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Sketch the absolute-value reflection of the cubic.

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

TOPIC 01

2026 JM02

f(|x|) copies only the original right half, not the original left half.

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

01 / Standard#Your turn

Sketch the absolute-value reflection of the cubic.

y=−f(∣x∣),−2≤x≤2y=-f(|x|),\quad -2\le x\le2

Official paper · jm02-2026 · 2(a)(v) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 9

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Retain the right-hand part of f and reflect it in the y-axis.
Hint 2
The leading minus then reflects the result in the x-axis.
Worked solution
  1. Write an even expression for the transformed curve.

    y=−∣x∣3+x2+∣x∣y=-|x|^3+x^2+|x|
  2. Use the right half of the old curve and its mirror; there are peaks at ±1 and a corner at the origin.

    (0,0), (±1,1), (±2,−2),x-intercepts:0, ±1+52(0,0),\ (\pm1,1),\ (\pm2,-2),\quad x\text{-intercepts}:0,\ \pm\frac{1+\sqrt5}2
y = −f(|x|)-2-1012-3-2-1012(0,0)(-1,1)(1,1)(-2,-2)(2,-2)y = −f(|x|)

The graph is even, with maxima at (±1,1) and endpoints (±2,−2).

Checks and common pitfalls: f(|x|) copies only the original right half, not the original left half.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Write an even expression for the transformed curve.
    y=−∣x∣3+x2+∣x∣y=-|x|^3+x^2+|x|
  • Use the right half of the old curve and its mirror; there are peaks at ±1 and a corner at the origin.
    (0,0), (±1,1), (±2,−2),x-intercepts:0, ±1+52(0,0),\ (\pm1,1),\ (\pm2,-2),\quad x\text{-intercepts}:0,\ \pm\frac{1+\sqrt5}2

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Curriculum and source notes ↗