Higher: scale and reflect quadratic 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 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
What is −2f(x) when f(x)=x²−3?
What you will learn
- Apply af(x) and f(ax), with nonzero 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 + (-2). Target: y = af(x), with a = 3.
The domain choice changes the original exercise deliberately. Every finite interval endpoint here is included; an infinite bound is never included.
Applied equation: y = (3)x² + (-6). Original domain: All real numbers; target domain: All real numbers. Original range: [-2, ∞); target range: [-6, ∞).
Point rule: (u,v) → (u, (3)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 | 6 |
| -1 | -1 | -3 |
| 0 | 0 | -6 |
| 1 | 1 | -3 |
| 2 | 2 | 6 |
| Original (u,v) | Image (x,y) |
|---|---|
| (0, -2) | (0, -6) |
| (1, -1) | (1, -3) |
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 heights and follow the turn
For y=af(x), keep x and multiply every height by a. A positive factor stretches or compresses vertically about the x-axis; a negative factor also reflects in that axis. The quadratic turn keeps its horizontal coordinate, but its height is multiplied. A minimum can become a maximum after reflection.
Multiply the complete quadratic, including its constant. The input domain is unchanged and the range is multiplied by a, with the bounds reordered when a<0. A turn at height zero stays at height zero; use a different, nonzero height when recovering the factor from correspondences.
Divide inputs before considering symmetry
For y=f(ax) with a≠0, use (u,v)→(u/a,v). In a quadratic, substituting ax into a square creates an a² coefficient; the horizontal scale is still 1/|a| rather than a². Negative a adds reflection in the y-axis. The heights and range remain the same because each allowed old input is reached.
A full even quadratic such as x²+1 has the same trace after input reflection because f(−x)=f(x). The labelled points still swap sides. A shifted square such as (x−1)² is not even about the y-axis: replacing x by −2x moves its turn as well as changing its width. Solve ax in the original interval to retain the correct restricted domain.
Draw the mapped curve, not only an expanded formula
Use the turn, supplied points and another allowed point to draw each smooth quadratic. When the full domain is allowed, include points on both sides of its turn. When the domain is finite, map both endpoints and stop there. A matching full trace does not justify a different point history or a changed restricted domain.
Recover a from a nonzero height ratio for af(x), or from u_old/x_new for f(ax). Write the substituted expression before expanding, then test the turn and a second point in the expansion. Exact fractions locate points precisely; a hand-drawn bend between them is approximate unless its coordinate has also been derived algebraically.
Sources for this lesson
Pause and explain
Where does the turn (1,0) of f(x)=(x−1)² move under f(−2x)?
Worked example
Start from f(x)=x²−2, with domain all real x and range [-2, ∞). Original landmarks: (0, -2), (1, -1). A uses y=af(x) with a=3: an output scale of magnitude 3. 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 3. Use (u,v) → (u,av) with a=3. The labelled points become (0, -2) → (0, -6); (1, -1) → (1, -3). B: a horizontal coordinate scale of 1/2. Use (u,v) → (u/a,v), with a≠0 with a=2. The labelled points become (0, -2) → (0, -2); (1, -1) → (1/2, -1).
- A: y=3x²−6. Domain all real x; range [-6, ∞). The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. B: y=4x²−2. Domain all real x; range [-2, ∞). 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, -3) 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=3x²−6; D=all real x, R=[-6, ∞). B: y=4x²−2; D=all real x, R=[-2, ∞).
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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), (2, 5). A uses y=af(x) with a=-2: an output scale of magnitude 2 followed by reflection in the x-axis. B uses y=f(ax) with a=-1: a horizontal coordinate scale of 1 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. B is unchanged as a full trace because x²+1 is even. Its labelled point (2,5) moves to (−2,5); an unchanged trace does not erase that mapping.
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 2 followed by reflection in the x-axis. Use (u,v) → (u,av) with a=-2. The labelled points become (0, 1) → (0, -2); (2, 5) → (2, -10). B: a horizontal coordinate scale of 1 together with reflection in the y-axis. Use (u,v) → (u/a,v), with a≠0 with a=-1. The labelled points become (0, 1) → (0, 1); (2, 5) → (-2, 5).
- A: y=−2x²−2. Domain all real x; range (-∞, -2]. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. B: y=x²+1. Domain all real x; range [1, ∞). Require (-1)x in the original domain; divide its endpoints by the parameter and reorder them when it is negative.
- Check the extra original point (-1, 2): A gives (-1, -4) 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. B is unchanged as a full trace because x²+1 is even. Its labelled point (2,5) moves to (−2,5); an unchanged trace does not erase that mapping.
A: y=−2x²−2; D=all real x, R=(-∞, -2]. B: y=x²+1; 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)=2x²−3, with domain all real x and range [-3, ∞). Original landmarks: (0, -3), (1, -1). A is known to have form y=af(x); its labelled correspondences are (0, -3) → (0, -3/2); (1, -1) → (1, -1/2). B is known to have form y=f(ax); its labelled correspondences are (0, -3) → (0, -3); (1, -1) → (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, -3) → (0, -3/2), a=y/v=(-3/2)/(-3)=1/2. Use (u,v) → (u,av) with a=1/2. The labelled points become (0, -3) → (0, -3/2); (1, -1) → (1, -1/2). B: Using (1, -1) → (2, -1), u=ax, so a=u/x=(1)/(2)=1/2. Use (u,v) → (u/a,v), with a≠0 with a=1/2. The labelled points become (0, -3) → (0, -3); (1, -1) → (2, -1).
- A: y=x²−3/2. Domain all real x; range [-3/2, ∞). The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. B: y=(1/2)x²−3. Domain all real x; range [-3, ∞). 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, -1): A gives (-1, -1/2) 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.
A: y=x²−3/2; D=all real x, R=[-3/2, ∞). B: y=(1/2)x²−3; D=all real x, R=[-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.
- 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), (3, 4). 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. Include the finite endpoints. Explain why extending a drawing or rewriting a formula does not restore an excluded original input. The original turn (1,0) moves to (1,0) for A and (−1/2,0) for B. This shifted quadratic is not even about the y-axis.
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, 0) → (1, 0); (3, 4) → (3, -12). 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, 0) → (-1/2, 0); (3, 4) → (-3/2, 4).
- A: y=−3x²+6x−3. Domain [-1, 3]; range [-12, 0]. The original inputs remain allowed; multiply the output interval and reorder its endpoints if the factor is negative. B: y=4x²+4x+1. Domain [-3/2, 1/2]; range [0, 4]. 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, 4): A gives (-1, -12) and B gives (1/2, 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, -12); (3, 4) → (3, -12). The endpoint images for B are (-1, 4) → (1/2, 4); (3, 4) → (-3/2, 4). The original turn (1,0) moves to (1,0) for A and (−1/2,0) for B. This shifted quadratic is not even about the y-axis.
A: y=−3x²+6x−3; D=[-1, 3], R=[-12, 0]. B: y=4x²+4x+1; D=[-3/2, 1/2], 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
- 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?
Why can a full graph of x²+1 hide reflection in the y-axis?
Compare with the explanation
Because (−x)²+1=x²+1, although each labelled nonzero input changes side.
The base is even. Equality of the full trace does not mean that every labelled point stayed in place.
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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.