The shape of the lesson
By the end of the lesson, learners can
- Intersections are solutions: In the lab, draw a curve and a line and read off where they cross: those \(x\)-values are exactly the solutions of the equation formed by setting the two expressions equal.
- Tangent, two points or none: For a line containing an unknown constant, substitute and look at the discriminant of the resulting quadratic.
- Lines and circles: The same method works for circles: substitute the line into the circle's equation.
- Substitute the line into the curve or circle.
- Collect everything on one side: \(ax^{2} + bx + c = 0\), where \(a\), \(b\), \(c\) may contain the constant.
- Decide: two points \(> 0\); tangent \(= 0\); no meeting \(< 0\).
- Solve for the constant, then find any point of contact from the repeated root.
- State the conclusion in words, e.g. 'so the line is a tangent'.
One picture every idea returns to
Substitute the line into the curve to get a quadratic. Its discriminant decides the picture: \(b^{2} - 4ac > 0\), the line cuts the curve at two points; \(= 0\), the line is a tangent; \(< 0\), the line does not meet the curve.
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, Cross, touch or miss.. Where two graphs meet, their equations are true together, so points of intersection are solutions of an equation. 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: Intersections are solutions | In the lab, draw a curve and a line and read off where they cross: those \(x\)-values are exactly the solutions of the equation formed by setting the two expressions equal. 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 Line meets parabola on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Tangent, two points or none | For a line containing an unknown constant, substitute and look at the discriminant of the resulting quadratic. Model the worked example A family of parallel lines on the board, one line at a time. Your turn: Find the value of \(k\) for which \(y = 5x + k\) is a tangent to \(y = x^{2} + x + 2\). | Pairs |
| 21 to 30 min | Teach and check: Lines and circles | The same method works for circles: substitute the line into the circle's equation. Model the worked example Tangents to a circle on the board, one line at a time. | Numbered Heads Together |
| 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, 22 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 Lines and circles, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Points of intersection need both coordinates. | After finding \(x\), substitute into the LINEAR equation for \(y\) (it is simpler and avoids pairing errors). |
| Collect EVERYTHING on one side before reading \(a\), \(b\), \(c\): in \(x^{2} + (2 - m)x + 4 = 0\) the coefficient \(b\) is the whole bracket \((2 - m)\), not 2 and not \(-m\). | Return to Tangent, two points or none: ask a pair to state the correct version and show it on the lab before moving on. |
| Showing a line is a tangent needs a stated conclusion: '\(b^{2} - 4ac = 0\), so there is a repeated root, so the line is a tangent'. | Finding a single root without saying why it means tangency loses the explanation mark. |
Intersections of lines and curves: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Find the coordinates of the points of intersection of \(y = x^{2} - 4x + 1\) and \(y = 2x - 4\).
The line \(y = 5x + k\) is a tangent to the curve \(y = x^{2} + x + 2\). Find \(k\) and the coordinates of the point of contact.
Find the set of values of \(m\) for which the line \(y = mx + 3\) does not meet the curve \(y = 2x^{2} + 5\).
The circle \(x^{2} + y^{2} - 6x + 4y + 8 = 0\) is given.
(a) Find its centre and radius.
(b) Find the values of \(c\) for which the line \(y = 2x + c\) is a tangent to the circle.
Show that the line \(y = 4 - x\) does not meet the curve \(y = x^{2} + x + 6\).
A ball is kicked from the foot of a hill. Taking the kick as the origin, with \(x\) and \(y\) in metres, the ball's path is \(y = x - \dfrac{x^{2}}{40}\) and the hillside is the line \(y = \dfrac{x}{4}\).
(a) Find where the ball lands on the hillside.
(b) Find the value of \(h\) for which the line \(y = \dfrac{x}{4} + h\) is a tangent to the path, and explain what \(h\) represents.
Answer key and marking notes
- \(x^{2} - 4x + 1 = 2x - 4 \Rightarrow x^{2} - 6x + 5 = 0\).
- \((x - 1)(x - 5) = 0 \Rightarrow x = 1\) or \(x = 5\).
- Points \((1, -2)\) and \((5, 6)\).
- \(x^{2} + x + 2 = 5x + k \Rightarrow x^{2} - 4x + (2 - k) = 0\).
- Tangent: \(16 - 4(2 - k) = 0 \Rightarrow 8 + 4k = 0 \Rightarrow k = -2\).
- \(x^{2} - 4x + 4 = 0 \Rightarrow x = 2\), \(y = 5(2) - 2 = 8\). Point \((2, 8)\).
- \(2x^{2} + 5 = mx + 3 \Rightarrow 2x^{2} - mx + 2 = 0\).
- No meeting: \(m^{2} - 4(2)(2) < 0 \Rightarrow m^{2} < 16\).
- \(-4 < m < 4\).
- (a) \((x - 3)^{2} + (y + 2)^{2} = 9 + 4 - 8 = 5\): centre \((3, -2)\), radius \(\sqrt{5}\).
- (b) \(x^{2} + (2x + c)^{2} - 6x + 4(2x + c) + 8 = 0\).
- \(5x^{2} + (4c + 2)x + (c^{2} + 4c + 8) = 0\).
- Tangent: \((4c + 2)^{2} - 20(c^{2} + 4c + 8) = 0 \Rightarrow -4c^{2} - 64c - 156 = 0 \Rightarrow c^{2} + 16c + 39 = 0\).
- \((c + 3)(c + 13) = 0\): \(c = -3\) or \(c = -13\).
- \(x^{2} + x + 6 = 4 - x \Rightarrow x^{2} + 2x + 2 = 0\).
- Discriminant \(= 2^{2} - 4(1)(2) = 4 - 8 = -4 < 0\).
- No real roots, so the line and the curve do not meet.
- (a) \(x - \dfrac{x^{2}}{40} = \dfrac{x}{4} \Rightarrow \dfrac{3x}{4} - \dfrac{x^{2}}{40} = 0 \Rightarrow x\left(30 - x\right) = 0\) (after multiplying by 40).
- \(x = 30\) (\(x = 0\) is the kick), \(y = 7.5\): the ball lands at \((30, 7.5)\).
- (b) \(x - \dfrac{x^{2}}{40} = \dfrac{x}{4} + h \Rightarrow \dfrac{x^{2}}{40} - \dfrac{3x}{4} + h = 0 \Rightarrow x^{2} - 30x + 40h = 0\).
- Tangent: \(900 - 160h = 0 \Rightarrow h = \dfrac{45}{8} = 5.625\), touching at \(x = 15\).
- \(h\) is the greatest vertical height of the ball above the hillside: 5.625 m, when \(x = 15\) m.
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.
Find the coordinates of the points where the line \(y = 6 - x\) meets the curve \(y = x^{2} - 4x + 2\).
- A\((4,\ 2)\) and \((-1,\ 7)\)
- B\((-4,\ 10)\) and \((1,\ 5)\)
- C\((4,\ 2)\) only
- D\((4,\ 10)\) and \((-1,\ 5)\)
The graph of \(y = x^{2} - 3x + 1\) has been drawn. Which straight line should be drawn on the same axes to solve the equation \(x^{2} - 5x + 3 = 0\)?
- A\(y = 2x - 2\)
- B\(y = -2x + 2\)
- C\(y = 2x + 2\)
- D\(y = 5x - 3\)
Find the value of \(k\) for which the line \(y = k\) is a tangent to the curve \(y = x^{2} + 4x + 9\).
- A\(5\)
- B\(-2\)
- C\(9\)
- D\(13\)
Find the set of values of \(k\) for which the line \(y = x + k\) meets the curve \(y = x^{2} - x + 3\) at two distinct points.
- A\(k > 2\)
- B\(k < 2\)
- C\(k \ge 2\)
- D\(k > -2\)
The line \(y = mx + 2\) is a tangent to the curve \(y = x^{2} + 3x + 6\). Find the possible values of \(m\).
- A\(m = -1\) or \(m = 7\)
- B\(m = 1\) or \(m = -7\)
- C\(m = 7\) only
- D\(m = 3 \pm 2\sqrt{6}\)
The line \(y = 2x + k\) is a tangent to the curve \(y = x^{2} - 4x + 7\). Find the coordinates of the point of contact.
- A\((3,\ 4)\)
- B\((-3,\ -8)\)
- C\((3,\ 8)\)
- D\((6,\ 10)\)
Find the values of \(c\) for which the line \(y = x + c\) is a tangent to the circle \(x^{2} + y^{2} = 18\).
- A\(c = \pm 6\)
- B\(c = \pm 3\sqrt{2}\)
- C\(c = 6\) only
- D\(c = \pm 36\)
Find the set of values of \(k\) for which the line \(y = kx - 3\) does not meet the curve \(y = x^{2} + 1\).
- A\(-4 < k < 4\)
- B\(k < -4\) or \(k > 4\)
- C\(-2 < k < 2\)
- D\(k < 4\)
Quiz key
\(x^{2} - 4x + 2 = 6 - x\) gives \(x^{2} - 3x - 4 = 0\), so \((x - 4)(x + 1) = 0\) and \(x = 4\) or \(x = -1\). From \(y = 6 - x\): \((4,\ 2)\) and \((-1,\ 7)\). The tempting \((-4,\ 10)\), \((1,\ 5)\) come from \((x + 4)(x - 1)\), the wrong signs in the factors. Stopping at one point misses an intersection, and \((4,\ 10)\), \((-1,\ 5)\) use \(y = 6 + x\).
\(x^{2} - 5x + 3 = 0 \iff x^{2} - 3x + 1 = 2x - 2\) (add \(2x - 2\) to both sides). So draw \(y = 2x - 2\); the \(x\)-coordinates of the intersections are the solutions. The tempting \(y = -2x + 2\) subtracts instead of adding. \(y = 2x + 2\) gets the constant wrong, and \(y = 5x - 3\) would go with the curve \(y = x^{2}\), not the curve that was drawn.
\(x^{2} + 4x + 9 = (x + 2)^{2} + 5\), so the vertex is \((-2,\ 5)\) and the tangent line is \(y = 5\). (Discriminant: \(16 - 4(9 - k) = 0\) gives \(k = 5\).) The tempting \(-2\) is the \(x\)-coordinate of the vertex. 9 is the \(y\)-intercept, and 13 comes from \((x + 2)^{2} - 4 + 9\) with \(+4\) instead of \(-4\).
\(x^{2} - x + 3 = x + k\) gives \(x^{2} - 2x + (3 - k) = 0\). Two distinct points: \(4 - 4(3 - k) > 0\), so \(4k > 8\) and \(k > 2\). The tempting \(k < 2\) is the condition for the line to miss the curve. \(k \ge 2\) includes \(k = 2\), where the line is a tangent (one point), and \(k > -2\) comes from \(4 - 4(3 + k)\), a sign slip with \(c\).
\(x^{2} + 3x + 6 = mx + 2\) gives \(x^{2} + (3 - m)x + 4 = 0\). Tangent: \((3 - m)^{2} - 16 = 0\), so \(3 - m = \pm 4\): \(m = -1\) or \(m = 7\). The tempting \(1\) or \(-7\) solves \(3 + m = \pm 4\), a sign slip when collecting terms. \(m = 7\) only loses a root, and \(3 \pm 2\sqrt{6}\) uses the constant 6 instead of \(6 - 2 = 4\).
\(x^{2} - 4x + 7 = 2x + k\) gives \(x^{2} - 6x + (7 - k) = 0\). Tangent: \(36 - 4(7 - k) = 0\), so \(k = -2\). Then \(x^{2} - 6x + 9 = (x - 3)^{2} = 0\), \(x = 3\), and \(y = 2(3) - 2 = 4\). Check on the curve: \(9 - 12 + 7 = 4\). The tempting \((3,\ 8)\) uses \(k = +2\). \((-3,\ -8)\) takes the root as \(+\frac{b}{2a}\), and \((6,\ 10)\) uses \(-b\) without dividing by \(2a\).
\(x^{2} + (x + c)^{2} = 18\) gives \(2x^{2} + 2cx + c^{2} - 18 = 0\). Tangent: \((2c)^{2} - 4(2)(c^{2} - 18) = 0\), so \(4c^{2} - 8c^{2} + 144 = 0\), \(c^{2} = 36\), \(c = \pm 6\). The tempting \(\pm 3\sqrt{2}\) is the radius \(\sqrt{18}\), not the intercept: the tangents are further from the origin along the \(y\)-axis than the radius. \(c = 6\) only misses the lower tangent, and \(\pm 36\) forgets the square root.
\(x^{2} + 1 = kx - 3\) gives \(x^{2} - kx + 4 = 0\). No intersection: \(k^{2} - 16 < 0\), so \((k - 4)(k + 4) < 0\) and \(-4 < k < 4\). The tempting outside region gives two intersection points. \(-2 < k < 2\) solves \(k^{2} < 4\), forgetting the 4 in \(4ac\), and \(k < 4\) ignores the lower critical value: \(k = -5\) gives a line that cuts the curve.
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.