IB Maths AI HL Transformations of Graphs Paper 1 & 2 x-axis, y-axis, asymptotes ~8 min read

Reflections of Graphs

A reflection flips the graph across an axis, keeping shape and size but swapping orientation. y = −f(x) flips vertically (across the x-axis), y = f(−x) flips horizontally (across the y-axis). Only one coordinate of each point changes sign.

📘 What you need to know

The two reflections side by side

Two simple rules cover this topic. Putting a minus sign outside the function — y = −f(x) — multiplies every output by −1, so the curve flips top-to-bottom across the x-axis. Putting a minus sign inside the bracket — y = f(−x) — feeds in the negative of the input, so the curve flips left-to-right across the y-axis. In both cases the shape is identical to the original: same widths, same heights, same maxima and minima — just relocated.

What changes (and what doesn’t)

For y = −f(x): each y-coordinate is negated. A maximum becomes a minimum (and vice versa). Any point that was on the x-axis stays where it is. Horizontal asymptotes flip sign (y = 3 becomes y = −3); vertical asymptotes don’t move. For y = f(−x): each x-coordinate is negated. A maximum stays a maximum (the curve isn’t tipped over, just mirrored sideways). Points on the y-axis stay put. Vertical asymptotes flip sign (x = 4 becomes x = −4); horizontal asymptotes are untouched.

y-axis reflection flips x; x-axis reflection flips y y = f(−x) — reflect in y-axis x y f(x) (2, −3) f(−x) (−2, −3) x → −x y = −f(x) — reflect in x-axis x y f(x) (2, −3) min −f(x) (2, 3) max y → −y
The vertex (2, −3) mirrors to (−2, −3) under y-axis reflection (still a minimum) and to (2, 3) under x-axis reflection (now a maximum).
Reflection at a glance y = −f(x)  ↔  reflect in x-axis  ·  y = f(−x)  ↔  reflect in y-axis point (p, q) → (p, −q) under −f; → (−p, q) under f(−x)

Reflecting equations and asymptotes

For an explicit function, the reflection is mechanical. To get y = −f(x), put a minus sign in front of the whole formula and expand. To get y = f(−x), substitute −x wherever x appears, then simplify. Asymptotes follow the same rules: only the asymptote of the matching orientation moves. Vertical asymptotes follow horizontal reflections (y-axis reflection); horizontal asymptotes follow vertical reflections (x-axis reflection).

One-axis rule: a reflection moves only the coordinate matching its name. x-axis reflection → y-coordinates flip. y-axis reflection → x-coordinates flip. Asymptotes obey the same rule.

🧭 Recipe — reflecting a graph

  1. Identify the axis of reflection: minus outside → x-axis; minus inside → y-axis.
  2. Apply to the equation: y = −f(x) or y = f(−x); expand or simplify.
  3. Reflect every key point: negate the matching coordinate (y for x-axis; x for y-axis).
  4. Reflect the asymptotes: horizontal asymptotes flip under x-axis reflection; vertical asymptotes flip under y-axis reflection.
  5. Sketch: the new curve is identical in shape, mirrored across the chosen axis.

Worked examples

WE 1

Identify the reflection type

For the graph of y = f(x): (a) Write the equation of the graph reflected in the x-axis. (b) Write the equation of the graph reflected in the y-axis. (c) State which coordinate changes for each reflection.

(a) x-axis reflection → minus outside y = −f(x) (b) y-axis reflection → minus inside y = f(−x) (c) which coordinate changes −f(x): y-coordinates change sign, x stays f(−x): x-coordinates change sign, y stays the axis of reflection’s coordinate is the one that flips.
WE 2

Image of a point

The point Q(−2, 7) lies on the graph of y = f(x). Find the corresponding point on the graph of: (a) y = −f(x); (b) y = f(−x).

(a) negate y-coordinate (−2, 7) → (−2, −7) (−2, −7) (b) negate x-coordinate (−2, 7) → (2, 7) (2, 7)
WE 3

Reflect a quadratic

Given g(x) = x2 − 4x + 5. (a) Find y = −g(x) expanded. (b) Find y = g(−x) expanded. (c) State the vertex of each new graph.

find vertex of g first x² − 4x + 5 = (x − 2)² + 1, vertex (2, 1) (a) y = −g(x) = −(x² − 4x + 5) = −x² + 4x − 5; vertex (2, −1) (b) y = g(−x) = (−x)² − 4(−x) + 5 = x² + 4x + 5 = (x + 2)² + 1 = x² + 4x + 5; vertex (−2, 1) x-axis reflection flipped y; y-axis reflection flipped x.
WE 4

Exponential with asymptote

Given f(x) = 2x + 3 with horizontal asymptote y = 3. For each reflection, find the new asymptote and the y-intercept: (a) y = −f(x); (b) y = f(−x).

f(0) = 1 + 3 = 4 (a) y = −f(x) = −2ᵛ − 3 asymp: y = 3 → y = −3 (horizontal flips) y-intercept: −f(0) = −4 asymp y = −3; (0, −4) (b) y = f(−x) = 2⁻ᵛ + 3 asymp: y-axis reflection doesn’t change horizontal asymptote y = 3 (unchanged); y-intercept f(0) = 4 asymp y = 3; (0, 4)
WE 5

Rational function with both asymptotes

Given f(x) = 1x − 4 + 2 with vertical asymptote x = 4 and horizontal asymptote y = 2. State the asymptotes of: (a) y = f(−x); (b) y = −f(x).

(a) y-axis reflection vertical asymp x = 4 → x = −4 horizontal asymp y = 2 (unchanged) x = −4; y = 2 (b) x-axis reflection vertical asymp x = 4 (unchanged) horizontal asymp y = 2 → y = −2 x = 4; y = −2 the reflection’s name tells you which asymptote moves.
WE 6

Comprehensive: features under both reflections

The graph of y = f(x) has a maximum at (4, 9), a minimum at (−1, −3) and a vertical asymptote x = 2. State the new max/min and asymptote for: (a) y = −f(x); (b) y = f(−x).

(a) x-axis reflection: y flips, max↔min (4, 9) max → (4, −9) min (−1, −3) min → (−1, 3) max vertical asymp x = 2 (unchanged) min (4, −9), max (−1, 3), asymp x = 2 (b) y-axis reflection: x flips, max stays max (4, 9) max → (−4, 9) max (−1, −3) min → (1, −3) min vertical asymp x = 2 → x = −2 max (−4, 9), min (1, −3), asymp x = −2

💡 Top tips

⚠ Common mistakes

Next up: Stretches of Graphs — multiplying the function or its input by a constant to stretch (or compress) the curve along an axis.

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