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Which statement holds?

Read the idea, work independently, then explain what changed.

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高一必修 第一册(A版).pdf · 2.3 · PDF 57 / printed page 50

TOPIC 01

Vertex form and completing the square

Explain turning points, symmetry and constrained optima through a square.

What you will be able to explain

  • Complete a square without changing the function.
  • Distinguish the extremum value, attaining input and admissible domain.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Your turn

Which statement holds?

y=−2(x−1)2+3y=-2(x-1)^2+3
  • All inputs are real unless a smaller domain is stated.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. Minimum 3
  2. Maximum 3
  3. Maximum −2
  4. No extremum
Working and explanation
BUILD THE REASONING
Hint 1
Check the square coefficient.
Hint 2
A negative square term is nonpositive.
Worked solution
  1. Bound the changing term.
    −2(x−1)2≤0-2(x-1)^2\le0
  2. Equality is feasible at x=1.
    y≤3y\le3
B: maximum 3.
Checks and common pitfalls: Opening direction decides the type of extremum.
Think first. Reveal a hint when the class is ready.
Focus on one question
Teacher preparation and assessment
40-minute lesson
MinutesFocusTeaching and student action
5Diagnose signsUse both diagnostics and ask where the square vanishes.
8Build a squareTrack the added and subtracted constant in x²−6x+5.
10Compare formsPairs recover a quadratic from its vertex and a point, then explain a symmetric table.
12Defend a choiceGroups solve the pack problem and challenge arbitrary rounding.
5Exit checkSeparate the optimum value from the attaining input.
Question sequence
  • Why must the added square be subtracted too?
  • Can two distinct allowed inputs tie?
Board development
  • Preserve equality.
    x2−6x+5=(x−3)2−4x^2-6x+5=(x-3)^2-4
  • Compare allowed distances.
    C=(n−4.5)2+9.75C=(n-4.5)^2+9.75
Expected correct and incorrect responses
  • “The square is zero at 3, so the value is −4” connects input and output.
  • “Round to 5” overlooks the tied solution 4.
Follow-up questions
  • What changes if the leading coefficient is negative?
Remediation
  • For axis-sign errors, substitute both proposed inputs into the square.
  • For compensation errors, expand the proposed vertex form.
Exit check
  • For f=2(x+2)²−1 and integer x≥0, justify the minimum. Expected f(0)=7.
Assessment checklist and sample responses
  • 2 marks: preserve equality while completing.
    Full: 2[(x+2)²−4]+3; partial: missing −8.
  • 2 marks: justify admissibility and ties.
    Full: n=4 or 5, cost 10; incomplete: n=4.5.
No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.
Curriculum and source notes ↗