IB Maths AA HL Topic 2 — Functions Paper 1 & 2 HL only ~8 min read

Square Transformations

The square transformation y = [f(x)]2 keeps every x-coordinate the same and squares each y-value. Negatives flip up (like with |f|), but with a key difference: the result is smooth at the x-intercepts, not a cusp. Squaring also stretches values with |y| > 1 away from the axis and pulls values with |y| < 1 toward it. The whole transformation reduces to another feature-mapping exercise.

📘 What you need to know

The basic idea — square the y-coordinate

Square transformation y = f(x)   →   y = [f(x)]2

Every point (x, y) on the original transforms to (x, y2). Since y2 is always non-negative, the resulting graph never dips below the x-axis — similar to y = |f(x)|, but with an important geometric difference at the x-intercepts.

Two regions matter: where the original |y| > 1, squaring pushes the height further from the axis (e.g., y = 3 → y2 = 9). Where |y| < 1, squaring pulls the height closer to the axis (e.g., y = 0.5 → y2 = 0.25). At y = 0 and y = 1, nothing changes.

How key features transform

On y = f(x)becomes on y = [f(x)]2
y-intercept (0, c)y-intercept at (0, c2)
x-intercept (root) at (a, 0)smooth minimum at (a, 0) — touches the x-axis
vertical asymptote at x = avertical asymptote at x = a (unchanged)
local max at (x1, y1), y1 > 0local max at (x1, y12) — type preserved
local max at (x1, y1), y1 ≤ 0local min at (x1, y12) — flipped up
local min at (x1, y1), y1 ≥ 0local min at (x1, y12) — type preserved
local min at (x1, y1), y1 < 0local max at (x1, y12) — flipped up
horizontal asymptote y = khorizontal asymptote y = k2
The single rule for extrema: a max or min with negative y1 flips to the opposite type when squared (just like with |f|). With positive or zero y1, the type stays the same. Think of the squaring as “reflect-then-rescale” — but only for negative values.

Square vs absolute value — the key difference

Both y = [f(x)]2 and y = |f(x)| produce graphs that never go below the x-axis. But they differ in two important ways:

[f(x)]2
smooth minimum at x-intercepts
curve touches axis tangentially — no cusp
|f(x)|
cusp at x-intercepts
sharp corner — gradient changes sign abruptly

Squaring also stretches and compresses heights non-uniformly — heights with |y| > 1 get pushed further from the axis (often dramatically), while heights with |y| < 1 get pulled in. The absolute value just reflects without rescaling.

🧭 Recipe — sketch y = [f(x)]2 from y = f(x)

  1. Mark every key feature on the original — intercepts, asymptotes, extrema (with their signs).
  2. Square each y-coordinate (keep x-coordinates unchanged).
  3. Decide max-or-min type for each extremum based on the sign of y1: positive ⇒ same type, negative ⇒ swap.
  4. Convert x-intercepts to smooth minima on the x-axis. Draw smoothly, not as cusps.
  5. Keep VAs in place; convert any HA at y = k to HA at y = k2.
  6. Sketch confirming the range is y ≥ 0 everywhere.

Worked examples

WE 1

Square of a quadratic

Sketch the graph of y = (x2 − 4)2, identifying all key features.

Step 1: Identify features of f(x) = x² − 4 parabola; roots ±2; min at (0, −4); f → +∞ as x → ±∞ Step 2: Apply feature map roots ±2 → smooth minima at (±2, 0) min (0, −4) with y₁ = −4 < 0 → local max at (0, 16) f → +∞ → [f]² → +∞ (range y ≥ 0) Step 3: Y-intercept y(0) = (0 − 4)² = 16 → (0, 16) smooth minima at (±2, 0); local max (0, 16); symmetric about y-axis curve touches x-axis tangentially at ±2 — no cusps, unlike |x² − 4| which would have sharp corners there
WE 2

Square of a linear function

Sketch the graph of y = (2x − 6)2, identifying the vertex and y-intercept.

Step 1: Identify features of f(x) = 2x − 6 linear; root at x = 3; y-int (0, −6) Step 2: Apply feature map root x = 3 → smooth minimum at (3, 0) y-int (0, −6) → y-int (0, 36) Step 3: Confirm shape (2x − 6)² = 4(x − 3)² — parabola opening up with vertex (3, 0) vertex (3, 0); y-int (0, 36); parabola opening upward the square of a linear function is always a parabola — root becomes the vertex
WE 3

Track a local minimum with negative y-value

The function y = f(x) has a local minimum at (4, −3). State the position and type of the corresponding feature on y = [f(x)]2.

Step 1: x-coordinate is unchanged x = 4 Step 2: y-coordinate becomes y² y = −3 → y² = 9 Step 3: Sign of y₁ determines type y₁ = −3 < 0 → min flips to max local maximum at (4, 9) a min that sits below the axis becomes a max above it — squaring flips negatives up
WE 4

Track both a positive max and a negative min

The function y = f(x) has a local maximum at (1, 5) and a local minimum at (−2, −3). State the position and type of each feature on y = [f(x)]2.

Step 1: (1, 5) — positive max y₁ = 5 > 0 → max stays a max new point: (1, 25) Step 2: (−2, −3) — negative min y₁ = −3 < 0 → min flips to max new point: (−2, 9) local max (1, 25); local max (−2, 9) both points are now maxima — when y₁ < 0, the squaring inverts the role of extrema
WE 5

Square vs absolute value — direct comparison

For f(x) = (x − 1)(x + 2), state the position and type of features on (a) y = [f(x)]2, (b) y = |f(x)|.

Identify features of f(x) = x² + x − 2 roots x = 1 and x = −2; vertex (−1/2, −9/4); y-int (0, −2) (a) y = [f(x)]² roots → smooth minima at (1, 0) and (−2, 0) vertex (−1/2, −9/4) → max at (−1/2, 81/16) y-int → (0, 4) (a) smooth minima at (1, 0) and (−2, 0); local max (−1/2, 81/16); y-int (0, 4) (b) y = |f(x)| roots → cusps at (1, 0) and (−2, 0) vertex (−1/2, −9/4) → max at (−1/2, 9/4) y-int → (0, 2) (b) cusps at (1, 0) and (−2, 0); local max (−1/2, 9/4); y-int (0, 2) same range and roots, but [f]² is smooth and stretches |y| > 1 values further (compare 9/4 vs 81/16)
WE 6

Comprehensive transformation

The function y = f(x) has:
• local maximum at A(−3, 4)
• y-intercept at B(0, 2)
• x-intercept at C(5, 0)
• vertical asymptote at x = 8
• horizontal asymptote at y = −2.
Describe the position and type of each feature on y = [f(x)]2.

Apply the feature map to each in turn A(−3, 4): y₁ = 4 > 0 → max stays a max → (−3, 16) B(0, 2): y-int → y-int (0, 4) C(5, 0): root → smooth minimum at (5, 0) VA at x = 8 → still VA at x = 8 HA at y = −2 → HA at y = (−2)² = 4 local max (−3, 16); y-int (0, 4); smooth min (5, 0); VA at x = 8; HA at y = 4 five features, five clean mappings — squaring kept the VA and the max type because y₁ was positive at A

💡 Top tips

⚠ Common mistakes

And that closes Section 2.10 — Modulus & Further Transformations, the final section of Topic 2 — Functions. You now have the full toolkit: linear and quadratic functions, exponentials and logs, polynomials, transformations, inequalities, modulus, reciprocal, and square. Functions are the foundation everything else in the IB AA HL course builds on — calculus, complex numbers, and probability all lean heavily on what you’ve covered here.

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