The shape of the lesson
By the end of the lesson, learners can
- Half the coefficient, squared: In the lab, press Make the square and watch \(x^{2} + bx\) rearranged into a square: there is always a small corner missing, of area \(\left(\frac{b}{2}\right)^{2}\).
- When \(a\) is not 1: For \(ax^{2} + bx + c\), take the factor \(a\) out of the \(x^{2}\) and \(x\) terms only, complete the square inside the bracket, then multiply back out carefully.
- Reading the graph: A square is never negative.
- Factor \(a\) out of the \(x^{2}\) and \(x\) terms if \(a \ne 1\).
- Halve the coefficient of \(x\) inside the bracket: \(x^{2} + bx = \left(x + \frac{b}{2}\right)^{2} - \frac{b^{2}}{4}\).
- Multiply back by \(a\), including the subtracted square, then collect constants.
- Read off \(a(x + p)^{2} + q\): vertex \((-p, q)\), minimum if \(a > 0\), maximum if \(a < 0\).
- Check by expanding: you must get the original quadratic back.
One picture every idea returns to
\(x^{2} + bx + c = \left(x + \dfrac{b}{2}\right)^{2} - \left(\dfrac{b}{2}\right)^{2} + c\). In the form \(a(x + p)^{2} + q\), the vertex is \((-p,\ q)\): a minimum if \(a > 0\), a maximum if \(a < 0\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Every parabola has a vertex.. Rewrite any quadratic as a square plus a constant and its graph gives up its secrets: the vertex, the line of symmetry, the least or greatest value and the range, all without a single derivative. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Half the coefficient, squared | In the lab, press Make the square and watch \(x^{2} + bx\) rearranged into a square: there is always a small corner missing, of area \(\left(\frac{b}{2}\right)^{2}\). Project tab 4 of the lab (Algebra lab, completing the square) and let a student drive it while the class predicts. Model the worked example Integer half on the board, one line at a time. Your turn: Express \(x^{2} + 8x + 3\) in the form \((x + a)^{2} + b\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: When \(a\) is not 1 | For \(ax^{2} + bx + c\), take the factor \(a\) out of the \(x^{2}\) and \(x\) terms only, complete the square inside the bracket, then multiply back out carefully. Model the worked example Positive a on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Reading the graph | A square is never negative. Model the worked example Vertex and sketch on the board, one line at a time. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 20 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Reading the graph, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The sign inside the bracket matches the sign of \(b\): \(x^{2} - 6x\) gives \((x - 3)^{2}\), not \((x + 3)^{2}\). | And the square you subtract is always positive: \(-(-3)^{2} = -9\). |
| Writing \(2x^{2} - 12x + 7 = 2(x - 3)^{2} - 9 + 7\) is the classic slip: the 9 sits INSIDE the bracket that is multiplied by 2, so it must become 18. | Expand your answer to catch it. |
| The vertex of \(a(x + p)^{2} + q\) is at \(x = -p\), not \(x = p\): \((x + 1)^{2}\) is zero when \(x = -1\). | Read the \(x\)-coordinate as the value that makes the bracket zero. |
Completing the square: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
Express \(x^{2} - 14x + 50\) in the form \((x - a)^{2} + b\).
(a) Express \(3x^{2} - 18x + 32\) in the form \(a(x + b)^{2} + c\).
(b) State the coordinates of the vertex of the curve \(y = 3x^{2} - 18x + 32\).
By completing the square, solve \(x^{2} + 4x - 8 = 0\), giving your answers in the form \(p \pm q\sqrt{3}\).
The function f is defined by \(f(x) = 5 + 8x - 2x^{2}\) for \(x \in \mathbb{R}\).
(a) Express \(f(x)\) in the form \(a - b(x - c)^{2}\).
(b) State the range of f and the coordinates of the stationary point of \(y = f(x)\).
(a) Show that \(x^{2} - 6x + 13 > 0\) for all real values of \(x\).
(b) Hence find the greatest value of \(\dfrac{1}{x^{2} - 6x + 13}\).
A ball is thrown upwards. Its height above the ground \(t\) seconds later is \(h = 1.5 + 14t - 5t^{2}\) metres.
(a) Express \(h\) in the form \(a - b(t - c)^{2}\).
(b) Find the greatest height of the ball and the time at which it is reached.
(c) Find the time at which the ball hits the ground, giving your answer to 3 significant figures.
Answer key and marking notes
- Half of \(-14\) is \(-7\): \((x - 7)^{2} - 49 + 50\).
- \(= (x - 7)^{2} + 1\), so \(a = 7,\ b = 1\).
- (a) \(3(x^{2} - 6x) + 32 = 3\big[(x - 3)^{2} - 9\big] + 32\).
- \(= 3(x - 3)^{2} - 27 + 32 = 3(x - 3)^{2} + 5\).
- (b) The bracket is zero at \(x = 3\), so the vertex is \((3,\ 5)\), a minimum.
- \(x^{2} + 4x - 8 = (x + 2)^{2} - 4 - 8 = (x + 2)^{2} - 12\).
- \((x + 2)^{2} = 12 \Rightarrow x + 2 = \pm\sqrt{12} = \pm 2\sqrt{3}\).
- \(x = -2 \pm 2\sqrt{3}\).
- (a) \(f(x) = 5 - 2(x^{2} - 4x) = 5 - 2\big[(x - 2)^{2} - 4\big]\).
- \(= 5 + 8 - 2(x - 2)^{2} = 13 - 2(x - 2)^{2}\).
- (b) \(2(x - 2)^{2} \ge 0\), so \(f(x) \le 13\). Range: \(f(x) \le 13\).
- The stationary point is the maximum point \((2,\ 13)\).
- (a) \(x^{2} - 6x + 13 = (x - 3)^{2} - 9 + 13 = (x - 3)^{2} + 4\).
- \((x - 3)^{2} \ge 0\) for all real \(x\), so \((x - 3)^{2} + 4 \ge 4 > 0\).
- (b) The denominator is least, 4, when \(x = 3\), so the fraction is greatest there: \(\dfrac{1}{4}\).
- (a) \(h = 1.5 - 5(t^{2} - 2.8t) = 1.5 - 5\big[(t - 1.4)^{2} - 1.96\big]\).
- \(= 1.5 + 9.8 - 5(t - 1.4)^{2} = 11.3 - 5(t - 1.4)^{2}\).
- (b) Greatest height \(11.3\) m, when \(t = 1.4\) s.
- (c) \(h = 0\): \(5(t - 1.4)^{2} = 11.3 \Rightarrow (t - 1.4)^{2} = 2.26\).
- \(t - 1.4 = \pm\sqrt{2.26} = \pm 1.503...\); the time must be positive, so \(t = 1.4 + 1.503... = 2.90\) s.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Express \(x^{2} - 10x + 7\) in the form \((x - a)^{2} + b\).
- A\((x - 5)^{2} - 18\)
- B\((x - 5)^{2} + 32\)
- C\((x + 5)^{2} - 18\)
- D\((x - 10)^{2} - 93\)
Express \(3x^{2} + 12x + 5\) in the form \(a(x + b)^{2} + c\).
- A\(3(x + 2)^{2} - 7\)
- B\(3(x + 2)^{2} + 1\)
- C\(3(x + 6)^{2} - 103\)
- D\(3(x - 2)^{2} - 7\)
Find the greatest value of \(7 + 6x - x^{2}\) and the value of \(x\) at which it occurs.
- AGreatest value 16, when \(x = 3\)
- BGreatest value 16, when \(x = -3\)
- CGreatest value \(-2\), when \(x = 3\)
- DGreatest value 7, when \(x = 0\)
The function f is defined by \(\text{f}(x) = x^{2} + 4x + 9\) for \(x \in \mathbb{R}\). Find the range of f.
- A\(\text{f}(x) \ge 5\)
- B\(\text{f}(x) > 5\)
- C\(\text{f}(x) \ge 9\)
- D\(\text{f}(x) \ge -2\)
Solve \(x^{2} + 8x + 11 = 0\) by completing the square, giving your answers in exact form.
- A\(x = -4 \pm \sqrt{5}\)
- B\(x = 4 \pm \sqrt{5}\)
- C\(x = -4 \pm 3\sqrt{3}\)
- D\(x = -4 \pm 5\)
The function f is defined by \(\text{f}(x) = 2x^{2} - 8x + 3\) for \(x \ge 3\). By first expressing \(\text{f}(x)\) in the form \(a(x + b)^{2} + c\), find the range of f.
- A\(\text{f}(x) \ge -3\)
- B\(\text{f}(x) \ge -5\)
- C\(\text{f}(x) \ge 3\)
- D\(\text{f}(x) \le -3\)
The curve \(y = 4x^{2} - 12x + k\) has a minimum value of 1. By completing the square, find the value of the constant \(k\).
- A\(k = 10\)
- B\(k = 3.25\)
- C\(k = -8\)
- D\(k = 37\)
By completing the square, find the set of values of the constant \(c\) for which \(x^{2} - 6x + c > 0\) for all real values of \(x\).
- A\(c > 9\)
- B\(c \ge 9\)
- C\(c > 3\)
- D\(c < 9\)
Quiz key
Half of \(-10\) is \(-5\): \(x^{2} - 10x = (x - 5)^{2} - 25\), so \(x^{2} - 10x + 7 = (x - 5)^{2} - 25 + 7 = (x - 5)^{2} - 18\). The tempting \(+32\) ADDS the 25 instead of taking it away. \((x + 5)^{2}\) has the wrong sign (it expands to \(+10x\)), and \((x - 10)^{2}\) uses the whole coefficient instead of half of it.
\(3x^{2} + 12x + 5 = 3(x^{2} + 4x) + 5 = 3\big[(x + 2)^{2} - 4\big] + 5 = 3(x + 2)^{2} - 12 + 5 = 3(x + 2)^{2} - 7\). The tempting \(+1\) forgets that the \(-4\) sits inside the bracket multiplied by 3, so it must become \(-12\). \(3(x + 6)^{2}\) halves 12 without first taking out the 3, and \(3(x - 2)^{2}\) has the wrong sign inside the bracket.
\(7 + 6x - x^{2} = 7 - (x^{2} - 6x) = 7 - \big[(x - 3)^{2} - 9\big] = 16 - (x - 3)^{2}\). Since \((x - 3)^{2} \ge 0\), the greatest value is 16, when \(x = 3\). The tempting \(-2\) comes from \(7 - 9\): the minus sign in front of the bracket turns \(-9\) into \(+9\). \(x = -3\) reads the sign inside the bracket the wrong way round, and 7 is only the \(y\)-intercept.
\(x^{2} + 4x + 9 = (x + 2)^{2} + 5\). The square is at least 0 and equals 0 when \(x = -2\), so the least value 5 IS reached: the range is \(\text{f}(x) \ge 5\). The tempting \(\text{f}(x) > 5\) excludes a value the function actually takes. 9 is only \(\text{f}(0)\), and \(-2\) is the \(x\)-coordinate of the vertex, not a value of f.
\(x^{2} + 8x + 11 = (x + 4)^{2} - 16 + 11 = (x + 4)^{2} - 5\). So \((x + 4)^{2} = 5\), \(x + 4 = \pm\sqrt{5}\), and \(x = -4 \pm \sqrt{5}\). The tempting \(4 \pm \sqrt{5}\) moves the 4 across without changing its sign. \(-4 \pm 3\sqrt{3}\) comes from \((x + 4)^{2} = 16 + 11 = 27\), adding the 16 instead of subtracting it, and \(-4 \pm 5\) forgets the square root.
\(2x^{2} - 8x + 3 = 2\big[(x - 2)^{2} - 4\big] + 3 = 2(x - 2)^{2} - 5\). The vertex is at \(x = 2\), which is NOT in the domain. For \(x \ge 3\) the function is increasing, so the least value is \(\text{f}(3) = 2(1)^{2} - 5 = -3\), and the range is \(\text{f}(x) \ge -3\). The tempting \(-5\) is the minimum over all real \(x\), which ignores the restricted domain. 3 is \(\text{f}(0)\), and \(\le\) reverses a curve that increases without limit.
\(4x^{2} - 12x = (2x - 3)^{2} - 9\), so \(y = (2x - 3)^{2} + k - 9\). The minimum value is \(k - 9 = 1\), giving \(k = 10\). The tempting \(3.25\) uses \(4\big(x - \frac{3}{2}\big)^{2} - \frac{9}{4} + k\), forgetting to multiply the \(\frac{9}{4}\) by 4. \(-8\) solves \(k - 9 = 1\) with a sign slip, and 37 halves 12 without first taking out the factor 4.
\(x^{2} - 6x + c = (x - 3)^{2} + c - 9\). Its least value is \(c - 9\), at \(x = 3\). For the expression to be positive for EVERY \(x\), this least value must be positive: \(c - 9 > 0\), so \(c > 9\). The tempting \(c \ge 9\) allows \(c = 9\), when the expression equals 0 at \(x = 3\), which is not \(> 0\). \(c > 3\) confuses the bracket number with its square, and \(c < 9\) reverses the condition.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.