01 · Transform a point
(6, −2) lies on f. Find its image on y = 3f(2(x + 1)) + 5.
Hint
Solve 2(x + 1) = 6 and transform the output.
Worked solution
x = 2 and y = 3(−2) + 5 = −1, giving (2, −1).
Understand · explore · practise
Combined graph transformations
Combine graph translations, stretches and reflections. Track points, domains, ranges, asymptotes and stationary points, and compare transformation order interactively.
Before you startBasic graph transformations, function notation and domain/range; derivative section is optional
01 / Track a labelled point
Solve for the new input instead of guessing horizontal direction.
Suppose (u, v) lies on y = f(x). For the transformed graph y = a f(b(x − c)) + d, with nonzero a and b, make the inside input equal to u: b(x − c) = u.
(u, v) → (u/b + c, av + d)
The horizontal coordinate is divided by b and then shifted by c. The vertical coordinate is multiplied by a and then shifted by d. Negative scale factors introduce reflections.
If (2, −1) is on fWorked example
New graph: y = −3f(2(x − 1)) + 4
Here a = −3, b = 2, c = 1 and d = 4.
New x = 2/2 + 1 = 2
Solve the inside input condition.
New y = −3(−1) + 4 = 7
The transformed point is (2, 7).
02 / Horizontal order
Factor the input before naming the shift.
The graph y = f(2(x − 3)) can be obtained by a horizontal scale factor 1/2 followed by a translation 3 units right. Its point mapping is u → u/2 + 3.
If you translate 3 units right first and then scale horizontally by 1/2, the mapping becomes u → (u + 3)/2. The final rule is f(2x − 3), equivalent to f(2(x − 3/2)). The final shift is 3/2, not 3.
A horizontal scale factor of 1/2 is often described as a compression. State its direction and factor explicitly to avoid confusing it with multiplying input coordinates by 2.
03 / Vertical order
A later stretch also stretches an earlier translation.
Starting from y = f(x), moving up 2 gives f(x) + 2. Stretching that whole graph vertically by 2 then gives 2(f(x) + 2) = 2f(x) + 4.
Reversing those steps gives 2f(x) + 2. The same named operations can produce different graphs when performed in a different order.
Does the order matter?Explore
Dashed: f(x) = x² − 1. Gold ring: original point. Green: final graph.
Final rule: y = 2(f(x) + 2) = 2x² + 2.
x = 1: original 0 → after up 2: 2 → after ×2: 4.
Watch the same point take two different routes
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Include reflections
A negative factor changes orientation as well as size.
Interpret y = −2f(−(x + 1)) + 3Worked example
Inside input: −(x + 1) = u ⇒ x = −u − 1
Reflect horizontally, then move 1 unit left.
Output: y = −2v + 3
Reflect vertically and stretch by factor 2, then move 3 units up.
Point (3, 4) → (−4, −5)
The complete mapping is more reliable than reversing signs by eye.
Horizontal and vertical operations act on different coordinates and can be interleaved without changing this point mapping. Two operations on the same coordinate may not commute.
05 / Transform domains and ranges
Apply the coordinate rules to the allowed sets.
Suppose f has domain [−2, 6] and range [−1, 5]Worked example
g(x) = −3f(−2(x − 3)) + 4
Keep the original domain and range restrictions.
−2 ≤ −2(x − 3) ≤ 6 ⇒ 0 ≤ x ≤ 4
A negative horizontal factor reverses the ordering of the endpoints.
−1 ≤ f ≤ 5 ⇒ −11 ≤ g ≤ 7
The vertical map sends −1 to 7 and 5 to −11. Range: [−11, 7].
Track endpoint inclusion too. A reflected open endpoint remains open at its new location. Do not assume every transformed graph has domain all real numbers.
06 / Asymptotes and intercepts
Transform a feature only when the relevant coordinate rule applies.
If x = u is a vertical asymptote of f, then x = u/b + c is its image. A horizontal asymptote y = v becomes y = av + d. Extrema and labelled points follow the same coordinate mapping.
y = 2/(3(x − 4)) − 5, from f(x) = 1/xWorked example
Vertical asymptote: x = 4
The original x = 0 maps to 0/3 + 4.
Horizontal asymptote: y = −5
The original y = 0 maps to 2(0) − 5.
x-intercept: 2/(3(x − 4)) = 5 ⇒ x = 62/15
Find the new intercept by solving; it does not come from an original zero because 1/x has none.
y-intercept: y = −1/6 − 5 = −31/6
Substitute x = 0 only if it belongs to the domain.
An original x-intercept (u, 0) maps to height d. It remains an x-intercept only when d = 0. Likewise, translating horizontally can move the original y-intercept away from the y-axis.
07 / Exponential, logarithm and trigonometric graphs
Use a landmark and the domain before drawing.
Three familiar familiesWorked example
y = e^(2(x − 1)) + 3
From eˣ: horizontal scale 1/2, right 1, up 3. Point (0, 1) becomes (1, 4). Domain all real; range y > 3; horizontal asymptote y = 3.
y = −ln(x + 2) + 1
From ln x: left 2, reflect vertically, up 1. Point (1, 0) becomes (−1, 1). Domain x > −2; range all real; vertical asymptote x = −2.
y = 2 sin(3(x − π/6)) − 1
In radians: amplitude 2, period 2π/3, horizontal shift π/6, range [−3, 1]. The original point (0, 0) maps to (π/6, −1).
08 / Stationary points
Optional calculus connection: the inside input still identifies the point.
For differentiable f, g(x) = a f(b(x − c)) + d has derivative g′(x) = ab f′(b(x − c)). Since a and b are nonzero, a stationary input u of f maps to x = u/b + c.
If f has a local maximum at (2, 3), then g(x) = −3f(2(x − 1)) + 4 has a local minimum at (2, −5). The negative vertical factor swaps maximum and minimum; a horizontal reflection alone does not swap them.
This does not say every stationary point is an extremum: a stationary point of inflection remains one under these nonzero affine coordinate changes.
09 / Your turn
Write a coordinate mapping before naming several operations.
For questions about unknown f, use only the information supplied.
02 · Factor the input
Describe the horizontal transformation in y = f(4x − 8).
Hint
Write 4x − 8 as 4(x − 2).
Worked solution
Scale horizontally by factor 1/4, then translate 2 units right. Point mapping: u → u/4 + 2.
03 · Reverse the vertical order
Compare moving up 3 then stretching vertically by 2, with stretching vertically by 2 then moving up 3.
Hint
Apply each operation to the whole expression obtained so far.
Worked solution
First order: 2(f(x) + 3) = 2f(x) + 6. Reverse order: 2f(x) + 3. The final graphs differ by a vertical translation of 3.
04 · Domain and range
f has domain (1, 5] and range [−2, 4). Find the domain and range of g(x) = −2f(x + 3) + 1.
Hint
Translate the input restriction and reverse output endpoints.
Worked solution
1 < x + 3 ≤ 5 gives −2 < x ≤ 2. Output −2 maps to 5 (included), while 4 maps to −7 (excluded). Range: (−7, 5].
05 · Asymptotes
From f(x) = 1/x, find the asymptotes of y = −4f(2(x + 3)) + 7.
Hint
Transform x = 0 and y = 0 separately.
Worked solution
Vertical asymptote x = −3; horizontal asymptote y = 7.
06 · A moved zero
f(4) = 0. What point does this give on y = f(x − 2) + 3? Is it an x-intercept?
Hint
The original zero moves up.
Worked solution
The point is (6, 3), so it is not an x-intercept. To find new zeros you would need inputs where f equals −3.
07 · Stationary point
f has a local minimum at (−2, 5). Find the corresponding point and type for g(x) = −f(−2x) + 4.
Hint
The negative vertical factor changes the type.
Worked solution
−2x = −2 gives x = 1; y = −5 + 4 = −1. The point is a local maximum at (1, −1).
08 · Trigonometric features
State the range, period and horizontal shift of y = 3 cos(2(x + π/4)) − 2.
Hint
Separate amplitude, input multiplier and translation.
Worked solution
Range [−5, 1], period π, shift π/4 left. The input is 2(x − (−π/4)).
09 · Logarithmic domain
Find the domain and vertical asymptote of y = ln(5 − 2x) + 4.
Hint
The logarithm argument must be positive.
Worked solution
5 − 2x > 0 gives x < 5/2. The vertical asymptote is x = 5/2, approached from the left.
10 / Recap
One coordinate rule keeps a long transformation sequence under control.
- For y = a f(b(x − c)) + d, map (u, v) to (u/b + c, av + d).
- Factor the input and state transformation order.
- Negative factors reverse coordinate order; preserve open endpoints.
- Transform asymptotes and stationary points with the same mapping.
- Find new intercepts from their definitions when translations move the axes relative to the graph.
Section 1 of 10 · Track a labelled point