Higher: scale and reflect linear function graphs
Separate output scaling from input scaling, recover signed factors from labelled landmarks, and draw the valid transformed graphs.
Section 3.3 · Graphs (teaching linked; review pending)
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
- Evaluate the original linear rule and plot exact coordinate pairs.
- Read function notation and solve simple equations with signed fractions.
- Use interval domains and ranges.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a≠0, which mapping describes y=f(ax)?
What you will learn
- Apply af(x) and f(ax), with nonzero a to linear graphs using exact point mappings.
- Recover unknown parameters and transformed equations from known forms and labelled correspondences.
- Draw the original and both targets, retaining exact domains and ranges.
Original: f(x) = (2)x + (3). Target: y = af(x), with a = 2.
The domain choice changes the original exercise deliberately. Every finite interval endpoint here is included; an infinite bound is never included.
Applied equation: y = (4)x + (6). Original domain: All real numbers; target domain: All real numbers. Original range: (-∞, ∞); target range: (-∞, ∞).
Point rule: (u,v) → (u, (2)v).
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 = 6. 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 | 2 |
| 0 | 0 | 6 |
| 1 | 1 | 10 |
| 2 | 2 | 14 |
| Original (u,v) | Image (x,y) |
|---|---|
| (0, 3) | (0, 6) |
| (1, 5) | (1, 10) |
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.
Scale complete outputs, including the constant
In y=af(x), the mapping is (u,v)→(u,av). The horizontal coordinate is unchanged. The positive magnitude |a| scales vertical distances from the x-axis; a negative a also reflects in the x-axis. Multiply the whole expression: a(mx+c)=amx+ac, including the intercept.
For nonzero a, a full nonhorizontal line still has all real outputs. A restricted line segment keeps its input domain, while its range is multiplied by a; reverse the endpoint order when a is negative. A point at height zero stays on the x-axis and cannot, by itself, reveal an output scale factor.
Scale inputs by dividing horizontal coordinates
In y=f(ax), match the old input u by solving ax=u. For a≠0, the new point is (u/a,v). The horizontal scale magnitude is 1/|a|, so |a|>1 compresses the horizontal spacing. A negative input factor also reflects the graph in the y-axis. It does not multiply the output coordinate.
For a linear rule, f(ax)=m(ax)+c=ma x+c. Compare this with af(x): the input-scaled line keeps the old intercept c, whereas output scaling multiplies c. The new domain consists of x with ax in the old domain; divide finite endpoints by a and reorder them after a negative division. Its range stays unchanged.
Recover factors and keep the domain visible
To recover an output factor use a=v_new/v_old with a nonzero old height. To recover an input factor use a=u_old/x_new with a nonzero new horizontal coordinate. Check a second labelled correspondence. An unlabelled line alone may admit several histories; the supplied form and labels identify the calculation being requested.
Draw the original and both target lines or segments through mapped points, choosing scales for fractional coordinates as well as whole numbers. Include finite endpoints and an extra check point. Equation-derived points are exact; an intersection read from a drawing is approximate. These nonzero-factor tasks do not permit dividing by zero; a=0 requires a separate domain check.
Sources for this lesson
Pause and explain
If f has domain [−2,6], what is the domain of f(−2x)?
Worked example
Start from f(x)=2x+3, with domain all real x and range all real x. Original landmarks: (1, 5), (0, 3). A uses y=af(x) with a=2: an output scale of magnitude 2. B uses y=f(ax) with a=2: a horizontal coordinate scale of 1/2. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point.
Show the worked solution
- A: an output scale of magnitude 2. Use (u,v) → (u,av) with a=2. The labelled points become (1, 5) → (1, 10); (0, 3) → (0, 6). B: a horizontal coordinate scale of 1/2. Use (u,v) → (u/a,v), with a≠0 with a=2. The labelled points become (1, 5) → (1/2, 5); (0, 3) → (0, 3).
- A: y=4x+6. Domain all real x; range all real x. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. B: y=4x+3. Domain all real x; range all real x. Require (2)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative.
- Check the extra original point (-1, 1): A gives (-1, 2) and B gives (-1/2, 1). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate.
Answer A: y=4x+6; D=all real x, R=all real x. B: y=4x+3; D=all real x, R=all real x.
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.
Start from f(x)=−x+2, with domain all real x and range all real x. Original landmarks: (1, 1), (0, 2). A uses y=af(x) with a=-3: an output scale of magnitude 3 followed by reflection in the x-axis. B uses y=f(ax) with a=-2: a horizontal coordinate scale of 1/2 together with reflection in the y-axis. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point.
Give me a hint
Track one coordinate at a time. For an input change set the new argument equal to the old input, then solve for the new horizontal coordinate. Apply the same rule to the domain endpoints.
Compare my reasoning
- A: an output scale of magnitude 3 followed by reflection in the x-axis. Use (u,v) → (u,av) with a=-3. The labelled points become (1, 1) → (1, -3); (0, 2) → (0, -6). B: a horizontal coordinate scale of 1/2 together with reflection in the y-axis. Use (u,v) → (u/a,v), with a≠0 with a=-2. The labelled points become (1, 1) → (-1/2, 1); (0, 2) → (0, 2).
- A: y=3x−6. Domain all real x; range all real x. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. B: y=2x+2. Domain all real x; range all real x. Require (-2)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative.
- Check the extra original point (-1, 3): A gives (-1, -9) and B gives (1/2, 3). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate.
A: y=3x−6; D=all real x, R=all real x. B: y=2x+2; D=all real x, R=all real x.
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Explain the direction, scale or reflection through the coordinate mapping.
Start from f(x)=3x−1, with domain all real x and range all real x. Original landmarks: (2, 5), (0, -1). A is known to have form y=af(x); its labelled correspondences are (2, 5) → (2, 5/2); (0, -1) → (0, -1/2). B is known to have form y=f(ax); its labelled correspondences are (2, 5) → (6, 5); (0, -1) → (0, -1). Recover each unknown a from those correspondences and explain why the given form matters. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point.
Give me a hint
Use the known form to decide which coordinate is preserved. Recover an output shift by a difference, an input shift from u=x+a, an output scale from y=av, or an input scale from u=ax; choose a nonzero coordinate for division.
Compare my reasoning
- A: Using (2, 5) → (2, 5/2), a=y/v=(5/2)/(5)=1/2. Use (u,v) → (u,av) with a=1/2. The labelled points become (2, 5) → (2, 5/2); (0, -1) → (0, -1/2). B: Using (2, 5) → (6, 5), u=ax, so a=u/x=(2)/(6)=1/3. Use (u,v) → (u/a,v), with a≠0 with a=1/3. The labelled points become (2, 5) → (6, 5); (0, -1) → (0, -1).
- A: y=(3/2)x−1/2. Domain all real x; range all real x. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. B: y=x−1. Domain all real x; range all real x. Require (1/3)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative.
- Check the extra original point (-1, -4): A gives (-1, -2) and B gives (-3, -4). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate.
A: y=(3/2)x−1/2; D=all real x, R=all real x. B: y=x−1; D=all real x, R=all real x.
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Recover each unknown parameter from labelled correspondences and the stated form.
Start from f(x)=2x−1, with domain [-2, 4] and range [-5, 7]. Original landmarks: (2, 3), (-2, -5). A uses y=af(x) with a=-1/2: an output scale of magnitude 1/2 followed by reflection in the x-axis. B uses y=f(ax) with a=-1/2: a horizontal coordinate scale of 2 together with reflection in the y-axis. Draw the original and both targets yourself on labelled axes, choosing scales that show the landmarks and the valid portions. State each mapping, exact equation, domain and range; map the supplied points and verify one additional point. Include the finite endpoints. Explain why extending a drawing or rewriting a formula does not restore an excluded original input.
Give me a hint
Track one coordinate at a time. For an input change set the new argument equal to the old input, then solve for the new horizontal coordinate. Apply the same rule to the domain endpoints.
Compare my reasoning
- A: an output scale of magnitude 1/2 followed by reflection in the x-axis. Use (u,v) → (u,av) with a=-1/2. The labelled points become (2, 3) → (2, -3/2); (-2, -5) → (-2, 5/2). B: a horizontal coordinate scale of 2 together with reflection in the y-axis. Use (u,v) → (u/a,v), with a≠0 with a=-1/2. The labelled points become (2, 3) → (-4, 3); (-2, -5) → (4, -5).
- A: y=−x+1/2. Domain [-2, 4]; range [-7/2, 5/2]. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. B: y=−x−1. Domain [-8, 4]; range [-5, 7]. Require (-1/2)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative.
- Check the extra original point (-1, -3): A gives (-1, 3/2) and B gives (2, -3). Substitute each image into its target equation and keep it in the allowed domain. Draw the original and both target traces through the mapped points, using a straight line for linear portions and a smooth curve for the other families. Label A and B separately even if traces coincide. Exact coordinates come from these rules; readings between drawn points are approximate. The endpoint images for A are (-2, -5) → (-2, 5/2); (4, 7) → (4, -7/2). The endpoint images for B are (-2, -5) → (4, -5); (4, 7) → (-8, 7).
A: y=−x+1/2; D=[-2, 4], R=[-7/2, 5/2]. B: y=−x−1; D=[-8, 4], R=[-5, 7].
Look for these in your work
- Draw and label the original plus both targets, with all supplied and one extra point mapped correctly.
- Give both exact equations and their actual domains and ranges, preserving restricted endpoints.
- Explain the direction, scale or reflection through the coordinate mapping.
Common mistakes
- Multiplying horizontal coordinates by a for f(ax) instead of dividing by nonzero a.
- Dropping a constant under output scaling, or failing to reverse finite bounds for negative input factors.
- Treating an unchanged even trace as an unchanged point mapping.
What can you explain now?
For f(x)=4x+5, compare 2f(x) and f(2x).
Compare with the explanation
2f(x)=8x+10; f(2x)=8x+5.
Output scaling doubles the constant as well as the variable term; input scaling replaces only the input.
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 the input and output mappings, recover a parameter from a labelled point, and state how a restricted domain survives the change.
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.