = 2x2 − 8x − 10. y = a(x + 1)(x − 5); at x = 0, −5a = −10, a = 2. 1 mark for the factored form, 1 for a = 2, 1 for the expansion. |
| 8 | (a) k = 14 or k = −14. (b) −14 < k < 14. The discriminant is k2 − 196. 1 mark for the discriminant, 1 for both values, 1 for the range. |
| 9 | (a) (5, 6) (b) (2, −5) (c) (1, 5). 1 mark each. For (c) the input is doubled, so x halves. |
| 10 | Stretch first: y = 4x2 + 1. Translation first: y = 4(x2 + 1) = 4x2 + 4. Not the same. 1 mark for each equation, 1 for the conclusion that the order matters. Two translations (one horizontal, one vertical), or two stretches in different directions, can be done in either order. |
| 11 | y = −(3 ÷ 16)x(x − 8); 2.25 m high. Zeros at 0 and 8, so the vertex is at x = 4 with y = 3. 1 mark for the zeros, 1 for the vertex at x = 4, 1 for a = −3 ÷ 16 (from 3 = a × 4 × (−4)), 1 for 2.25 m, with the unit. Vertex form y = −(3 ÷ 16)(x − 4)2 + 3 is the same. |
| 12 | Yes. If a and c have opposite signs then ac < 0, so −4ac > 0 and b2 − 4ac is greater than 0 whatever b is: two solutions. 1 mark for the discriminant, 1 for the sign argument. Testing a few examples is not a proof. |
| Marks to a level | Total 32. 1 to 7 marks: 1–2. 8 to 15: 3–4. 16 to 25: 5–6. 26 to 32, with at least 5 of Section 3's 9 marks: 7–8. Check the profile against the clarification before settling the level. |
Exemplar answers
| Task A2 response | What it looks like |
|---|---|
| Level 5–6 profile | Section 1 all correct (11 marks). Section 2: question 6 gives (4 plus or minus √40) ÷ 6 not simplified (2); question 7 right (3); question 8 finds k = 14 and k = −14 but not the range (2); question 9 right (3): 10 marks. Section 3: question 10 gets both equations but no conclusion (2). 23 marks, and the profile matches 5–6: Sections 1 and 2 mostly secure, a start on Section 3. |
| Level 7–8 answer to question 12 | “Right. The discriminant is b2 − 4ac. If a and c have opposite signs, ac is negative, so −4ac is positive. b2 is never negative, so the discriminant is always positive, and a positive discriminant means two solutions. For example 2x2 + x − 3 has discriminant 1 + 24 = 25.” |
Differentiation
A way in for anyone stuck
- Read Section 3 aloud on request: reading is not what is being assessed.
- Offer the three forms of a quadratic on a reference card, and a blank pair table for questions 3 and 4.
- For question 11, give the sketch of a parabola on axes marked 0 to 8, with the nozzle and the landing point labelled.
Stretch
- Question 11 with a second jet that lands 12 m away and peaks at 6 m. Find its equation, then the distances from the nozzle at which it is 4.5 m high. (y = −x(x − 12) ÷ 6; x = 3 and x = 9.)
- Show that x2 + 6x + 9 touches the x-axis at one point, and describe the transformation of y = x2 that gives it. (It is (x + 3)2: a translation of 3 left, touching at x = −3.)
Students learning in English as an additional language
| Word in task A2 | Plain meaning |
|---|---|
| discriminant | the part of the quadratic formula under the square root, b2 − 4ac |
| vertex form | a quadratic written as a(x − h)2 + k, so the vertex is (h, k) |
| image | the point or graph you get after a transformation |
| translation | a move with no turning or stretching |
| scale factor | the number a length is multiplied by in a stretch |
Sentence starters for task A2:
- The discriminant is …, so there are … solutions because …
- The zeros are … and …, so y = a(x …)(x …)
- The order matters here because …
Task B1: Straight lines through a parabola
Why set it. A real pattern hunt that builds on lessons 2 and 3 (factorised form, the roots of a quadratic) without the unit having taught the pattern. The sheet never names it: students organise their results, find it, describe it as a rule, test it on other parabolas and prove it with letters.
When: Mid-unit, after lesson 4. Time: 50 minutes. Format: In class. Students may collect results in pairs, but each writes up their own findings.
The student sheet, with its task-specific clarification
In MYP Year 5, support you give (a prompt, a table outline) is allowed at 1 to 2 only; from 3 to 4 upwards the pattern-finding must be the student's own. Note any support on the student's work.
When: Mid-unit, after lesson 4. Time: 50 minutes. Calculator: no calculator. Criterion: B, investigating patterns.
Take two points on the parabola y = x2, for example the points with x = 1 and x = 2. They are (1, 1) and (2, 4). A straight line goes through both. Its equation y = mx + c depends on the two x-values you chose. This task is about how.
Part 1. Collect results
- For each pair of x-values, give the two points on y = x2 and then the equation of the straight line through them: (1 and 2), (1 and 3), (2 and 3), (2 and 4). Record every pair, its points and the values of m and c in a table you design yourself.
- Do the same for two pairs that include a negative x-value: (−1 and 3) and (−2 and 2).
Part 2. Describe what you find
- Look at the values of m and c in your table. Describe what you see, in words, as a rule another student could use to write down the line for any pair of x-values.
- Use your rule to predict the line through the points at x = 5 and x = 7. Then work the line out from the two points, to check.
Part 3. Test your rule
- Without finding the points, predict m and c for the line through x = −4 and x = 6. Then check by finding the points.
- Choose a different parabola, such as y = x2 + 3 or y = 2x2. Collect enough lines to test whether your rule, or one like it, works. Say what is the same and what is different.
Part 4. Explain
- Take the points with x = a and x = b on y = x2. Use algebra to show why your rule must be true for any a and b.
- For one of your lines, set x2 equal to mx + c and solve the equation. What do you find, and why does it have to happen?
How your work is judged
| Level | Criterion B in task B1: what it looks like |
|---|---|
| 1–2 | With help, you find the points for some pairs and the equation of some lines, and say something you see in your table. |
| 3–4 | You complete the table and suggest a rule for the gradient or for the intercept that fits most of your lines. |
| 5–6 | You organise your results, describe a rule for both m and c that fits all your lines, and use it to predict and check the line for x = 5 and x = 7. |
| 7–8 | You organise your results, describe a rule that fits all your findings, use it to predict correctly for x = −4 and x = 6, test it on a second parabola saying what stays the same and what changes, and justify the rule with algebra. |
Markscheme
| Task B1 question | Answer and marking |
|---|---|
| The pattern (do not share it before the task) | For y = x2, the line through the points at x = a and x = b has gradient a + b and y-intercept −ab. For y = x2 + 3 the gradient is the same and the intercept is −ab + 3. For y = 2x2 the gradient is 2(a + b) and the intercept is −2ab. |
| 1 | (1, 2): (1, 1) and (2, 4), y = 3x − 2. (1, 3): (1, 1) and (3, 9), y = 4x − 3. (2, 3): (2, 4) and (3, 9), y = 5x − 6. (2, 4): (2, 4) and (4, 16), y = 6x − 8. |
| 2 | (−1, 3): (−1, 1) and (3, 9), y = 2x + 3. (−2, 2): (−2, 4) and (2, 4), y = 4 (m = 0). The second is a horizontal line: the rule gives m = −2 + 2 = 0 and c = 4. |
| 3 | Any correct description for both m and c, for example “m is the two x-values added, c is minus the two x-values multiplied”. A rule for m only is evidence for 3–4. A rule that works for positive pairs only is a 3–4 answer until it is checked on question 2. |
| 4 | y = 12x − 35. m = 5 + 7 = 12, c = −35. Check: (5, 25) and (7, 49) give m = 24 ÷ 2 = 12 and c = 25 − 60 = −35. |
| 5 | m = 2 and c = 24. m = −4 + 6 = 2, c = −(−4 × 6) = 24. Check: (−4, 16) and (6, 36) give m = 20 ÷ 10 = 2 and c = 16 + 8 = 24. |
| 6 | For y = x2 + 3 the gradient is unchanged and every intercept moves up by 3 (the whole graph moved up 3). For y = 2x2 both m and c are doubled. Credit a test that collects at least four lines and says clearly what the rule needs to change. At 7–8 expect the link to the transformations of lessons 6 and 7. |
| 7 | m = (b2 − a2) ÷ (b − a) = (b − a)(b + a) ÷ (b − a) = a + b, valid as a ≠ b. Then c = a2 − (a + b)a = −ab. 1 mark for the gradient, 1 for the cancellation, 1 for c. Expect this only at 7–8. |
| 8 | The solutions are exactly the two x-values. For y = 3x − 2: x2 − 3x + 2 = 0, which factorises as (x − 1)(x − 2), so x = 1 or x = 2. It has to: the line goes through the two points, so those x-values satisfy the equation, and a quadratic has at most two solutions. In general x2 − (a + b)x + ab = (x − a)(x − b). |
Exemplar answers
| Task B1 response | What it looks like |
|---|---|
| Level 3–4 | “The lines are y = 3x − 2, y = 4x − 3, y = 5x − 6, y = 6x − 8. The gradient is the two numbers added. For 5 and 7 I think m = 12.” The table is right and the gradient rule works, but the intercept is not described, c = −35 is not predicted, and the negative pairs are missed. |
| Level 7–8 | “For x = a and x = b, the gradient is a + b and the intercept is −ab. For −4 and 6: m = 2, c = 24, which I checked from (−4, 16) and (6, 36). For y = x2 + 3 the gradient was the same but every c was 3 more, because the graph is moved up 3. Proof: m = (b2 − a2) ÷ (b − a) = (b − a)(b + a) ÷ (b − a) = a + b, and c = a2 − (a + b)a = −ab. Then x2 = (a + b)x − ab gives (x − a)(x − b) = 0, so the line meets the parabola at exactly the two points I started with.” |
Differentiation
A way in for anyone stuck
- Give a table outline with columns for the pair, the two points, the gradient and the intercept, and let the student write the headings.
- Suggest finding the gradient first from the two points, then c by putting one point into y = mx + c.
- Prompt with a question, never the rule: “What do you get for m when the two x-values are 1 and 2? And for 1 and 3?”
Stretch
- What happens to the line when the two x-values are the same, say 3 and 3? Which line does your rule give, and what is it called? (m = 6, c = −9: the tangent y = 6x − 9.)
- Find a rule for y = x2 − 4x, using the same approach with its own pairs of x-values. (m = a + b − 4, and c = −ab.)
Students learning in English as an additional language
| Word in task B1 | Plain meaning |
|---|---|
| parabola | the curved graph of a quadratic function |
| gradient | how steep a line is; the amount y changes when x goes up by 1 |
| intercept | where a line crosses an axis; the y-intercept is the value of c in y = mx + c |
| predict | say what will happen before you check |
| rule | a sentence that tells you what happens every time |
| prove | show with a chain of reasons that something is true for every case |
Sentence starters for task B1:
- For the pair … and … I found m = … and c = …
- My rule is: …
- I tested my rule on … and it worked / did not work because …
Task C1: Explain it to someone who missed it
Why set it. Criterion C on its own: notation, a labelled graph and a line of reasoning a classmate can follow, built around two real errors from lessons 3 and 5.
When: After lesson 6. Time: 35 minutes. Format: Individual, written, in class or as homework. One page of words, equations and a sketch.
The student sheet, with its task-specific clarification
When: After lesson 6. Time: 35 minutes. Calculator: no calculator. Criterion: C, communicating.
Mei was away for lessons 3 to 5. She copied this working from a classmate, Zara, and it has mistakes in it.
Zara’s working
(a) y = 2x2 + 12x + 5 = 2(x + 6)2 + 5 − 36 = 2(x + 6)2 − 31, so the vertex is (6, −31).
(b) Solve x2 = 4x. Divide both sides by x: x = 4.
Write a one-page explanation for Mei. She should be able to follow it on her own, without asking anyone.
Part 1. Put it right
- Find every mistake in Zara’s working. For each one, say what she did, why it is wrong, and write the correct working.
Part 2. Teach the idea
- Explain to Mei how to write a quadratic with a leading coefficient above 1 in vertex form. Use Zara’s function, one step on each line, and say which idea each step uses.
- Draw a sketch of y = 2x2 + 12x + 5 that supports your corrected answer to (a): label the vertex, the axis of symmetry and the y-intercept.
Part 3. Check your explanation
- Make up one equation of your own where dividing both sides by x would lose a solution. Solve it correctly, and say which solution would be lost and why.
- Read your page as if you were Mei. Check that every equals sign joins equal things, that you used the words vertex, factor and solution correctly, and that your sketch matches your working.
How your work is judged
| Level | Criterion C in task C1: what it looks like |
|---|---|
| 1–2 | Some of your explanation can be followed, but steps or words are missing, and your sketch or notation is confusing. |
| 3–4 | You explain both of Zara’s calculations with some correct words and notation; your steps are complete and in a sensible order, so Mei could follow most of it. |
| 5–6 | You correct every mistake and explain why each is wrong; your sketch is labelled and agrees with your working, you usually use the right words and notation, and the page is mostly well ordered. |
| 7–8 | Every step is explained, your sketch and working match your words, you always use correct notation (an equals sign only between equal values, brackets and squares in the right places) and the key words, and the page is ordered so Mei could learn from it alone. |
Markscheme
| Task C1 question | Answer and marking |
|---|---|
| The mistakes in (a) | Zara took half of 12 instead of half of 6 inside the bracket, did not multiply the number she subtracted by the 2 in front, and read the vertex as (6, −31) with the sign of h wrong. Correct: 2(x2 + 6x) + 5 = 2((x + 3)2 − 9) + 5 = 2(x + 3)2 − 13, vertex (−3, −13). Her first equals sign is also false, since 2(x + 6)2 is not equal to 2x2 + 12x + 5. |
| The mistake in (b) | Dividing by x loses the solution x = 0 (and would divide by zero there). Correct: x2 − 4x = 0, so x(x − 4) = 0, giving x = 0 or x = 4. |
| The sketch | A parabola opening upward, vertex marked (−3, −13), the axis x = −3 as a dashed line, and the y-intercept (0, 5). Accept zeros not shown; they are not needed here. |
| Own equation (Part 3) | Any equation of the form x2 = kx, or x(x − p) = qx, with both solutions given: for example x2 = 9x gives x = 0 or x = 9. The student should say that x = 0 is the lost solution. |
| How to level | Level the page as a whole against the clarification, not by counting corrections. Notation, an agreeing sketch and ordering decide between 5–6 and 7–8 more often than the corrections do. Mathematical slips count only where they break the line of reasoning Mei has to follow; accuracy itself is criterion A. |
Exemplar answers
| Task C1 response | What it looks like |
|---|---|
| Level 3–4 extract | “In (a) Zara is wrong, you should use 3 not 6. 2(x + 3)2 = 5 − 18 = −13. Vertex (−3, −13). (b) x = 4 and x = 0.” The corrections are found, but the equals signs join unequal things, no reason is given for dividing by x, and there is no sketch. |
| Level 7–8 extract | “Step 1: take out the 2 from the x terms, 2(x2 + 6x) + 5. Step 2: half of 6 is 3, so x2 + 6x = (x + 3)2 − 9. Step 3: put that back, 2((x + 3)2 − 9) + 5 = 2(x + 3)2 − 18 + 5 = 2(x + 3)2 − 13. Zara halved 12 and forgot that the 9 is inside a bracket with 2 in front. The vertex is where the bracket is 0, so (−3, −13), as my sketch shows with its axis x = −3.” Each step on its own line, each reason named, equals signs only between equal values. |
Differentiation
A way in for anyone stuck
- Give the completing-the-square steps from lesson 3 as a reference card with the leading coefficient case marked.
- Let students explain aloud to a partner first, then write down what they said.
- Offer the frame: “Zara wrote …. This is wrong because …. The correct working is ….”
Stretch
- Explain why the vertex of a(x − h)2 + k is (h, k) and not (−h, k), using one value of x.
- Write a second version of the page for a student who has not met completing the square. What did you change, and why?
Students learning in English as an additional language
| Word in task C1 | Plain meaning |
|---|---|
| axis of symmetry | the line that cuts the parabola into two matching halves |
| lose a solution | forget an answer that is correct |
| explain | say what you did and why |
Sentence starters for task C1:
- Zara wrote …, but …
- First you …, because …
- This solution would be lost because …
Task D1: A vegetable plot against a wall
Why set it. Quadratic modelling, the vertex, and solving an inequality by factorising, inside one real decision with a real limit on the model. An alternative to D2: set one of them.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, in class. A written solution with explanations and a short note to the grower.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: no calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A grower in a hill village has 40 m of fencing. She wants a rectangular vegetable plot with a long straight wall as one side, so the fence is needed on the three other sides only. The two sides that meet the wall are each x m long.
Part 1. The model
- Write the length of the side parallel to the wall in terms of x. Then write the area A of the plot in terms of x, and say which values of x make sense.
- Find the greatest possible area and the dimensions that give it. Show your method.
Part 2. What she needs
- The crop needs an area of at least 150 m2. Find the values of x that give at least that area.
Part 3. Does it make sense?
- The wall is only 18 m long. What does that change? Find the best plot she can make.
- A neighbour says: “A square is always the best shape, so make it a square.” Test this claim for this plot, with the wall as one side.
- Write a short note to the grower with your recommendation, so that someone who has not seen your working can understand it.
How your work is judged
| Level | Criterion D in task D1: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example the area for a value of x, but no full model or recommendation. |
| 3–4 | You write the area as a quadratic, find its greatest value and the dimensions, and say whether the answer makes sense. |
| 5–6 | You choose suitable methods for every part (the vertex, an inequality solved by factorising), reach correct answers, and explain what the 18 m wall changes and whether the neighbour is right. |
| 7–8 | You choose and use suitable methods successfully throughout, reach correct answers, explain why the maximum moves when the wall is 18 m, test the neighbour’s claim with numbers, and say what the model leaves out about a real plot. |
| Level | Criterion C in task D1: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you reached the area or the range of x. |
| 3–4 | Your working for the area and the range can be followed, with some correct notation and units (m, m2). |
| 5–6 | Your working is clear and complete, usually with correct units and notation, and your note to the grower is mostly well organised. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your note to the grower makes sense on its own. |
Markscheme
| Task D1 question | Answer and marking |
|---|---|
| 1 | Parallel side 40 − 2x m; A = x(40 − 2x) = 40x − 2x2. x must be positive and 40 − 2x must be positive, so 0 < x < 20. |
| 2 | x = 10, so the plot is 10 m by 20 m and the area is 200 m2. A = −2(x − 10)2 + 200, or the vertex is halfway between the zeros x = 0 and x = 20. A complete answer says why 200 is the greatest, not just that it works. |
| 3 | 5 ≤ x ≤ 15. 40x − 2x2 ≥ 150 gives x2 − 20x + 75 ≤ 0, and (x − 5)(x − 15) ≤ 0. Both ends give exactly 150 m2: 5 × 30 and 15 × 10. |
| 4 | The parallel side must be at most 18 m, so 40 − 2x ≤ 18 and x ≥ 11. The area falls as x grows past 10, so the best plot has x = 11: 11 m by 18 m, area 198 m2, a little under 200. It still gives more than 150 m2. |
| 5 | The neighbour is wrong. A square needs 40 − 2x = x, so x = 40 ÷ 3, about 13.3 m, and area about 177.8 m2, which is less than 198 m2 (and 200 m2). The best shape here is a rectangle twice as long as it is wide, because the wall saves one side. |
| 6 (criterion C) | The note gives the plot size, the area, the reason (the 18 m wall limit and the 150 m2 need), with units, in sentences that make sense without the working. |
Exemplar answers
| Task D1 response | What it looks like |
|---|---|
| Level 3–4 (D) | “The area is x(40 − 2x). I tried 10 and got 200, and I tried 11 and got 198, so 10 by 20 is the best. For at least 150, x = 5 to 15. The wall is 18 so use 11.” Correct values, but no reason that 200 is the greatest, no handling of the 18 m limit as an inequality and nothing on the neighbour. |
| Level 7–8 (D and C) | Vertex form shows 200 is the greatest and why; inequality solved by factorising; “with an 18 m wall, x must be at least 11; the area drops after 10, so 11 m by 18 m gives 198 m2. A square gives about 178 m2, so the claim is false here. In real life the ground will not be level, so I would leave a little spare fence.” The note to the grower is four clear sentences with units. |
Differentiation
A way in for anyone stuck
- Give a table of x from 0 to 20 in steps of 5 for the area, and let students plot it before writing the formula.
- Offer a labelled rectangle with the wall drawn as one side and x on the two short sides.
- For Part 3, ask: “What is the longest the parallel side can be? What value of x gives that?”
Stretch
- With a 30 m wall, which of the 40 m fence plans is best? (The wall does not limit the plot: x = 10 still gives 20 m, area 200 m2.)
- Suppose she also wants a fence across the middle, parallel to the two short sides. Find the plot with the greatest area. (A = x(40 − 3x), maximum at x = 20 ÷ 3, area about 133.3 m2, plot about 6.7 m by 20 m.)
Students learning in English as an additional language
| Word in task D1 | Plain meaning |
|---|---|
| fencing | wire or wood that goes round the edge of a plot |
| parallel | going in the same direction and never meeting |
| at least | this number or more |
| greatest | the biggest |
| recommend | say what someone should do, and why |
Sentence starters for task D1:
- The area is … because …
- x must be at least … because the wall is …
- I recommend … because …
Task D2: Pricing a river tour
Why set it. Quadratic modelling, the vertex, and testing a model against a decision about price, inside one real situation. An alternative to D1: set one of them.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, in class, one lesson. A written solution with explanations and a short reply to the operator.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: no calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A river tour operator takes 40 passengers on each trip at 300 coins a ticket. A survey suggests that for every increase of 20 coins in the ticket price, 2 fewer passengers come. Let n be the number of 20 coins increases.
Part 1. Income
- Write the ticket price and the number of passengers in terms of n, and so the income R from one trip.
- Expand your expression. Then find the number of increases (a whole number) that gives the greatest income, and the income it gives.
- For which values of n is the income greater than the 12 000 coins at the current price? Show your method.
Part 2. Profit
- Each trip costs 4000 coins for the boat and 50 coins for every passenger. Write the profit P in terms of n and find the whole number of increases that makes the profit greatest.
- Is the best price for income the same as the best price for profit? Explain why, using your numbers.
Part 3. A recommendation
- Write a short reply to the operator: the price you recommend, the number of passengers, the profit, and how far the model can be trusted.
How your work is judged
| Level | Criterion D in task D2: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example the income at one price, but no full model or decision. |
| 3–4 | You write the income as a quadratic, find the best whole number of increases and the profit, and say whether your answers make sense. |
| 5–6 | You choose suitable methods for every part (expanding, the vertex, an inequality solved by factorising), reach correct answers, compare income with profit and discuss how far the model can be trusted. |
| 7–8 | You choose and use suitable methods successfully throughout, reach correct answers, explain why the whole-number answer is not at the vertex, explain why the best income and the best profit differ, and explain what the model assumes about passengers and prices. |
| Level | Criterion C in task D2: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you reached the income or the profit. |
| 3–4 | Your working for the income and the profit can be followed, with some correct notation and units (coins). |
| 5–6 | Your working is clear and complete, usually with correct units and notation, and your reply to the operator is mostly well organised. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your reply to the operator makes sense on its own. |
Markscheme
| Task D2 question | Answer and marking |
|---|---|
| 1 | Price 300 + 20n, passengers 40 − 2n, R = (300 + 20n)(40 − 2n). At n = 0 this is 300 × 40 = 12 000. |
| 2 | R = 12 000 + 200n − 40n2. The vertex is at n = 2.5, so n = 2 and n = 3 both give 12 240 coins. At n = 2: 340 × 36 = 12 240; at n = 3: 360 × 34 = 12 240. The vertex value 12 250 is not reachable with whole numbers. A student who gives only n = 2 or only n = 3 has not recognised the symmetry about 2.5. |
| 3 | 0 < n < 5. 200n − 40n2 > 0 gives 40n(5 − n) > 0. At n = 5, R = 12 000 again: the income is the same as now. |
| 4 | P = 6000 + 300n − 40n2; n = 4 gives 6560 coins. The costs are 4000 + 50(40 − 2n) = 6000 − 100n, so P = R − 6000 + 100n. The vertex is at n = 3.75, so n = 4: 380 coins a ticket, 32 passengers, profit 6560 coins. At n = 3 the profit is 6540 coins. |
| 5 | No. Income is greatest at n = 2.5 (price 350 coins); profit is greatest at n = 3.75 (price 375 coins). Each passenger costs the operator 50 coins, so a higher price with fewer passengers saves 100 coins for every increase. |
| 6 (criteria C and D) | A strong reply recommends 380 coins (32 passengers, profit 6560 coins), says that 6560 is only 20 coins above the 6540 coins at 360 coins, so the difference is within what the model can promise, and that the straight-line link between price and passengers is a guess from a survey. Credit any reasoned judgement either way. |
Exemplar answers
| Task D2 response | What it looks like |
|---|---|
| Level 3–4 (D) | “R = (300 + 20n)(40 − 2n). If n = 2 it is 340 × 36 = 12 240. n = 3 gives 12 240 too. The profit is more at n = 4, about 6560.” Right values, but no expanded form, no reason for the two equal values, no range of n and no comment on the model. |
| Level 7–8 (D and C) | “R = 12 000 + 200n − 40n2; its greatest value is at n = 2.5, so 2 and 3 give the same income, 12 240 coins. The income is higher than now for 0 < n < 5. Profit is greatest at n = 3.75, so n = 4: 380 coins, 32 passengers, 6560 coins. The model assumes a steady fall of 2 passengers per 20 coins, which a survey could not promise; I would test 360 coins for a month.” Every answer correct, units throughout. |
Differentiation
A way in for anyone stuck
- Give a table of n from 0 to 5 with columns for price, passengers and income, and let students fill it before writing the expression.
- Allow the expression to be expanded with a grid before simplifying.
- For Part 2, ask: “What happens to the cost of the boat when fewer passengers come?”
Stretch
- If the cost per passenger is 100 coins instead of 50, which n gives the greatest profit? (P = 4000 + 400n − 40n2: vertex n = 5, so n = 5.)
- Find the number of increases at which the income is 11 000 coins. Is there more than one? (n2 − 5n − 25 = 0 gives n about 8.1 or about −3.1; only the first is a price increase.)
Students learning in English as an additional language
| Word in task D2 | Plain meaning |
|---|---|
| ticket price | the amount one passenger pays |
| income | the money received before any costs are taken away |
| profit | income minus costs |
| increase | a change to a bigger number |
| survey | a way of collecting people’s answers to find out what they do or think |
Sentence starters for task D2:
- The price is … and the number of passengers is …
- The income is greatest when n is … because …
Practicalities
Every task is original and written for this unit; none is copied or adapted from a published book.
The student sheets, with their task-specific clarifications, are on this teacher page only, so every task stays unseen until you set it. Share a sheet's clarification with the class beforehand if you want them to know what each level looks like. Markschemes and exemplars are only on this page.
To print a sheet, open it on the unit page, then print the page. All six tasks are written for no calculator: every number is chosen to be exact by hand.
If a student may already have met A2 (a sibling, a previous class), use this parallel version (same marks, same markscheme notes): Q1 h(x) = 1 ÷ (x + 2) (domain x ≠ −2, range y ≠ 0). Q2 f(x) = x2 − 3, find f(3a + 1) (9a2 + 6a − 2). Q3 factorise 6x2 − 11x − 10 ((3x + 2)(2x − 5)). Q4 solve 3x2 + 5x − 28 = 0 (x = 7 ÷ 3 or x = −4). Q5 x2 − 6x + 10 in vertex form ((x − 3)2 + 1, vertex (3, 1)). Q6 solve 2x2 + 2x − 3 = 0 exactly (discriminant 28, x = (−1 + √7) ÷ 2 or (−1 − √7) ÷ 2). Q7 zeros at −3 and 5 and passes through (0, 30) (y = −2x2 + 4x + 30). Q8 x2 + kx + 81 = 0 (k = 18 or −18 for one solution; −18 < k < 18 for none). Q9 the point (1, −3) is on y = f(x): the image under f(x + 2) − 4 is (−1, −7), under −f(x) is (1, 3), under f(x ÷ 2) is (2, −3). Q10 stretch by scale factor 3 parallel to the y-axis then translate 2 down (y = 3x2 − 2) against translate 2 down then stretch (y = 3x2 − 6). Q11 a jet lands 6 m away and peaks 4.5 m up (y = −x(x − 6) ÷ 2; 2.5 m high at 1 m from the nozzle). Q12 “If a and c have the same sign, ax2 + bx + c = 0 never has a solution” (false: x2 + 5x + 1, discriminant 21). For A1, change the numbers the same way and rework the markscheme.
The Year 5 descriptor bands on this page are band summaries written for this site; check them against the current IB MYP Mathematics guide before relying on them for reporting.