At a glance. One lesson of about 60 minutes. It assumes the three forms (Lesson 3) and solving two linear equations. Students find a quadratic from three points, a vertex and a point, or zeros and a point, and decide when the data is a line.
Calculator: GDC optional. Everything can be done by hand. The optional section shows quadratic regression on a GDC and warns that with more than three points it gives a best fit, not an exact one.
What students will learn
Know
- A quadratic has three unknowns, so it needs three independent facts.
- The vertex and zeros forms need only a to be found.
- Collinear points give a = 0.
Understand
- The number of facts needed matches the number of unknowns.
- Choosing the form makes the algebra shorter.
- A fitted model is trusted near its data.
Do
- Solve for a, b and c from three points.
- Fit from a vertex and a point, or zeros and a point.
- Use the model to answer a question in context.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: forming and solving the equations (questions 1 to 7).
- Criterion B. Investigating patterns: the what-if. Students find the condition for a = 0 and explain it.
- Criterion C. Communicating: showing each equation and the check with a point not used (worked examples 1 and the faded example).
- Criterion D. Applying mathematics in real-life contexts: the cable model (question 5), and the warning card's caution on how far a fitted model can be trusted.
The one thing to do with the figure
Ask for a prediction before every slide. With the heights 3, −1 and 3 on the screen, ask: “If I raise the right height, which way does the vertex move?” Then check with the readout.
Then ask for three heights that give a line, and ask why a straight line is the only curve through three points in line. Students should see that a parabola would have to bend.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the salt basin. Collect predictions for the number of measurements. |
| 12 min | The figure with predictions; the key idea; settle the hook. |
| 10 min | Warning card and worked examples; pairs finish the faded example. |
| 5 min | Question 1, the check. If many choose A or D, return to using all three points. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 8. |
The answers
| Warm-up | a = 4, b = 3; (3, 5); 2x² − 14x + 12. |
|---|---|
| Hook | Three, because y = ax² + bx + c has three unknowns. |
| Worked examples | y = 2x² − 3x + 4; y = 2x² − 8x + 2; y = −0.5(x + 2)(x − 4), vertex (1, 4.5). |
| Faded example | y = 3x² − 2x − 1. |
| 1. Equation through three points | B. A fits only two of the points; C has c = −5; D fits only two. |
| 2. Two equations for a and b | 3. a + b = 5, a − b = 1. |
| 3. Zeros and one point | −8. a = 2. |
| 4. Vertex and a point to the y-intercept | 29. a = 2. |
| 5. Modelling a cable | 0.375. a = 0.025. |
| 6. Two equal heights | 9. b = −6. |
| 7. Three points, the y-intercept | 6. a = 1, b = −5. |
| 8. When three points do not give a parabola | C. The three points are collinear: y falls 2 for every 1 across. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A, C or D in question 1 | Used only two of the three points, or lost the constant. | Substitute each point into every option. |
| 2 or 5 in question 2 | Gave b or the first equation. | Ask: what does the question ask for? |
| 2, −4 or 8 in question 3 | Gave a, left a out, or lost the sign. | Find a first, then evaluate with it. |
| 13, −3 or 35 in question 4 | Left out a, gave k, or added the 3. | Find a from the extra point first. |
| 1, 0.625 or 1.625 in question 5 | Gave a measurement, or mishandled the drop from the point. | Draw the parabola and mark the 25a drop. |
| 4, −6 or 3 in question 6 | Gave the height, b or the axis. | Find b from the axis, then c from a point. |
| 2, 1 or −5 in question 7 | Gave a given height, a or b. | Ask which letter is the y-intercept. |
| A, B or D in question 8 | Believed any three points make a parabola. | Check the differences in y for equal steps in x. |
Questions worth asking
- Why is three the number of points needed? Three unknowns need three independent equations.
- Why do three collinear points give a = 0? The line is the only curve through them that fits; the x² term is not needed.
- Which form do you choose when you know the vertex? Vertex form, because it already uses h and k.
- How far can you trust a fitted model? Near the data; outside it, check against the situation.
Differentiation
A way in for anyone stuck
- Give a three-line template: write each point as an equation, then subtract, then solve.
- Start with the point on the y-axis, which gives c at once.
- Provide the vertex and zeros forms on a reference card.
Stretch
- Show that for points at x = −1, 0 and 1 with heights p, q and r, a = (p + r − 2q) ÷ 2 and b = (r − p) ÷ 2.
- Four points are given. Describe a way to test whether they lie on one parabola.
Students learning in English as an additional language
“Fit” means find the equation that passes through the points. “Unknown” means a letter whose value we do not know yet. “Collinear” means in a straight line. “Model” means an equation that describes a situation.
| Word | Plain meaning | Example |
|---|---|---|
| unknown | a letter whose value must be found | a, b and c in ax² + bx + c |
| independent facts | facts that each give new information | the heights at x = 0 and at x = 2, not the height at x = 0 measured twice |
| collinear | lying on one straight line | (0, 1), (2, 5) and (4, 9) |
| fit | to find the equation that passes through given points | a quadratic fitted to three measurements |
| model | an equation that describes a real situation | the height of a thrown ball |
| check | put a point back in to test an answer | substitute x = 3 and compare y |
Sentence starters for explaining aloud or in writing:
- I put the point (…) into the equation to get …
- I used vertex form because …
- The points are collinear because …
Two things not to say
“Two points fix a parabola.” Two points fix a line; a parabola needs three facts.
“Any three points lie on a parabola.” Three points in a straight line do not.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- Find cA parabola y = x² + bx + c passes through (0, 7). Find c.
7. - Vertex formWrite the equation with vertex (1, 3) and a = 2.
y = 2(x − 1)² + 3.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA thrown ball is at height 2 m at t = 0, 7 m at t = 1 and 6 m at t = 2. Write h(t).
h = −3t² + 8t + 2: c = 2, a + b = 5, 4a + 2b = 4, so a = −3 and b = 8. - Earlier work (Year 9)Find the gradient of the line through (1, 2) and (3, 8).
3.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningFind the quadratic with zeros 3 and −1 passing through (1, 8).
y = −2(x − 3)(x + 1), because a(1 − 3)(1 + 1) = −4a = 8 gives a = −2. - ModellingHeights 10, 4 and 6 are measured at x = 0, 2 and 4. Find the quadratic.
c = 10; 2a + b = −3 and 4a + b = −1, so a = 1 and b = −5: y = x² − 5x + 10.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
The figure works with the keyboard (tab to a slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.
The With a GDC section is closed by default and optional. Students without a GDC lose nothing; every answer is reached by hand.