At a glance. One lesson of about 60 minutes. It opens the unit and assumes Year 9 function notation and the vertex of a parabola. Students evaluate f of an expression, find a natural domain from a denominator or a root, and find a range on all reals and on a restricted domain.
Calculator: GDC optional. Everything can be done by hand. The optional section shows how a GDC table reports a refused input and how a graph confirms a range, and warns that a screen window can mislead.
What students will learn
Know
- f(expression) means the whole expression replaces x.
- The natural domain excludes values that make a denominator 0 or a square root's content negative.
- The range of a quadratic comes from its vertex and, on a restricted domain, its end points.
Understand
- A function is a rule together with the set of inputs it accepts.
- Changing the input before the rule is applied is different from changing the output afterwards.
- A range is reasoned from the rule, and a picture only checks it.
Do
- Simplify f(a + b) for a polynomial f.
- State the natural domain of a root, a fraction, or both.
- Find the greatest or least value on a restricted domain.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: evaluating, simplifying f of an expression and finding domains and ranges (questions 1 to 7).
- Criterion B. Investigating patterns: the what-if. Students compare the ends of two root domains and generalize which part of the rule blocks which end.
- Criterion C. Communicating: stating a domain and range in correct inequality notation, and writing the reason for each refused input (worked examples 2 and 3).
- Criterion D. Applying mathematics in real-life contexts: the spreadsheet hook, where a refused row is a real error a stallholder would see.
The one thing to do with the figure
Ask for a prediction before every slide. With function 2 on the screen, ask: “Where will the graph have a gap?” before moving the input. Then slide x slowly through the refused value and let students say what the readout does.
Before you switch to function 4, ask: “Which function has a range that stops at both ends?” Then switch. Students should link the ends of the range to the maximum value of the number under the root.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the spreadsheet. Collect predictions for the failing row. |
| 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 C, test x = −8 and x = 0; if B, test x = −5. |
| 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 | x² + 10x + 25; 11; 2. |
|---|---|
| Hook | x = 3 is refused, because the denominator is 0; the outputs never include 0. |
| Worked examples | −m² + 7m − 6; x ≥ 3 and x ≠ 8; y ≤ 10, then 1 ≤ y ≤ 10. |
| Faded example | x ≠ 3 and x ≠ −3. |
| 1. Natural domain of a square root | D. A turns the inequality; C loses the sign of 10; B drops the end point. |
| 2. Evaluating a function | −13. 5 − 18. |
| 3. Value refused by a denominator | −4. x + 4 = 0. |
| 4. Function of an expression | −1. t² − t + 2. |
| 5. Range on a restricted domain | 9. f(−1) = 9. |
| 6. Counting inputs in a domain | 9. −4 to 4. |
| 7. Greatest value of a reciprocal | 5. 15 ÷ 3. |
| 8. Natural domain with a strict end point | C. A keeps x = 5; B turns the inequality; D refuses only 5. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A, B or C in question 1 | Turned the inequality, dropped the end point, or lost the sign of 10. | Test one value in each region, such as x = −8 and x = 0. |
| −31, 11 or 23 in question 2 | Squared the product, forgot to square, or took (−3)² as −9. | Write g(−3) = 5 − 2(−3)², then (−3)² = 9 on its own line. |
| 4, 12 or 0 in question 3 | Solved the wrong sign, gave the numerator, or gave a harmless input. | Ask: what makes the bottom 0? |
| 2, −3 or 5 in question 4 | Collected only one of the two t terms, or mishandled a minus sign. | Expand each bracket on its own line before collecting. |
| −3, −7 or 2 in question 5 | Gave the wrong end, the vertex, or the y-intercept. | Draw the parabola and mark both ends and the vertex. |
| 5 or 11 in question 6 | Counted only the non-negative whole numbers, or took the root as 5. | Test x = 4 and x = 5 in 20 − x². |
| 15 or 3 in question 7 | Gave the numerator or the denominator's least value. | Ask what makes a fraction with a fixed top as large as possible. |
| A, B or D in question 8 | Kept the end point, turned the inequality, or refused one value only. | Test x = 5, x = 0 and x = 6 in each rule. |
Questions worth asking
- Why does a square root cause a domain restriction but a square does not? A negative number has no real square root, while every real number can be squared.
- Is the domain of f(x + 2) the same as the domain of f(x)? Not always: if f accepts only x ≥ 0, then f(x + 2) accepts only x ≥ −2. The input expression must land in the domain.
- Where can the greatest value of a quadratic on a restricted domain be? At the vertex if it is inside the domain and opens downward, otherwise at an end point.
- How do you know a range read from a screen is complete? The rule is reasoned out, with the vertex and the end points, so no part of the graph is hidden off the screen.
Differentiation
A way in for anyone stuck
- Give a card with the two refusal rules: a denominator cannot be 0; the content of a square root cannot be negative.
- Write f(…) with an empty bracket first, then fill the bracket with the expression.
- For range, sketch the parabola and shade the allowed values of x on the axis before reading off the y values.
Stretch
- Find the natural domain and range of the function y = √(x − 1) + √(5 − x).
- Show that f(x) = x² on the domain x ≥ 0 has a different graph from f(x) = x² on all reals, and explain why the range is the same.
Students learning in English as an additional language
“Domain” is the set of inputs and “range” is the set of outputs. “Refused” or “not allowed” is used here for an input that gives no real value. Say “f of x plus one” for f(x + 1), and “the square root of…” for the radical.
| Word | Plain meaning | Example |
|---|---|---|
| domain | the set of inputs a function accepts | x ≥ 7 for √(x − 7) |
| range | the set of outputs a function can give | y ≥ 0 for a square root |
| natural domain | all real numbers except those that break the rule | all x except x = 6 for 1 ÷ (x − 6) |
| function notation | f(x) names the output of the rule f at the input x | f(x) = 4x − 1 |
| restricted domain | a domain limited on purpose, not by the rule | x² for 0 ≤ x ≤ 3 |
| refused input | an input that gives no real output | x = 6 in 3 ÷ (x − 6) |
Sentence starters for explaining aloud or in writing:
- The rule refuses x = … because …
- The domain is … and the range is …
- To find f(a + 1) I put … in place of …
Two things not to say
“f(x + 2) means add 2 to f(x).” The 2 is added to the input, inside the bracket, before the rule is used.
“Read the range off the screen.” A window shows only part of a graph; use the vertex and the ends.
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.
- Evaluatef(x) = 7 − 3x. Find f(−2).
13. - DomainState the value that 6 ÷ (x − 9) refuses.
x = 9.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA taxi in Accra charges (60 ÷ n) per person when n people share. Which values of n make sense?
Whole numbers n ≥ 1: n = 0 would divide by 0, and a negative number of people is not possible. - Earlier work (Extended Unit 1, Lesson 3)Write √48 in the form a√3.
4√3.
3Challenge
Reasoning, generalizing and working backwards.
- Reasoningf(x) = x² + 4x + 7 on −5 ≤ x ≤ 1. Find the range.
The vertex is (−2, 3); f(−5) = 12 and f(1) = 12, so 3 ≤ y ≤ 12. - NotationIf f(x) = 2x + 1, show that f(a + b) ≠ f(a) + f(b), and say when equality holds.
f(a + b) = 2a + 2b + 1 but f(a) + f(b) = 2a + 2b + 2. They are never equal.
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.