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

Reciprocal Transformations

The reciprocal transformation y = 1/f(x) keeps every x-coordinate the same and replaces each y-value with its reciprocal. The geometry is striking: roots become vertical asymptotes, vertical asymptotes become roots, and maxima and minima swap roles. Once you know the feature-mapping rules, sketching y = 1/f(x) from a graph of y = f(x) is a fast, mechanical process.

📘 What you need to know

Definition: y = 1/f(x) replaces each y on the original graph with 1/y. x-coordinates are unchanged.
Sign is preserved: where f > 0, 1/f > 0; where f < 0, 1/f < 0.
Increasing/decreasing flip: where f is increasing, 1/f is decreasing (and vice versa).
Magnitude flip: large y-values become small; small y-values become large.
Fixed points: any point with y = ±1 stays exactly the same.
Roots ↔ Vertical asymptotes: a root of f at x = a becomes a VA of 1/f at x = a, and vice versa.
Maxima ↔ Minima: a local max of f at (x₁, y₁) with y₁ ≠ 0 becomes a local min of 1/f at (x₁, 1/y₁), and vice versa.
Horizontal asymptotes: HA at y = k (with k ≠ 0) becomes HA at y = 1/k; if f → ±∞, then 1/f → 0 (HA at y = 0).

The basic idea — flip the y-coordinate

Reciprocal transformation y = f(x) → y = 1f(x)

Every point (x, y) on the original graph (where y ≠ 0) transforms to (x, 1/y). That single rule generates everything else: large heights become small heights close to the x-axis, and small heights become large heights pushed away from the x-axis. Where the original graph crosses zero, the reciprocal blows up.

Geometric reading: think of the x-axis as a “ceiling” for the reciprocal — points far from it on the original become close to it on the reciprocal, and vice versa. This explains why roots (closest to the axis) become asymptotes (furthest from it).

How key features transform

The whole transformation reduces to a small mapping table. Memorise it once; reuse forever:

On y = f(x)	becomes on y = 1/f(x)
y-intercept (0, c), c ≠ 0	y-intercept at (0, 1/c)
x-intercept (root) at (a, 0)	vertical asymptote at x = a
vertical asymptote at x = a	discontinuity at (a, 0) — looks like a root
local max at (x₁, y₁), y₁ ≠ 0	local min at (x₁, 1/y₁)
local min at (x₁, y₁), y₁ ≠ 0	local max at (x₁, 1/y₁)
horizontal asymptote y = k, k ≠ 0	horizontal asymptote y = 1/k
f(x) → ±∞ (no HA)	horizontal asymptote y = 0
HA at y = 0	1/f(x) → ±∞ (no HA)
point with y = ±1	same point (fixed)

A useful symmetry: roots and vertical asymptotes are mutual mirrors under this transformation. So is the relationship between f → ∞ and 1/f → 0 (HA at zero). The roles flip but the picture stays connected.

Special cases worth memorising

If f(a) = 0

VA at x = a on 1/f

the reciprocal of zero blows up — every root becomes an asymptote

If f has VA at x = a

discontinuity (a, 0) on 1/f

looks like a root — but technically there’s a hole

🧭 Recipe — sketch y = 1/f(x) from y = f(x)

Mark every key feature on the original — intercepts, asymptotes, extrema.
Apply the feature map: roots → VAs, VAs → roots, maxes ↔ mins, etc.
Convert each y-coordinate to its reciprocal (keep the x-coordinate unchanged).
Determine the sign in each region — same as the original: positive where f > 0, negative where f < 0.
Sketch each region’s behaviour approaching its asymptotes correctly.
Mark fixed points if the original passed through any (x, ±1) — they stay put.

Worked examples

WE 1

Sketch the reciprocal of a quadratic

Sketch the graph of y = 1/(x² − 4), identifying all asymptotes and 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 → vertical asymptotes x = ±2 min (0, −4) → local max at (0, −1/4) f → +∞ → HA at y = 0 Step 3: Sign analysis x < −2: f > 0 → 1/f > 0 (above x-axis) −2 < x < 2: f < 0 → 1/f < 0 (below x-axis, max −1/4) x > 2: f > 0 → 1/f > 0 (above x-axis) VAs at x = ±2; HA at y = 0; local max (0, −1/4); symmetric about y-axis three regions, each between consecutive asymptotes; outer regions hug the x-axis from above; middle region sits below it

WE 2

Reciprocal of a linear function

Sketch the graph of y = 1/(2x + 4), identifying all asymptotes and intercepts.

Step 1: Identify features of f(x) = 2x + 4 linear; root at x = −2; y-int (0, 4); f → ±∞ as x → ±∞ Step 2: Apply feature map root x = −2 → vertical asymptote x = −2 y-int (0, 4) → y-int (0, 1/4) f → ±∞ → HA at y = 0 Step 3: Sign analysis x < −2: f < 0 → 1/f < 0 x > −2: f > 0 → 1/f > 0 hyperbola; VA at x = −2; HA at y = 0; y-int (0, 1/4) reciprocal of a linear function is always a hyperbola — root becomes the only VA, HA at zero

WE 3

Track a local minimum through the transformation

The function y = f(x) has a local minimum at (3, −2). State the position and type of the corresponding feature on y = 1/f(x).

Step 1: x-coordinate is unchanged x = 3 Step 2: y-coordinate becomes 1/y y = −2 → 1/y = −1/2 Step 3: Min ↔ max swap (since y₁ ≠ 0) local min becomes local max local maximum at (3, −1/2) type swaps because reciprocal flips magnitudes — the smallest negative value of f gives the closest-to-zero (so largest) negative value of 1/f

WE 4

Track multiple features through the transformation

The function y = f(x) has a y-intercept at (0, 5), an x-intercept at (−2, 0), and a horizontal asymptote at y = 3. Find the corresponding features of y = 1/f(x).

Step 1: y-intercept (0, 5) y = 5 → 1/y = 1/5 → y-int (0, 1/5) Step 2: x-intercept (−2, 0) root → vertical asymptote at x = −2 Step 3: Horizontal asymptote y = 3 y = 3 → 1/y = 1/3 → HA at y = 1/3 y-int (0, 1/5); VA at x = −2; HA at y = 1/3 the three features map cleanly via the table — no surprises when none of the y-values are zero

WE 5

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 = 1/f(x).

Apply the feature map to each in turn A(−3, 4) — local max → local min at (−3, 1/4) B(0, 2) — y-int → y-int at (0, 1/2) C(5, 0) — root → VA at x = 5 VA at x = 8 → discontinuity at (8, 0) (looks like a root) HA at y = −2 → HA at y = −1/2 local min (−3, 1/4); y-int (0, 1/2); VA at x = 5; root-like point at (8, 0); HA at y = −1/2 five features, five clean mappings — work systematically through the table

WE 6

Reverse engineering — features of f from features of 1/f

The graph of y = 1/f(x) has a vertical asymptote at x = 4, a root at x = −1, a local maximum at (2, 3), and a horizontal asymptote at y = 0. Find the corresponding features of y = f(x).

Step 1: VA at x = 4 of 1/f → root at x = 4 of f f has x-intercept at (4, 0) Step 2: Root at x = −1 of 1/f → VA at x = −1 of f f has vertical asymptote at x = −1 Step 3: Local max (2, 3) of 1/f → local min (2, 1/3) of f Step 4: HA at y = 0 of 1/f → f → ±∞ (no HA) x-int (4, 0); VA at x = −1; local min (2, 1/3); no HA the reciprocal map is its own inverse — applying it twice gets you back to the original

💡 Top tips

Memorise the feature map. Six clean rules cover almost every exam scenario.
Sign is preserved — positive regions stay positive, negative stay negative. Use this as a sanity check after sketching.
Roots and VAs are mutual mirrors. Whenever you see one, expect the other under the reciprocal.
Max ↔ Min always (when y ≠ 0). The type swaps even though the x-coordinate stays put.
Watch for the HA at zero special case: f → 0 means 1/f → ±∞, so no horizontal asymptote on the reciprocal.
Fixed points at y = ±1 give you free anchor points — useful for orienting your sketch.
The reciprocal map is its own inverse: applying it twice gives you back the original function.

⚠ Common mistakes

Forgetting that max ↔ min. Reciprocal flips the role of extrema even when the x-coordinate is the same.
Reciprocating the x-coordinate. Only y-values get flipped — x-coordinates stay exactly the same.
Treating an HA at zero like an HA at non-zero. y = 1/k only works for k ≠ 0; k = 0 means the reciprocal blows up.
Getting the sign wrong in some region. Sign is preserved — if f < 0 there, 1/f < 0 there.
Missing the sign-flip at a VA: the reciprocal will jump from +∞ to −∞ (or vice versa) across the asymptote, just like the original did across the root.
Drawing through the discontinuity at (a, 0) where the original had a VA. There’s a hole in the graph there, even if it looks like a root.
Forgetting that increasing → decreasing. The monotonicity in each region flips under the transformation.

Reciprocal transformation is a feature-mapping exercise once you’ve memorised the rules. The next note, Square Transformations, takes the same approach but with y = [f(x)]²: every y-value gets squared, so negatives flip up (like with |f|), but the squaring also flattens small values and exaggerates large ones. A different feature map, same structured approach.

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