Higher: translate quadratic function graphs
Translate outputs and inputs, recover the signed shifts from labelled landmarks, and retain the correct graph domains.
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 quadratic 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
Which mapping applies to y=f(x+5)?
What you will learn
- Apply f(x)+a and f(x+a) to quadratic 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) = (1)x² + (-3). Target: y = f(x)+a, 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 = (1)x² + (-1). Original domain: All real numbers; target domain: All real numbers. Original range: [-3, ∞); target range: [-1, ∞).
Point rule: (u,v) → (u, v + (2)).
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 = -1. The selected target point is ringed in white.
Show transformation values and mapped landmarks
| Target x | Original argument | Target output |
|---|---|---|
| -2 | -2 | 3 |
| -1 | -1 | 0 |
| 0 | 0 | -1 |
| 1 | 1 | 0 |
| 2 | 2 | 3 |
| Original (u,v) | Image (x,y) |
|---|---|
| (0, -3) | (0, -1) |
| (1, -2) | (1, 0) |
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.
Move a parabola without changing its shape
For y=f(x)+a, keep each horizontal coordinate and add a to its height. The turn moves vertically with the rest of the graph. If a quadratic has minimum k, the output-shifted quadratic has minimum k+a. Its original domain stays the same, and its range moves by the same output amount.
In a shifted-square form, keep the whole square intact before adding the output change. A change outside the function is added once; it is not inserted into the squared input. Draw the original turn and at least one point on each side when those inputs are allowed. Map endpoints as well when the parabola is restricted.
Shift the input, then expand carefully
For y=f(x+a), solve x+a=u to get x=u−a. The turn and every other point move horizontally by −a, while their heights stay fixed. Replace every occurrence of x in the original formula, putting brackets around x+a. Only then expand if an expanded equation is needed.
The domain must satisfy x+a in the original domain, so a finite interval [l,r] becomes [l−a,r−a]. The range does not change under an input translation. The minimum height alone cannot locate the new turn: its horizontal coordinate must also be mapped. Check expansion by substituting the new turn and one other mapped point.
Use landmarks rather than an unlabelled trace
Draw the original and both target curves through mapped turns and check points, using smooth parabolic traces. Add a point on the other side of the turn if the given landmarks show only one side. A restricted domain may remove that side, in which case stop at the valid endpoint rather than completing a whole parabola.
When a form is known, an output correspondence gives a=v_new−v_old, while an input correspondence gives a=u_old−x_new. Label the corresponding points rather than guessing from a similar-looking curve. Exact equations and endpoints are mathematical results; positions read from a sketch between them are estimates.
Sources for this lesson
Pause and explain
For f(x)=x²+2, what is f(x−1)?
Worked example
Start from f(x)=x²−3, with domain all real x and range [-3, ∞). Original landmarks: (0, -3), (1, -2). A uses y=f(x)+a with a=2: a vertical translation up 2 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.
Show the worked solution
- A: a vertical translation up 2 output units. Use (u,v) → (u,v+a) with a=2. The labelled points become (0, -3) → (0, -1); (1, -2) → (1, 0). B: a horizontal translation right 2 input units. Use (u,v) → (u−a,v) with a=-2. The labelled points become (0, -3) → (2, -3); (1, -2) → (3, -2).
- A: y=x²−1. Domain all real x; range [-1, ∞). The input domain is unchanged; add the parameter to both range endpoints. B: y=x²−4x+1. Domain all real x; range [-3, ∞). The allowed input condition is x+(-2) in the original domain; subtract that parameter from both finite endpoints.
- Check the extra original point (-1, -2): A gives (-1, 0) and B gives (1, -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.
Answer A: y=x²−1; D=all real x, R=[-1, ∞). B: y=x²−4x+1; D=all real x, R=[-3, ∞).
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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)=x²+1, with domain all real x and range [1, ∞). Original landmarks: (0, 1), (1, 2). A uses y=f(x)+a with a=-4: a vertical translation down 4 output units. B uses y=f(x+a) with a=3: a horizontal translation left 3 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 4 output units. Use (u,v) → (u,v+a) with a=-4. The labelled points become (0, 1) → (0, -3); (1, 2) → (1, -2). B: a horizontal translation left 3 input units. Use (u,v) → (u−a,v) with a=3. The labelled points become (0, 1) → (-3, 1); (1, 2) → (-2, 2).
- A: y=x²−3. Domain all real x; range [-3, ∞). The input domain is unchanged; add the parameter to both range endpoints. B: y=x²+6x+10. Domain all real x; range [1, ∞). The allowed input condition is x+(3) in the original domain; subtract that parameter from both finite endpoints.
- Check the extra original point (-1, 2): A gives (-1, -2) and B gives (-4, 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=x²−3; D=all real x, R=[-3, ∞). B: y=x²+6x+10; D=all real x, R=[1, ∞).
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)=2x²−1, with domain all real x and range [-1, ∞). Original landmarks: (0, -1), (1, 1). A is known to have form y=f(x)+a; its labelled correspondences are (0, -1) → (0, 2); (1, 1) → (1, 4). B is known to have form y=f(x+a); its labelled correspondences are (0, -1) → (1/2, -1); (1, 1) → (3/2, 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 (0, -1) → (0, 2), a=2−(-1)=3. Use (u,v) → (u,v+a) with a=3. The labelled points become (0, -1) → (0, 2); (1, 1) → (1, 4). B: Using (0, -1) → (1/2, -1), u=x+a, so a=0−(1/2)=-1/2. Use (u,v) → (u−a,v) with a=-1/2. The labelled points become (0, -1) → (1/2, -1); (1, 1) → (3/2, 1).
- A: y=2x²+2. Domain all real x; range [2, ∞). The input domain is unchanged; add the parameter to both range endpoints. B: y=2x²−2x−1/2. Domain all real x; range [-1, ∞). The allowed input condition is x+(-1/2) in the original domain; subtract that parameter from both finite endpoints.
- Check the extra original point (-1, 1): A gives (-1, 4) 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.
A: y=2x²+2; D=all real x, R=[2, ∞). B: y=2x²−2x−1/2; D=all real x, R=[-1, ∞).
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)=(x−1)², with domain [-1, 3] and range [0, 4]. Original landmarks: (1, 0), (0, 1). A uses y=f(x)+a with a=-1: a vertical translation down 1 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. 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: a vertical translation down 1 output units. Use (u,v) → (u,v+a) with a=-1. The labelled points become (1, 0) → (1, -1); (0, 1) → (0, 0). B: a horizontal translation left 2 input units. Use (u,v) → (u−a,v) with a=2. The labelled points become (1, 0) → (-1, 0); (0, 1) → (-2, 1).
- A: y=x²−2x. Domain [-1, 3]; range [-1, 3]. The input domain is unchanged; add the parameter to both range endpoints. B: y=x²+2x+1. Domain [-3, 1]; range [0, 4]. The allowed input condition is x+(2) in the original domain; subtract that parameter from both finite endpoints.
- Check the extra original point (-1, 4): A gives (-1, 3) 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. The endpoint images for A are (-1, 4) → (-1, 3); (3, 4) → (3, 3). The endpoint images for B are (-1, 4) → (-3, 4); (3, 4) → (1, 4).
A: y=x²−2x; D=[-1, 3], R=[-1, 3]. B: y=x²+2x+1; D=[-3, 1], R=[0, 4].
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?
A quadratic has turn (2,−5). Where is the turn of y=f(x−3)?
Compare with the explanation
(5,−5).
Here a=−3, so the new horizontal coordinate is 2−(−3)=5; the height is unchanged.
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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.