Higher: analyse transformation domains, equivalence and order
Recover algebra while distinguishing equivalent traces, allowed domains, zero-factor cases and the order of successive changes.
Exam scope and assessment notes
Additional Higher only. Higher includes the shared Foundation content. The Graphs route has nine shared lessons, two Higher line-equation lessons, four Higher graph-family lessons, then these nine transformation lessons. H:3.3B requires all eight core transformation lessons: all four forms f(x)+a, f(x+a), af(x), f(ax) across linear, quadratic, degree sine and degree cosine graphs, with worked construction and every practice stage. H:3.3C requires equation recovery and analysis in all eight core lessons plus the ninth domain/equivalence/order synthesis lesson. This nine-lesson split is StudyForA’s teaching design, not a syllabus-prescribed lesson count. Map landmarks, write equations and draw the original and both target graphs on labelled axes; an automatic trace alone does not replace these constructions. Finite interval endpoints in these exercises are included and infinite bounds are not. Retain input preimages and exclusions; zero factors, even or periodic formulas do not restore invalid arguments or establish a unique history. The bounded interactive examples support exploration, not a general expression or composition parser. All nine Foundation and seven additional Higher Graphs source rows have teaching and staged-practice links. These links do not establish exhaustive subskill approval, mastery or examination readiness; specialist review and formal assessment validation remain pending. Written work is not automatically marked; explained answers support self-checking, not exam marks or proof of mastery. Specialist review and formal assessment validation remain unestablished. Pearson chapter-map and uncorrected-proof navigation remain separate evidence; final full-book and Teaching Hub alignment has not been verified.
Before you begin
- Use the four single-transformation mappings for nonzero factors.
- Write function compositions in the correct order and evaluate signed fractions.
- Use degree sine/cosine landmarks, interval domains and even/odd/periodic identities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which statement about a final unlabelled full graph is justified?
What you will learn
- Explain why a full trace need not identify a unique transformation history.
- Analyse zero factors through the original domain without using 1/0.
- Recover and compare ordered transformations, keeping domains and degree-trigonometric identities explicit.
Original: f(x) = (2)x + (-3). Target: y = f(x)+a, with a = 5.
The domain choice changes the original exercise deliberately. Every finite interval endpoint here is included; an infinite bound is never included.
Applied equation: y = (2)x + (2). Original domain: All real numbers; target domain: All real numbers. Original range: (-∞, ∞); target range: (-∞, ∞).
Point rule: (u,v) → (u, v + (5)).
Blue dashed: original. Red solid: target. Filled points mark supplied landmarks or included endpoints; grey dotted joins show point correspondences, not additional curves. If traces coincide, the red trace can cover the blue. The window is x from -6 to 6, y from -12 to 18; it does not change either mathematical domain.
At target input x=0, use original argument 0. Exact target output = 2. The selected target point is ringed in white.
Show transformation values and mapped landmarks
| Target x | Original argument | Target output |
|---|---|---|
| -2 | -2 | -2 |
| -1 | -1 | 0 |
| 0 | 0 | 2 |
| 1 | 1 | 4 |
| 2 | 2 | 6 |
| Original (u,v) | Image (x,y) |
|---|---|
| (0, -3) | (0, 2) |
| (1, -1) | (1, 4) |
Exact fractions and standard-angle expressions may remain unsimplified. Other sine/cosine readings are marked approximate. The curve is a numerical illustration; your own drawn graph readings are approximate. Use the point correspondences, equation and domain together to check your construction.
An equation and a history are different pieces of information
A final graph can have several valid transformation descriptions. For a line, changing the output intercept may also be achieved by shifting the input. For even functions an input reflection can leave the full trace unchanged; for periodic sine and cosine, adding a whole period leaves the full-rule values unchanged. These facts do not determine which route was actually used.
Use a stated form and labelled original-to-image points when recovering a parameter. Check the formula and the mapping separately. Two expressions represent the same function only when they have the same permitted inputs as well as equal outputs there. Preserve an established domain when expanding, using an even identity or adding a period inside a formula.
Zero needs its own argument
The rule y=0f(x) has output zero wherever the original f is defined. Its domain remains the original domain: multiplying an undefined value by zero does not define it. For a nonempty original domain its range is {0}. Draw the allowed part of the horizontal axis, including finite endpoints only when they are permitted.
The rule y=f(0x) asks for the single original value f(0) at every new x. If 0 belongs to the original domain, the target has all real inputs and constant output f(0). If 0 was excluded, there is no valid new input. There is no horizontal scale 1/0. A constant full-domain function has every positive period and no least positive period; do not assign it the least period of a nonconstant wave.
Apply changes in order and keep domains
Track a point after the first change, then after the second. A horizontal shift followed by an input scale generally differs from the same operations in reverse order, because the scale can act on the earlier shift too. Write the intermediate function with its own brackets before substituting into it, then compare a landmark and a numerical input in the final results.
For trigonometric analysis use degrees and retain amplitude, period and allowed input intervals. Cosine is even, sine is odd, and both repeat after 360° on their full real degree domains. These identities justify alternative formulas with equal values; they do not restore inputs excluded by a restricted original function. Draw and label the original plus the compared targets; exact landmarks and algebra support the conclusions, while other readings from a sketch are approximate.
Sources for this lesson
Pause and explain
A constant function y=4 on all real inputs has which period property?
Worked example
The full line y=2x+7 is obtained from f(x)=2x+3. Recover one vertical-translation description and one horizontal-translation description. Draw the original and both target descriptions, map the labelled point (1,5) under each, and give exact domains and ranges. Explain why the final trace cannot choose a unique history.
Show the worked solution
- For the known vertical form f(x)+a, compare 2x+3+a with 2x+7: a=4. Its mapping is (u,v)→(u,v+4), so (1,5)→(1,9). The equation is y=f(x)+4=2x+7.
- For the horizontal form f(x+a), expand 2(x+a)+3=2x+2a+3. Matching the constant 7 gives 2a=4 and a=2. Its mapping is (u,v)→(u−2,v), so (1,5)→(−1,5). Both targets and the original have all real domains and ranges.
- Draw the original line through (0,3) and (1,5). Both target descriptions lie on y=2x+7 through (0,7), (1,9) and (−1,5), so draw that line once with both labels. A vertical shift up 4 and a horizontal shift left 2 give the same full function but different labelled point histories; the final trace alone cannot select one.
Answer Vertical: f(x)+4=2x+7, with (1,5)→(1,9). Horizontal: f(x+2)=2x+7, with (1,5)→(−1,5). All domains and ranges are all real. The two full targets coincide, so their history is not uniquely inferable.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For each full-domain base f(x)=3x−2, f(x)=x²+4, f(x)=sin x and f(x)=cos x, draw the original and both targets A=0f(x) and B=f(0x). Use degrees for trigonometric inputs. Give exact equations, domains, ranges and mappings from (0,f(0)) and (1,f(1)) where useful; explain why there is no ordinary horizontal point mapping for B. Then repeat the domain analysis and draw the valid sets for f(x)=x² restricted to [1,3], including endpoint images. State the period property of full-domain constants.
Give me a hint
Check whether the original input zero is allowed before evaluating f(0). An output factor zero keeps the original input domain; a zero input factor asks for the same old input at every new x.
Compare my reasoning
- The original ranges are ℝ for the line, [4,∞) for x²+4, and [−1,1] for each trigonometric base; their original domains are all real inputs. For every full-domain base, A is y=0 on all real x, with range {0} and mapping (u,v)→(u,0). For B, substitute the allowed old input 0: the four constants are respectively −2, 4, 0 and 1. Every B has all real domain and the corresponding singleton range. An arbitrary old point cannot be mapped by division by zero: B repeats only the old value at input 0, at every new x.
- Draw each original in its own panel and label its two constant targets. For the line original, (0,−2) and (1,1) map under A to (0,0),(1,0); B is y=−2. For x²+4, (0,4),(1,5) map to (0,0),(1,0); B is y=4. Sine and cosine give B=0 and B=1, respectively, with degree axes; the originals have amplitude 1 and full period 360°. These constant targets have amplitude 0 when described as waves, every positive period and no least positive period.
- For restricted x² on [1,3], A has equation y=0 only for x∈[1,3], range {0}; (1,1)→(1,0) and (3,9)→(3,0). Draw this closed segment. For B, 0x=0 never belongs to [1,3], so its domain and range are empty and no target trace is drawn. The original domain/range are [1,3] and [1,9]. Finite segments have no global repetition domain.
Full-domain A: y=0, D=ℝ, R={0}. Full-domain B: constants −2,4,0,1 for the line, quadratic, sine and cosine, each D=ℝ and singleton R. For restricted x², A has D=[1,3], R={0}; B has empty domain and range. No 1/0 mapping is available. Full-domain constants have every positive period and no least positive period.
Look for these in your work
- Draw each original and both valid targets, distinguishing an empty target from a zero-height segment.
- Keep output-zero and input-zero domains separate and avoid division by zero.
- Give all exact constants and distinguish constant-period behaviour from nonconstant trigonometric periods.
Start with the full-domain f(x)=x². Two separately known operations have forms S[f](x)=f(x−h) and C[f](x)=f(kx), with h and nonzero k withheld. For S, the labelled correspondence is (0,0)→(2,0); for C, it is (1,1)→(1/3,1). Recover h and k, then apply S followed by C and C followed by S. Draw the original and both final curves. Give exact equations, mappings, turns, domains and ranges; map (1,1) in each order and test both final rules at x=0.
Give me a hint
Recover each parameter from its known form and the labelled input correspondence. Track a generic input u through both operations in the stated order before simplifying either equation.
Compare my reasoning
- S maps u to u+h, so 0+h=2 gives h=2. C maps u to u/k, so 1/k=1/3 gives k=3. Thus S is a shift right 2, and C is a horizontal compression of coordinate factor 1/3. The known correspondences determine these signed parameters; they are not inferred from an unlabelled even trace.
- S then C: u→u+2→(u+2)/3, with height v unchanged. The intermediate rule is (x−2)² and replacing its input by 3x gives (3x−2)². Its turn is (2/3,0); (1,1)→(1,1); at x=0 its height is 4. C then S: u→u/3→u/3+2. The intermediate rule is 9x² and shifting it right 2 gives 9(x−2)². Its turn is (2,0); (1,1)→(7/3,1); at x=0 its height is 36.
- The original and both targets have domain all real x and range [0,∞). Draw x² and both narrower target parabolas, marking their exact turns and the mapped point in each. For extra checks, (−1,1) maps to (1/3,1) in the first order and (5/3,1) in the second. The different turns and different values at x=0 prove that these operations do not commute; sketch readings elsewhere are approximate.
h=2, k=3. S then C: y=(3x−2)², mapping (u,v)→((u+2)/3,v), turn (2/3,0), (1,1)→(1,1), y(0)=4. C then S: y=9(x−2)², mapping (u,v)→(u/3+2,v), turn (2,0), (1,1)→(7/3,1), y(0)=36. All domains are ℝ and ranges [0,∞). The operations do not commute.
Look for these in your work
- Recover h and k from the stated forms and correspondences before using their values.
- Track both ordered mappings and give exact equations, turns and check values.
- Draw the original and both targets and justify the different results.
A full degree wave f(x)=cos(2x) is shifted right 30°. Draw the original and its shifted target, labelling an equivalent even or periodic formula on that same target trace. Recover an equation, mapping, amplitude, midline, period, domain, range and consecutive maxima; check a nonmaximum point too. Compare the full-domain identities sin(x+360°)=sin x, cos(−x)=cos x and sin(−x)=−sin x. In a separate panel take r(x)=cos x restricted to [0°,180°], and construct g(x)=r(x−60°). Draw r and g, label the alternative formula cos(x+300°) on g, and compare its retained domain with the domain of the different composition r(x+300°).
Give me a hint
Apply the right shift to the entire input 2x, giving 2(x−30°). For the restricted panel solve the input inequalities before using periodicity; a formula rewrite retains the target domain already found.
Compare my reasoning
- The full wave mapping is (u,v)→(u+30°,v), so the target is cos(2(x−30°))=cos(2x−60°). The original maxima (0°,1),(180°,1) become (30°,1),(210°,1). A check point (45°,0) becomes (75°,0). The original and target have all real degree domains, range [−1,1], amplitude 1, midline y=0 and least positive period 180°. Draw both waves smoothly and label the equivalent cos(60°−2x) or cos(2x+300°) on the same target trace.
- The identities sin(x+360°)=sin x, cos(−x)=cos x and sin(−x)=−sin x hold at every full real degree input. They justify equal values or coincident traces with specified domains; they do not select a unique transformation history. For example, the shifted wave’s equivalent even description does not negate the known right-shift mapping.
- For restricted r, require 0≤x−60≤180, giving g domain [60°,240°], range [−1,1] and formula cos(x−60°). Map (0°,1),(90°,0),(180°,−1) to (60°,1),(150°,0),(240°,−1); draw the original arc and this target arc with closed endpoints. The rewritten cos(x+300°) has the same values on the retained [60°,240°]. The separate composition r(x+300°) instead requires 0≤x+300≤180, so its domain is [−300°,−120°]; its corresponding points are (−300°,1),(−210°,0),(−120°,−1). Both restricted arcs have range [−1,1] and underlying full cosine period 360°, but no global repetition domain. Equal unrestricted formulas do not make these different restricted domains equal.
Full wave: cos(2x−60°), mapping (u,v)→(u+30°,v), D=ℝ in degrees, R=[−1,1], amplitude 1, midline 0, period 180°, maxima (30°,1),(210°,1); (75°,0) is a check point. Restricted g: cos(x−60°)=cos(x+300°) only on retained D=[60°,240°], R=[−1,1]. The separately constructed r(x+300°) has D=[−300°,−120°], so it is a different restricted function.
Look for these in your work
- Draw and label the full original and shifted wave with exact maxima, a check point, amplitude and degree period.
- Use even/odd/periodic identities with their correct full-domain meanings.
- Draw the restricted arcs, retain g’s domain through its rewrite and distinguish the different restricted composition.
Common mistakes
- Using a horizontal factor 1/0 or assigning a least positive period to a full-domain constant.
- Treating a periodic or even formula rewrite as permission to replace an inherited restricted domain.
- Reversing successive operations without changing the brackets, or claiming a unique history from an unlabelled trace.
What can you explain now?
If f(x)=x² is defined only on [1,3], is f(0x) the zero function on [1,3]?
Compare with the explanation
No. It has empty domain.
Every proposed new input asks for f(0), but 0 is outside the original domain. By contrast, 0f(x) is zero on [1,3].
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain how a known transformation form and labelled points recover a parameter, then give one domain reason that prevents two equivalent formulas from representing the same restricted function.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.