The shape of the lesson
By the end of the lesson, learners can
- Roots are on the x-axis: Open the solving tab in the lab.
- A line meets a curve: When a straight line and a curve are drawn on the same grid, each crossing point satisfies both equations at once, so its \(x\)-coordinate solves \(\text{curve} = \text{line}\).
- Cubic and reciprocal curves: Extended.
- Roots: \(f(x) = 0\) is solved where \(y = f(x)\) meets the \(x\)-axis.
- Equal to a number: \(f(x) = k\) is solved where the curve meets the horizontal line \(y = k\).
- New equation: add or subtract the same terms on both sides until one side is exactly the drawn curve.
- The other side is the line to draw; plot it with two or three points.
- Answer with the \(x\)-coordinates of every crossing, to the accuracy of the graph (usually 1 d.p.).
One picture every idea returns to
The solutions of \(\text{curve} = \text{line}\) are the \(x\)-coordinates of the points where the two graphs cross. To solve a new equation with a curve you already have, rearrange it so one side is exactly the curve's equation; the other side is the line to draw.
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, Where the graphs meet.. Some equations are hard to solve by algebra, but easy to solve with a graph: draw the right line, find where it crosses the curve, and read down to the \(x\)-axis. 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: Roots are on the x-axis | Open the solving tab in the lab. Project tab 3 of the lab (Graphs of functions lab, solving equations) and let a student drive it while the class predicts. Model the worked example Roots from the graph on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: A line meets a curve | When a straight line and a curve are drawn on the same grid, each crossing point satisfies both equations at once, so its \(x\)-coordinate solves \(\text{curve} = \text{line}\). Model the worked example Reading intersections on the board, one line at a time. Your turn: Using the same curve, how many solutions does \(x^2 - 2x - 2 = k\) have when (a) \(k = 1\) (b) \(k = -3\) (c) \(k = -5\)? | Pairs |
| 21 to 30 min | Teach and check: Cubic and reciprocal curves | Extended. Model the worked example A reciprocal and a line (Extended) on the board, one line at a time. | Corners |
| 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, 17 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 Cubic and reciprocal curves, 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.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
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 |
|---|---|
| Give the \(x\)-values, not the \(y\)-value. | For \(x^2 - 2x - 2 = 3\), the answer is NOT \(y = 3\); it is the two \(x\)-coordinates where the line \(y = 3\) cuts the curve. And give BOTH solutions. |
| When you rearrange, change BOTH sides by the same amount. | Students often move terms across and write \(y = 3x + 1\) for this question, but the line must come from keeping the equation balanced: \(y = x - 1\). |
| Only read solutions inside the drawn range. | If a line leaves the grid before meeting the curve again, say there are no more solutions in that range rather than guessing. |
Solving equations graphically: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
(a) Complete a table of values for \(y = x^2 - x - 6\) for \(x = -3, -2, \dots, 4\).
(b) Use your graph to solve \(x^2 - x - 6 = 0\).
The graph of \(y = x^2 - x - 6\) is drawn. Write down the equation of the line you would draw to solve (a) \(x^2 - x - 6 = 4\) (b) \(x^2 - 3x - 2 = 0\).
The line \(y = 2x - 2\) is drawn on the graph of \(y = x^2 - x - 6\). Find, by algebra, the coordinates of the points where they meet, so that you can check your graph.
The lowest point of the curve \(y = x^2 - x - 6\) is \((0.5, -6.25)\).
Find the value of \(k\) for which \(x^2 - x - 6 = k\) has exactly one solution, and state the number of solutions when \(k = -7\).
(Extended) The graphs of \(y = \dfrac{4}{x}\) and \(y = 5 - x\) are drawn for \(x > 0\).
Show that at the points where they meet, \(x^2 - 5x + 4 = 0\). Hence find the coordinates of these points.
A ball is thrown upwards. Its height, \(h\) metres, after \(t\) seconds is \(h = 20t - 5t^2\).
(a) Complete a table of values for \(t = 0, 1, 2, 3, 4\) and draw the graph.
(b) Draw the line \(h = 15\) and use it to find for how long the ball is more than 15 m high.
(c) When does the ball land?
Answer key and marking notes
- (a) \(y\): \(6, 0, -4, -6, -6, -4, 0, 6\).
- (b) The curve crosses the \(x\)-axis at \(x = -2\) and \(x = 3\).
- (a) The left side is the curve, so draw \(y = 4\).
- (b) Add \(2x - 4\) to both sides: \(x^2 - 3x - 2 + 2x - 4 = 2x - 4\), so \(x^2 - x - 6 = 2x - 4\). Draw \(y = 2x - 4\).
- \(x^2 - x - 6 = 2x - 2\), so \(x^2 - 3x - 4 = 0\).
- \((x - 4)(x + 1) = 0\), so \(x = 4\) or \(x = -1\).
- \(y = 2(4) - 2 = 6\) and \(y = 2(-1) - 2 = -4\): the points are \((4, 6)\) and \((-1, -4)\).
- Exactly one solution when the line \(y = k\) just touches the lowest point: \(k = -6.25\).
- \(k = -7\) is below the lowest point, so there are no solutions.
- \(\dfrac{4}{x} = 5 - x\). Multiply by \(x\): \(4 = 5x - x^2\).
- Rearrange: \(x^2 - 5x + 4 = 0\).
- \((x - 1)(x - 4) = 0\), so \(x = 1\) or \(x = 4\).
- \(y = 5 - 1 = 4\) and \(y = 5 - 4 = 1\): the points are \((1, 4)\) and \((4, 1)\).
- (a) \(h\): \(0, 15, 20, 15, 0\). The graph is a ∩-shaped parabola with its highest point at \((2, 20)\).
- (b) The line \(h = 15\) meets the curve at \(t = 1\) and \(t = 3\) (check: \(5t^2 - 20t + 15 = 0\), \(t^2 - 4t + 3 = 0\), \((t - 1)(t - 3) = 0\)).
- The ball is above 15 m between these times: \(3 - 1 = 2\) seconds.
- (c) \(h = 0\) again at \(t = 4\), so the ball lands after 4 seconds.
6 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.
The graph of \(y = x^2 + x - 5\) is drawn. Use the graph to solve \(x^2 + x - 5 = 0\).
- A\(x \approx -2.8\) or \(x \approx 1.8\)
- B\(x = -5\)
- C\(x \approx -0.5\) only
- D\(x \approx -2.3\) or \(x \approx 1.3\)
Use the graph of \(y = x^2 + x - 5\) to solve \(x^2 + x - 5 = 2\).
- A\(x \approx -3.2\) or \(x \approx 2.2\)
- B\(x \approx -2.8\) or \(x \approx 1.8\)
- C\(y = 2\)
- D\(x \approx 2.2\) only
The graph of \(y = x^2 + x - 5\) has been drawn. Which straight line should be drawn on the same grid to solve \(x^2 + 3x - 2 = 0\)?
- A\(y = -2x - 3\)
- B\(y = 2x + 3\)
- C\(y = -2x + 3\)
- D\(y = 3x - 2\)
The lowest point of the graph of \(y = x^2 + x - 5\) is \((-0.5, -5.25)\). For which value of \(k\) does the equation \(x^2 + x - 5 = k\) have NO solutions?
- A\(k = -6\)
- B\(k = -5\)
- C\(k = -5.2\)
- D\(k = 0\)
The graphs of \(y = \dfrac{4}{x}\) and \(y = 5 - x\) are drawn on the same grid. Which equation is satisfied by the \(x\)-coordinates of the points where they meet?
- A\(x^2 - 5x + 4 = 0\)
- B\(x^2 + 5x - 4 = 0\)
- C\(x^2 - 5x - 4 = 0\)
- D\(x^2 - 4x + 5 = 0\)
The graph of \(y = x^3 - 4x\) and the line \(y = x + 1\) are drawn. Use the graph to solve \(x^3 - 5x - 1 = 0\).
- A\(x \approx -2.1\), \(-0.2\) or \(2.3\)
- B\(x = -2\), 0 or 2
- C\(x \approx -0.2\) only
- D\(x \approx -2.3\), \(0.2\) or \(2.1\)
Quiz key
\(y = 0\) is the \(x\)-axis. The curve crosses it twice, at \(x \approx -2.8\) and \(x \approx 1.8\) (exactly \(\dfrac{-1 \pm \sqrt{21}}{2} = -2.79\) and \(1.79\)). \(-5\) is where the curve meets the \(y\)-axis, which is the \(y\)-intercept, not a solution. \(-0.5\) is the line of symmetry. \(-2.3\) and \(1.3\) are the solutions of \(x^2 + x - 5 = -2\), read from the wrong horizontal line.
Draw \(y = 2\). It cuts the curve twice; reading down gives \(x \approx -3.2\) and \(x \approx 2.2\) (exactly \(\dfrac{-1 \pm \sqrt{29}}{2} = -3.19\) and \(2.19\)). \(y = 2\) is the line you draw, not the answer: the solutions are \(x\)-values. \(-2.8\) and \(1.8\) are where the curve crosses \(y = 0\), and giving only 2.2 misses the second crossing on the left.
Subtract \(2x\) and subtract 3 from both sides: \(x^2 + 3x - 2 - 2x - 3 = 0 - 2x - 3\), so \(x^2 + x - 5 = -2x - 3\). Draw \(y = -2x - 3\). The tempting \(y = 2x + 3\) has the signs the wrong way round: it comes from moving the terms without keeping the equation balanced (it would solve \(x^2 - x - 8 = 0\)). \(y = 3x - 2\) just copies terms from the new equation.
The line \(y = k\) misses the U-shaped curve only if it is below the lowest point, so \(k < -5.25\). Only \(k = -6\) is below. \(k = -5.2\) is the tempting choice, but \(-5.2\) is slightly ABOVE \(-5.25\), so the line still cuts the curve twice. \(k = -5\) and \(k = 0\) also give two solutions.
\(\dfrac{4}{x} = 5 - x\). Multiply by \(x\): \(4 = 5x - x^2\). Add \(x^2\) and subtract \(5x\): \(x^2 - 5x + 4 = 0\), which gives \(x = 1\) or \(x = 4\) (the points \((1, 4)\) and \((4, 1)\)). \(x^2 - 5x - 4 = 0\) forgets to change the sign of 4 when moving it across; \(x^2 + 5x - 4 = 0\) changes every sign on one side only.
Add \(x + 1\) to both sides: \(x^3 - 4x = x + 1\). So the solutions are the \(x\)-coordinates of the three crossings of the curve and the line: about \(-2.1\), \(-0.2\) and \(2.3\) (more accurately \(-2.13\), \(-0.20\), \(2.33\)). \(-2\), 0 and 2 are where the curve crosses the \(x\)-axis, which solves \(x^3 - 4x = 0\) instead. A straight line can cross a cubic three times, so one answer is not enough.
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.