Transformations and mapping rules
Every transformed function on this course is y = af[b(x − h)] + k. This page decodes the four parameters, gives the mapping rule that turns any point into its image, settles the order they are applied in, and works through the questions that use them — domain and range, invariant points, and inverses.
- 1. The general form
- 2. The mapping rule
- 3. Why order matters
- 4. Domain, range and asymptotes
- 5. Reflections and invariant points
- 6. Inverses
- 7. Saying it the way markers want
The general form
The idea: y = af[b(x − h)] + k. Outside the brackets, a and k do what they look like they do. Inside, b and h do the opposite.
| Parameter | Does | Direction | Example |
|---|---|---|---|
| a | vertical stretch by |a|; reflection in the x-axis if a < 0 | affects y — behaves as written | a = −2: twice as tall, flipped upside down |
| b | horizontal stretch by 1/|b|; reflection in the y-axis if b < 0 | affects x — reciprocal of what it looks like | b = 3: squashed to a third of the width |
| h | horizontal translation | x − h moves right h | x − 5: 5 units right; x + 5: 5 units left |
| k | vertical translation | +k moves up | +3: 3 units up |
Watch out: b must be factored out before you can read h. In y = f(2x − 6), rewrite as y = f(2(x − 3)): the stretch is 1/2 and the translation is 3 right, not 6.
The mapping rule
The idea: (x, y) → (x/b + h, ay + k). Feed a point in, get its image out. Every "find the corresponding point" question is this one line.
| Transformation | Mapping |
|---|---|
| y = f(x) + k | (x, y) → (x, y + k) |
| y = f(x − h) | (x, y) → (x + h, y) |
| y = af(x) | (x, y) → (x, ay) |
| y = f(bx) | (x, y) → (x/b, y) |
| y = af[b(x − h)] + k | (x, y) → (x/b + h, ay + k) |
Worked: (−6, 4) under y = ½f(3x) − 1
Here a = ½, b = 3, h = 0, k = −1, so the mapping is (x, y) → (x/3, ½y − 1).
x: −6 ÷ 3 = −2. y: ½(4) − 1 = 1. The image is (−2, 1).
The most common error is multiplying by 3 instead of dividing. Ask what b = 3 does to the graph: it squeezes it toward the y-axis, so x-coordinates get smaller.
Worked, in reverse: which equation maps (2, 5) to (8, −7)?
x went from 2 to 8, so x/b + h took 2 to 8 — a translation of 6 right with b = 1 fits: h = 6. y went from 5 to −7: if a = −1 then −5 − 2 = −7, so k = −2, giving y = −f(x − 6) − 2. Check it: (2, 5) → (2 + 6, −5 − 2) = (8, −7). ✓
Why order matters
The idea: stretches and reflections first, translations last. Stretching after a shift stretches the shift too, and lands you somewhere else.
Take the point (2, 3) and apply a vertical stretch by 2 and a translation 5 up, in each order:
| Order | Working | Image |
|---|---|---|
| Stretch, then translate (correct) | 3 × 2 = 6, then 6 + 5 = 11 | (2, 11) — matches y = 2f(x) + 5 |
| Translate, then stretch | 3 + 5 = 8, then 8 × 2 = 16 | (2, 16) — this is y = 2[f(x) + 5] |
The equations are genuinely different, which is why "describe the transformations in order" is worth marks. The safe phrasing: apply the stretch and reflection described by a and b, then translate by h and k.
The same trap inside the brackets
y = f(2x − 6) is not "6 left then stretch". Factor: y = f(2(x − 3)) — horizontal stretch by 1/2, then 3 right. Always factor b out before describing anything.
Domain, range and asymptotes
The idea: transform the endpoints, not the function. Whatever happens to a point happens to the boundary of the domain and range.
Worked: y = f(x) has domain −4 ≤ x ≤ 8. What is the domain of y = f(2(x + 1))?
Whatever sits inside f must still be between −4 and 8:
- −4 ≤ 2(x + 1) ≤ 8
- −2 ≤ x + 1 ≤ 4 (divide by 2)
- −3 ≤ x ≤ 3 (subtract 1)
Or use the mapping rule on the endpoints: x → x/2 − 1 sends −4 to −3 and 8 to 3. Same answer, less algebra.
- Horizontal changes (b, h) move the domain; vertical changes (a, k) move the range.
- Asymptotes travel with the graph. y = log₂(x + 4) − 3 has its vertical asymptote at x = −4; y = 1/(x − 3) + 2 has a vertical asymptote x = 3 and a horizontal one y = 2.
- A reflection swaps the ends of a range: if the range was y ≥ 2, then after y = −f(x) it is y ≤ −2.
Reflections and invariant points
The idea: an invariant point is one the transformation leaves exactly where it is. Find them by asking which points the rule cannot move.
| Transformation | Invariant points | Why |
|---|---|---|
| y = −f(x) | every x-intercept | −0 = 0, so points on the x-axis stay |
| y = f(−x) | every y-intercept | x = 0 is unchanged by the sign flip |
| y = af(x), a ≠ 1 | the x-intercepts only | a·0 = 0, but any other y changes |
| y = √f(x) | points where y = 0 or y = 1 | √0 = 0 and √1 = 1 |
| y = 1/f(x) | points where y = 1 or y = −1 | these are their own reciprocals |
Counting them is a standard exam question: if a parabola crosses the x-axis twice, then y = −f(x) has 2 invariant points, and y = √f(x) has 2 plus however many times the graph reaches height 1.
Inverses
The idea: an inverse swaps x and y, so it reflects the graph in the line y = x. Domain and range swap with them.
- Write y = f(x) — replace the function name with y.
- Swap x and y.
- Solve for y. That expression is f⁻¹(x), if it is a function at all.
Worked: f(x) = 2x³ − 5
y = 2x³ − 5 → x = 2y³ − 5 → x + 5 = 2y³ → y³ = (x + 5)/2 → f⁻¹(x) = ∛((x + 5)/2).
Check with a point: f(2) = 11, and f⁻¹(11) = ∛(16/2) = ∛8 = 2. ✓
- f⁻¹ is not 1/f. The −1 is notation for "inverse", not an exponent.
- An inverse is a function only if the original passes the horizontal line test. y = x² fails, which is why its inverse needs a restricted domain such as x ≥ 0.
- Points swap: if (3, 7) is on f, then (7, 3) is on f⁻¹. Domain and range trade places too.
Saying it the way markers want
The idea: diploma markers report that vague transformation language is one of the most common ways students lose written-response marks. Name each one in full.
| Instead of | Write |
|---|---|
| "stretched by 2" | "a vertical stretch about the x-axis by a factor of 2" |
| "squished sideways" | "a horizontal stretch about the y-axis by a factor of 1/3" |
| "flipped" | "a reflection in the x-axis" |
| "moved over 5" | "a translation of 5 units right" |
| "domain is −3 to 3" | "the domain is {x | −3 ≤ x ≤ 3, x ∈ R}" |
Say what it is (stretch, reflection, translation), about what (the x-axis, the y-axis), and by how much (a factor, a number of units, a direction). Then list them in the order they are applied.