Higher: translate linear function graphs
Translate outputs and inputs, recover the signed shifts from labelled landmarks, and retain the correct graph domains.
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
Under y=f(x−6), where does a labelled point (2,7) move?
What you will learn
- Apply f(x)+a and f(x+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 = 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.
Translate outputs by a fixed amount
A point (u,v) on y=f(x) means f(u)=v. On y=f(x)+a the input u has not changed, so its new height is v+a. The mapping is (u,v)→(u,v+a): positive a moves the line up and negative a moves it down. Add a to the complete output, including its constant term.
For a linear rule f(x)=mx+c, the shifted equation is y=mx+(c+a). The gradient m is unchanged. The input domain stays the same, while every output in the range moves by a. A restricted segment keeps its original horizontal endpoints; a full-domain line still extends in both directions.
Translate inputs by solving the argument
In y=f(x+a), an old input u occurs when x+a=u, so the new horizontal coordinate is x=u−a. The mapping is (u,v)→(u−a,v). A positive a therefore moves the graph left, and a negative a moves it right. Solve this little equation instead of copying the sign into a movement description.
Substitute x+a everywhere x appeared: m(x+a)+c=mx+ma+c. The domain is the set of x for which x+a was an allowed old input; [l,r] becomes [l−a,r−a]. Its output range is unchanged. Mark the old intercept as a labelled point and map it, even when it no longer lies on the vertical axis.
Construct and compare whole graphs
Choose scales covering the old and new points, draw the original line or segment, then draw each target through its mapped points. Two points determine each line, but their exact positions and any finite endpoints still matter. Use another input to check the recovered equation before relying on the drawing. A coordinate read only from the drawing is approximate.
A full line can have more than one transformation description. For a nonhorizontal line a vertical shift can produce the same final equation as a horizontal shift of another size. Labelled correspondences and a stated form tell you which mapping to recover. If the old domain is restricted, matching expanded formulas is insufficient: compare the allowed inputs too.
Sources for this lesson
Pause and explain
For f(x)=3x−1, which equation represents f(x)+2?
Worked example
Start from f(x)=2x−3, with domain all real x and range all real x. Original landmarks: (1, -1), (0, -3). A uses y=f(x)+a with a=5: a vertical translation up 5 output units. B uses y=f(x+a) with a=2: a horizontal translation left 2 input units. 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: a vertical translation up 5 output units. Use (u,v) → (u,v+a) with a=5. The labelled points become (1, -1) → (1, 4); (0, -3) → (0, 2). B: a horizontal translation left 2 input units. Use (u,v) → (u−a,v) with a=2. The labelled points become (1, -1) → (-1, -1); (0, -3) → (-2, -3).
- A: y=2x+2. Domain all real x; range all real x. The input domain is unchanged; add the parameter to both range endpoints. B: y=2x+1. Domain all real x; range all real x. The allowed input condition is x+(2) in the original domain; subtract that parameter from both finite endpoints.
- Check the extra original point (-1, -5): A gives (-1, 0) and B gives (-3, -5). 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=2x+2; D=all real x, R=all real x. B: y=2x+1; D=all real x, R=all real x.
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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)=−3x+4, with domain all real x and range all real x. Original landmarks: (2, -2), (0, 4). A uses y=f(x)+a with a=-2: a vertical translation down 2 output units. B uses y=f(x+a) with a=-1: a horizontal translation right 1 input units. 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: a vertical translation down 2 output units. Use (u,v) → (u,v+a) with a=-2. The labelled points become (2, -2) → (2, -4); (0, 4) → (0, 2). B: a horizontal translation right 1 input units. Use (u,v) → (u−a,v) with a=-1. The labelled points become (2, -2) → (3, -2); (0, 4) → (1, 4).
- A: y=−3x+2. Domain all real x; range all real x. The input domain is unchanged; add the parameter to both range endpoints. B: y=−3x+7. Domain all real x; range all real x. The allowed input condition is x+(-1) in the original domain; subtract that parameter from both finite endpoints.
- Check the extra original point (-1, 7): A gives (-1, 5) and B gives (0, 7). 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+2; D=all real x, R=all real x. B: y=−3x+7; 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)=(1/2)x−2, with domain all real x and range all real x. Original landmarks: (2, -1), (0, -2). A is known to have form y=f(x)+a; its labelled correspondences are (2, -1) → (2, 2); (0, -2) → (0, 1). B is known to have form y=f(x+a); its labelled correspondences are (2, -1) → (-2, -1); (0, -2) → (-4, -2). 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, -1) → (2, 2), a=2−(-1)=3. Use (u,v) → (u,v+a) with a=3. The labelled points become (2, -1) → (2, 2); (0, -2) → (0, 1). B: Using (2, -1) → (-2, -1), u=x+a, so a=2−(-2)=4. Use (u,v) → (u−a,v) with a=4. The labelled points become (2, -1) → (-2, -1); (0, -2) → (-4, -2).
- A: y=(1/2)x+1. Domain all real x; range all real x. The input domain is unchanged; add the parameter to both range endpoints. B: y=(1/2)x. Domain all real x; range all real x. The allowed input condition is x+(4) in the original domain; subtract that parameter from both finite endpoints.
- Check the extra original point (-1, -5/2): A gives (-1, 1/2) and B gives (-5, -5/2). 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=(1/2)x+1; D=all real x, R=all real x. B: y=(1/2)x; 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 [-1, 3] and range [-5, 3]. Original landmarks: (1, -1), (-1, 3). A uses y=f(x)+a with a=4: a vertical translation up 4 output units. B uses y=f(x+a) with a=-2: a horizontal translation right 2 input units. 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. The two expanded equations happen to agree, but their domains differ: the output-shifted segment and the input-shifted segment are different functions. Draw both valid segments rather than replacing them by one unrestricted line.
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: a vertical translation up 4 output units. Use (u,v) → (u,v+a) with a=4. The labelled points become (1, -1) → (1, 3); (-1, 3) → (-1, 7). B: a horizontal translation right 2 input units. Use (u,v) → (u−a,v) with a=-2. The labelled points become (1, -1) → (3, -1); (-1, 3) → (1, 3).
- A: y=−2x+5. Domain [-1, 3]; range [-1, 7]. The input domain is unchanged; add the parameter to both range endpoints. B: y=−2x+5. Domain [1, 5]; range [-5, 3]. The allowed input condition is x+(-2) in the original domain; subtract that parameter from both finite endpoints.
- Check the extra original point (0, 1): A gives (0, 5) and B gives (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. The endpoint images for A are (-1, 3) → (-1, 7); (3, -5) → (3, -1). The endpoint images for B are (-1, 3) → (1, 3); (3, -5) → (5, -5). The two expanded equations happen to agree, but their domains differ: the output-shifted segment and the input-shifted segment are different functions. Draw both valid segments rather than replacing them by one unrestricted line.
A: y=−2x+5; D=[-1, 3], R=[-1, 7]. B: y=−2x+5; D=[1, 5], R=[-5, 3].
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
- Moving f(x+a) in the direction of the written sign instead of solving x+a=u.
- Changing a stated domain after expanding or rewriting the target formula.
- Inferring a unique history from an unlabelled trace.
What can you explain now?
For a known input translation y=f(x+a), (4,9) is labelled as corresponding to (1,9). Find a.
Compare with the explanation
a=3.
The old input is u=x+a, so 4=1+a. The unchanged height confirms an input translation.
After trying it yourself, choose your next review. This is your self-assessment.
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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.