IB Maths AA HL Topic 2 — Functions Paper 1 & 2 ~6 min read

Reflections of Graphs

A reflection flips a graph across an axis. There are exactly two cases to learn — flip across the x-axis with y = −f(x), or flip across the y-axis with y = f(−x). Each one negates a coordinate; the other coordinate is untouched.

📘 What you need to know

The two reflection rules

Reflection in x-axis
y = −f(x)
(a, b) → (a, −b)
minus outside the bracket
Reflection in y-axis
y = f(−x)
(a, b) → (−a, b)
minus inside the bracket
Both reflections of the same curve
x y O y = f(x) (a, b) y = −f(x) (a, −b) y = f(−x) (−a, b)

Reflection in the x-axis — y = −f(x)

Negate every y-value Every point (a, b) on y = f(x) maps to (a, −b) on y = −f(x)
Changes
y-coordinates flip sign
horizontal asymptotes flip sign
max ↔ min, above ↔ below
Stays the same
x-coordinates
vertical asymptotes
x-intercepts
anything with y = 0 doesn’t move

Reflection in the y-axis — y = f(−x)

Negate every x-value Every point (a, b) on y = f(x) maps to (−a, b) on y = f(−x)
Changes
x-coordinates flip sign
vertical asymptotes flip sign
left ↔ right
Stays the same
y-coordinates
horizontal asymptotes
y-intercepts
anything with x = 0 doesn’t move
Even and odd functions: if f(−x) = f(x) the graph is its own reflection in the y-axis (even, like x2 or cos x). If f(−x) = −f(x), reflecting in both axes gives back the original (odd, like x3 or sin x).

🧭 Recipe — reflecting a graph

  1. Locate the minus sign: outside bracket → x-axis flip; inside bracket → y-axis flip.
  2. For each labelled point (a, b): apply (a, −b) for −f(x), or (−a, b) for f(−x).
  3. Flip the matching asymptotes: HAs flip under −f(x); VAs flip under f(−x).
  4. Sketch the new curve with the same shape, just mirrored. Label all transformed features.

Worked examples

WE 1

Find the equation after a reflection in the x-axis

The graph of f(x) = x2 − 4x + 7 is reflected in the x-axis. Find the equation of the new graph.

Multiply the entire function by −1 y = −f(x) = −(x² − 4x + 7) y = −x² + 4x − 7 y = −x² + 4x − 7 every term gets a sign flip — the parabola that opened up now opens down
WE 2

Find the equation after a reflection in the y-axis

The graph of g(x) = 2x3 − 5x + 1 is reflected in the y-axis. Find the equation of the new graph in simplified form.

Replace x with −x y = g(−x) = 2(−x)³ − 5(−x) + 1 Simplify each term 2(−x)³ = −2x³ −5(−x) = 5x y = −2x³ + 5x + 1 y = −2x³ + 5x + 1 odd powers of x flip sign, even powers don’t, constants stay the same
WE 3

Reflect the key points of a graph

The graph of y = f(x) has a maximum at P(−3, 4), a minimum at Q(2, −5), and crosses the y-axis at (0, 1). State the coordinates of the corresponding three points on:
(a) y = −f(x)   (b) y = f(−x)

(a) y = −f(x): negate y-coordinates P(−3, 4) → (−3, −4) — now a minimum Q(2, −5) → (2, 5) — now a maximum (0, 1) → (0, −1) (a) min (−3, −4); max (2, 5); y-int (0, −1) (b) y = f(−x): negate x-coordinates P(−3, 4) → (3, 4) — still a maximum Q(2, −5) → (−2, −5) — still a minimum (0, 1) → (0, 1) — unchanged (on the y-axis) (b) max (3, 4); min (−2, −5); y-int (0, 1) x-axis flip swaps max with min; y-axis flip keeps them as max and min — just on the other side
WE 4

Reflection of a rational function

The graph of y = 3x − 1 + 2 is reflected in the x-axis.
(a) Find the equation of the new graph. (b) State its vertical and horizontal asymptotes.

(a) Multiply the whole expression by −1 y = −[3/(x − 1) + 2] y = −3/(x − 1) − 2 (a) y = −3/(x − 1) − 2 (b) Asymptotes — VA stays, HA flips sign original VA: x = 1 → unchanged original HA: y = 2 → flipped to y = −2 (b) VA: x = 1; HA: y = −2 x-axis reflection only touches y-coordinates — including the y-value of the horizontal asymptote
WE 5

Identify the reflection from two equations

The graph of y = ex is transformed by a single reflection to give the graph of y = ex. State the axis of reflection.

Compare the two equations y = e^x   vs   y = e^(−x) x has been replaced by −x The minus sign is inside the function input that’s the form y = f(−x) reflection in the y-axis classic exam set-up: e^x and e^(−x) are mirror images across the y-axis — same y-intercept (0, 1)
WE 6

Track intercepts under both reflections

The graph of y = f(x) has y-intercept (0, 6) and x-intercepts at (−2, 0) and (5, 0). State the y-intercept and x-intercepts of:
(a) y = −f(x)   (b) y = f(−x)

(a) y = −f(x): y-coordinates flip, x-coordinates stay y-int: (0, 6) → (0, −6) x-ints: (−2, 0) and (5, 0) → unchanged (y = 0 stays 0) (a) y-int (0, −6); x-ints (−2, 0) and (5, 0) (b) y = f(−x): x-coordinates flip, y-coordinates stay y-int: (0, 6) → (0, 6) — unchanged (x = 0 stays 0) x-ints: (−2, 0) → (2, 0); (5, 0) → (−5, 0) (b) y-int (0, 6); x-ints (2, 0) and (−5, 0) x-intercepts sit on the x-axis so they’re fixed under x-axis reflection; y-intercept sits on the y-axis so it’s fixed under y-axis reflection

💡 Top tips

⚠ Common mistakes

Translations slide and reflections flip — both leave the shape intact. The next note tackles stretches, where the size of the graph changes too. The minus sign rules you just learned generalise: outside the bracket affects y; inside the bracket affects x.

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