Grade 11 · Pure Mathematics 1 · Cambridge A-Level 9709 · Age 16–17
Welcome to Graph Transformations!
Graph transformations are one of the most visual and versatile topics in A-Level Pure Mathematics. Instead of computing new functions from scratch, you learn to take any known graph and predict exactly how it moves, stretches, reflects, or folds under a family of rules. This skill connects directly to calculus, trigonometry, and every function you will ever sketch in A-Level.
y = f(x+a) — shift LEFT a | y = f(x)+a — shift UP a | y = f(kx) — x-stretch ×(1/k) | y = kf(x) — y-stretch ×k
Learning Objectives
Apply translations f(x+a) and f(x)+a and state coordinate effects
Apply stretches f(kx) and kf(x) and state coordinate effects
Apply reflections f(−x) and −f(x)
Combine two or more transformations in the correct order
Sketch y = |f(x)| and y = f(|x|) from a given graph
Given key points of y = f(x), find new coordinates after any transformation
Determine the effect of transformations on asymptotes
Recognise which transformation maps one graph to another
Translations
f(x+a), f(x)+a — horizontal and vertical shifts
Stretches
f(kx) parallel to x-axis, kf(x) parallel to y-axis
Reflections
f(−x) in y-axis, −f(x) in x-axis, combined order
Modulus
|f(x)| folds negatives up, f(|x|) mirrors right half
Key Points
Track coordinates and asymptotes under any transform
Visualiser
Interactive canvas to see transformations live
Learn 1 — Translations: f(x+a) and f(x)+a
A translation moves the entire graph without changing its shape or size. There are two basic directions: horizontal (replacing x with x+a) and vertical (adding a to the whole function).
Vertical Translation: y = f(x) + a
Replace every y-value with y + a. The whole graph slides up by a if a > 0, or down by |a| if a < 0.
Coordinate effect: Every point (x, y) maps to (x, y + a).
Example: If f(x) = x², then y = f(x) + 3 = x² + 3.
The vertex moves from (0, 0) to (0, 3). All y-intercepts increase by 3. x-intercepts shift (or disappear) accordingly.
f(x) + a : (x, y) → (x, y + a)
Horizontal Translation: y = f(x + a)
Why does f(x + a) move LEFT when a is positive?
Consider y = f(x+3). To get the same output as the original f(x) at x = 0, you now need x+3 = 0, which means x = −3. So the point that used to be at x = 0 is now at x = −3. Every point moves 3 units to the LEFT. This is counterintuitive — the +3 inside moves the graph in the negative direction.
Formally: if (p, q) is on y = f(x), then at x = p−a we get f((p−a)+a) = f(p) = q. So the new x-coordinate is p−a, meaning the graph shifts LEFT by a (when a > 0).
Coordinate effect: Every point (x, y) maps to (x − a, y).
f(x + a) : (x, y) → (x − a, y) [LEFT by a when a > 0]
Examples with Different Function Types
Quadratic: f(x) = x²
y = f(x − 2) = (x − 2)² — shifts RIGHT 2. Vertex moves from (0,0) to (2,0).
y = f(x) + 5 = x² + 5 — shifts UP 5. Vertex moves from (0,0) to (0,5).
y = f(x + 1) − 3 = (x+1)² − 3 — shifts LEFT 1, DOWN 3. Vertex: (−1, −3).
Trigonometric: f(x) = sin x
y = sin(x + π/2) — shifts LEFT by π/2. Since sin(x + π/2) = cos x, this maps the sine curve onto the cosine curve.
y = sin x + 1 — shifts UP 1. The entire wave sits between y = 0 and y = 2.
Exponential: f(x) = eˣ
y = eˣ + 2 — shifts UP 2. Asymptote moves from y = 0 to y = 2.
y = e^(x+3) — shifts LEFT 3. Asymptote remains at y = 0 (vertical shifts affect asymptotes; horizontal shifts do not for y = eˣ).
Effect on Asymptotes
For y = 1/x (with asymptotes x = 0 and y = 0):
y = 1/(x+2) + 3 : vertical asymptote shifts to x = −2, horizontal asymptote shifts to y = 3.
In general: f(x+a) shifts vertical asymptotes by −a; f(x)+a shifts horizontal asymptotes by +a.
The classic exam trap is direction confusion. Remember: f(x + 3) moves LEFT (the +3 is inside), while f(x) + 3 moves UP (the +3 is outside). A useful memory: the substitution x+3 reaches its target x-value 3 units earlier, so the graph arrives 3 units to the LEFT.
Learn 2 — Stretches: f(kx) and kf(x)
A stretch scales the graph by a factor but anchors it to one axis. The two types stretch parallel to different axes, and the factor works in opposite ways: the factor appears directly as a y-multiplier, but as a reciprocal for x.
Stretch Parallel to the y-axis: y = kf(x)
Every y-value is multiplied by k. The graph stretches vertically while all x-intercepts stay fixed.
Coordinate effect: (x, y) → (x, ky)
Example: f(x) = x² − 4. Then y = 3f(x) = 3x² − 12.
Vertex (0, −4) → (0, −12). The x-intercepts x = ±2 are unchanged (y = 0, so k×0 = 0). The graph is taller and narrower in appearance.
kf(x) : (x, y) → (x, ky) — stretch parallel to y-axis, scale factor k
Stretch Parallel to the x-axis: y = f(kx)
Every x-value is divided by k. The graph stretches horizontally by scale factor 1/k while all y-intercepts stay fixed.
Coordinate effect: (x, y) → (x/k, y)
Example: f(x) = sin x. Then y = f(2x) = sin(2x).
The period halves from 2π to π. The point (π, 0) maps to (π/2, 0). The y-intercept (0, 0) stays at (0, 0).
In y = f(kx), the scale factor for the x-stretch is 1/k, not k. If k = 2, distances from the y-axis are halved (compressed, not stretched). If k = 1/2, distances from the y-axis double (stretched). Many students write "scale factor 2" when the correct answer is "scale factor 1/2".
Worked Examples with Trigonometric Functions
y = 2 sin x: Amplitude doubles from 1 to 2. Period unchanged (2π). Every point (x, sin x) maps to (x, 2 sin x). The x-intercepts at 0, π, 2π, … remain fixed.
y = sin(3x): Period becomes 2π/3. Amplitude unchanged at 1. The point (π/2, 1) maps to (π/6, 1).
Asymptotes Under Stretch
For y = 1/x with asymptotes x = 0 and y = 0:
y = 3 · (1/x) = 3/x : the vertical asymptote x = 0 and horizontal asymptote y = 0 are unchanged (they lie on the axes which are fixed under a y-stretch).
y = 1/(2x) = (1/2)(1/x) or equivalently f(2x): the asymptotes remain x = 0, y = 0 but all points move to half their x-distance from the y-axis.
For y = eˣ + 2 with asymptote y = 2:
y = 3(eˣ + 2) = 3eˣ + 6: the asymptote y = 2 maps to y = 6 (multiply by k). This is because the asymptote is a y-value, so it scales too.
Always state the axis affected: "stretch parallel to the x-axis, scale factor 1/k" or "stretch parallel to the y-axis, scale factor k". Omitting the axis or writing the wrong scale factor are both penalised in marking.
Learn 3 — Reflections and Combined Transformations
Reflection in the y-axis: y = f(−x)
Replace x with −x. Every point moves to the mirror image across the y-axis. Coordinate effect: (x, y) → (−x, y)
Example: f(x) = 2ˣ. Then y = f(−x) = 2^(−x) = (1/2)ˣ.
The point (3, 8) maps to (−3, 8). The y-intercept (0, 1) stays fixed (it lies on the y-axis). The asymptote y = 0 remains unchanged.
f(−x) : (x, y) → (−x, y) — reflection in the y-axis
Reflection in the x-axis: y = −f(x)
Replace y with −y. Every point moves to the mirror image across the x-axis. Coordinate effect: (x, y) → (x, −y)
Example: f(x) = x² − 4. Then y = −f(x) = −x² + 4.
The vertex (0, −4) maps to (0, 4). The x-intercepts x = ±2 stay fixed (y = 0, so −0 = 0). A U-parabola becomes a ∩-parabola.
−f(x) : (x, y) → (x, −y) — reflection in the x-axis
Combined Transformations — Order Matters!
The order in which you apply transformations changes the result.
Applying a translation then a stretch gives a different curve from applying the stretch then the translation. When the equation is given, read it from the inside out to determine the correct order.
Reading order rule: For y = af(bx + c) + d, work from the innermost operation outward:
1. Horizontal shift: replace x with x + c/b (i.e. shift left by c/b)
2. Horizontal stretch: scale factor 1/b
3. Vertical stretch: scale factor a
4. Vertical shift: add d
Note: Steps 1 and 2 can be combined by factoring: bx + c = b(x + c/b), so the horizontal shift is −c/b and the stretch factor is 1/b.
Example: Find the effect of y = 2f(3x − 6) on the point (4, 5) on y = f(x).
Step 1: Factor inside bracket: 3x − 6 = 3(x − 2), so this is f(3(x−2)).
Step 2: Horizontal stretch x-coords by 1/3: (4) → (4/3)
Step 3: Horizontal shift right by 2: (4/3) → (4/3 + 2) = (10/3)
Step 4: Vertical stretch y-coords by 2: (5) → (10)
New point: (10/3, 10)
Alternatively: solve 3x − 6 = 4 (original x) → x = 10/3. Then y = 2×5 = 10. Same answer.
Example: Does order matter? Compare g(x) = f(x+2) then stretch by 3 vs. stretch by 3 then translate.
Start with point (1, 4).
Order A — translate first, then stretch:
Translate: (1, 4) → (−1, 4). Stretch ×3 y: (−1, 12).
Order B — stretch first, then translate:
Stretch ×3 y: (1, 12). Translate: (−1, 12).
Both give (−1, 12) here because the x-translation and y-stretch act on different coordinates — they commute. However:
Order C — f(x) → 3f(x) → 3f(x+2) vs. f(x) → f(x+2) → 3f(x+2): same result.
But f(x) → f(3x) then f(3x+2) is NOT the same as f(3(x+2)) = f(3x+6). The danger is when a horizontal stretch and horizontal translation are combined.
When asked to describe a combined transformation, always state each transformation separately with full precision: axis, direction/scale factor, amount. Never say "translated 2 units" without specifying direction. The order you describe matters — describe the one applied first.
Learn 4 — Modulus Graphs
The modulus (absolute value) function |x| makes all values non-negative. When applied to a function, it creates graphs with "folds" — either the negative outputs are reflected upward, or the left half of the domain is mirrored from the right.
y = |f(x)|: Reflect Negative Parts Above the x-axis
To sketch y = |f(x)| from y = f(x):
1. Draw y = f(x) as usual.
2. Any part of the graph that is below the x-axis (where f(x) < 0) is reflected up — it becomes f(x) with the y-values negated (i.e. −f(x) where f(x) < 0).
3. Parts of the graph above or on the x-axis remain unchanged.
Coordinate effect: (x, y) → (x, |y|). Points with y ≥ 0 are unchanged; points with y < 0 map to (x, −y).
Example: Sketch y = |x² − 4|
First sketch y = x² − 4: a parabola, vertex (0, −4), x-intercepts at x = ±2.
The section between x = −2 and x = 2 has y < 0. Reflect this part up: the vertex (0, −4) maps to (0, 4), and the "valley" becomes a "peak".
The rest of the parabola (where y ≥ 0) remains unchanged.
The resulting graph touches the x-axis at x = ±2 and has a peak at (0, 4).
y = f(|x|): Reflect the Right Half onto the Left
To sketch y = f(|x|):
1. For x ≥ 0, y = f(|x|) = f(x) — unchanged.
2. For x < 0, y = f(|x|) = f(−x) — this is the reflection of the right half in the y-axis.
So: keep the right half (x ≥ 0) exactly as is, then mirror it to create the left half. The left half of the original graph is discarded.
Coordinate effect: (x, y) where x ≥ 0 maps to (±x, y) — i.e. both (x, y) and (−x, y) appear.
Example: Sketch y = sin(|x|)
For x ≥ 0: same as y = sin x — positive wave from 0 to π, negative from π to 2π, etc.
For x < 0: mirror the right half. So x = −π/2 gives sin(π/2) = 1, meaning the point (−π/2, 1) appears.
The result is an even function symmetric about the y-axis. The graph looks like a "W" repeated.
Solving Equations Graphically with Modulus
Solve |x − 3| = 2
Case 1: x − 3 ≥ 0 → x − 3 = 2 → x = 5
Case 2: x − 3 < 0 → −(x − 3) = 2 → x − 3 = −2 → x = 1 x = 1 or x = 5
Graphically: draw y = |x − 3| and y = 2. The horizontal line intersects the V-shape at two points.
Solve |2x + 1| = x + 4
Case 1: 2x + 1 ≥ 0 → 2x + 1 = x + 4 → x = 3. Check: |7| = 7 ✓
Case 2: 2x + 1 < 0 → −(2x + 1) = x + 4 → −3x = 5 → x = −5/3. Check: |−1/3| = 1/3, −5/3 + 4 = 7/3 ✗ Wait: |2(−5/3)+1| = |−7/3| = 7/3 and (−5/3)+4 = 7/3 ✓ x = 3 or x = −5/3
When sketching y = |f(x)|, mark the x-intercepts of f(x) carefully — these are where the "fold" happens. The graph of |f(x)| never goes below the x-axis. For y = f(|x|), the resulting graph is always symmetric about the y-axis.
Learn 5 — Transforming Key Points and Asymptotes
In exam questions, you are often given a sketch of y = f(x) with a few labelled points, and asked to sketch a transformed version. The technique: apply the coordinate transformation rule to each labelled point, then sketch the new curve through the new points.
The General Strategy
1. Identify the transformation(s) applied.
2. Write down the coordinate rule: (x, y) → (new x, new y).
3. Apply the rule to every labelled point.
4. Note any fixed points (where the transformation leaves the point unchanged).
5. Update asymptotes using the same rule.
6. Sketch the new graph through the new points, preserving the overall shape.
Worked Example: Translation
f(x) has key points: (−2, 0), (0, 3), (1, 4), (3, 0). Find the key points of y = f(x − 2) + 1.
Asymptotes are transformed by the same coordinate rules as points.
Example: y = f(x) has a vertical asymptote at x = 2 and horizontal asymptote at y = 0.
For y = f(x + 3): vertical asymptote at x = 2 → x − 3 = 2 → x = −1. Horizontal asymptote y = 0 is unaffected (vertical translation of 0 applied).
Example: y = f(x) has asymptote y = 4.
For y = 2f(x) − 1: y-value → 2y − 1. Asymptote y = 4 → 2(4) − 1 = 7. New asymptote: y = 7.
Fixed Points
A fixed point is one that maps to itself under the transformation.
For f(x) + a: fixed points have y = y + a, impossible unless a = 0. No fixed points for translations.
For kf(x): fixed points have ky = y → y = 0. So points on the x-axis are fixed under y-stretches.
For f(kx): fixed points have x/k = x → x = 0. So the y-axis intercept is fixed under x-stretches.
For f(−x): fixed points have −x = x → x = 0. Points on the y-axis are fixed.
For −f(x): fixed points have y = 0. Points on the x-axis are fixed.
Always transform asymptotes as well as key points — forgetting to update an asymptote is a common dropped mark. If the original asymptote passes through a point you can track easily, use the coordinate rule on that point to find the new asymptote.
Worked Examples
8 fully worked examples covering all transformation types.
Example 1 — Translation of a Cubic
The graph of y = x³ − 3x passes through (1, −2). Find the equation of the graph after the translation vector (3, −5) and state the new coordinates of that point.
Step 1: Translation vector (3, −5) means right 3, down 5. This corresponds to y = f(x − 3) − 5. M1
Step 2: New equation: y = (x − 3)³ − 3(x − 3) − 5. A1
Step 3: Coordinate rule: (x, y) → (x + 3, y − 5). Point (1, −2) → (4, −7). A1
Example 2 — Stretch of a Trigonometric Function
Describe the single transformation that maps y = cos x onto y = cos(x/3), and state the new period.
Step 1: y = cos(x/3) = f(x/3) where f(x) = cos x. This is f(kx) with k = 1/3. M1
Step 2: Stretch parallel to the x-axis, scale factor 1/k = 3. A1
Step 3: Original period 2π → new period 2π × 3 = 6π. A1
Example 3 — Combined Transformation
f(x) = x². A point P(2, 4) lies on y = f(x). Find the image of P under y = 3f(2x − 4).
Step 1: Factor: 3f(2x − 4) = 3f(2(x − 2)). So: x-stretch by 1/2, then x-translate right 2, then y-stretch by 3. M1
Step 2: Apply to P(2, 4): x-stretch: x → x/2 = 1. Then x-translate: 1 → 1 + 2 = 3. Then y-stretch: y → 3×4 = 12. M1
Step 3: Alternatively, solve 2x − 4 = 2 → x = 3. Then y = 3×f(2) = 3×4 = 12. Image: (3, 12).A1
Example 4 — Modulus Sketch: y = |x² − 9|
Sketch y = |x² − 9|, labelling all intercepts.
Step 1: Sketch y = x² − 9 first: parabola, vertex (0, −9), x-intercepts at x = ±3. B1
Step 2: Between x = −3 and x = 3, y < 0 — reflect these upward. Vertex (0, −9) → (0, 9). The curve now has a peak at (0, 9) and touches the x-axis at x = ±3. M1 A1
Step 3: Outside x = ±3 the graph is unchanged (y ≥ 0). The graph is always ≥ 0. A1
Example 5 — Key Points Under Reflection
y = f(x) passes through (−3, 2), (0, −1), (4, 0). Sketch y = −f(−x) and state the images of these points.
Step 1: y = −f(−x) combines f(−x) (reflect in y-axis) and −f (reflect in x-axis). Rule: (x, y) → (−x, −y). M1
y = f(x) has a vertical asymptote at x = 1 and horizontal asymptote at y = 2. Find the asymptotes of y = 2f(x − 3) + 1.
Step 1: Horizontal translation right 3: vertical asymptote x = 1 → x = 1 + 3 = 4. B1
Step 2: y-stretch by 2, then up 1: horizontal asymptote y = 2 → 2(2) + 1 = 5. M1 A1
New asymptotes: x = 4 and y = 5.
Example 7 — Identifying a Transformation
The graph of y = √x is transformed to y = √(3x). Describe the transformation fully.
Step 1: y = √(3x) = f(3x) where f(x) = √x. This is f(kx) with k = 3. M1
Step 2: Stretch parallel to the x-axis, scale factor 1/3. All x-coordinates are multiplied by 1/3. A1
Example check: Point (4, 2) on y = √x maps to (4/3, 2). Verify: √(3 × 4/3) = √4 = 2 ✓
Example 8 — Writing the Equation After Two Transforms
Starting from y = x², apply: (i) reflect in the x-axis, (ii) translate by vector (2, 5). Write the final equation.
Step 1: Reflect in x-axis: y = −x². B1
Step 2: Translate right 2, up 5: replace x with (x − 2), add 5 to whole expression. y = −(x − 2)² + 5. M1 A1
Verify: Vertex of −x² is (0,0). After translation: (0+2, 0+5) = (2, 5). Check: −(2−2)² + 5 = 5 ✓
Common Mistakes
These are the most frequently penalised errors in Cambridge A-Level graph transformation questions.
Mistake 1 — Direction of Horizontal Translation
✗ "y = f(x + 3) shifts the graph 3 units to the right."
✓ y = f(x + 3) shifts the graph 3 units to the LEFT. The +3 is inside, so the graph moves in the negative x direction. Think: f(x+3) reaches the original output 3 steps earlier.
Mistake 2 — Stretch Factor Reciprocal Confusion
✗ "y = f(2x) is a stretch of scale factor 2 parallel to the x-axis."
✓ y = f(2x) is a stretch parallel to the x-axis with scale factor 1/2 (compression). The coordinates become (x/2, y), not (2x, y).
Mistake 3 — Wrong Order for Combined Horizontal Transforms
✗ For y = f(2x − 6), applying "translate left 6, then compress by 1/2" gives y = f(2(x+6)) = f(2x+12), which is wrong.
✓ Factor first: 2x − 6 = 2(x − 3). So the transformations are: compress x by 1/2, then translate right 3. Always factor before reading off the transforms.
Mistake 4 — Forgetting to Update Asymptotes
✗ Sketching y = f(x+2)+3 and leaving the asymptote of y = f(x) unchanged.
✓ Asymptotes transform by the same rules as coordinates. If y = f(x) has asymptote y = 0, then y = f(x+2)+3 has asymptote y = 3.
Mistake 5 — Modulus Sketch: Reflecting the Wrong Part
✗ For y = |f(x)|, reflecting the parts of the graph that are above the x-axis.
✓ Only the parts below the x-axis (where y < 0) are reflected upward. Parts above or on the x-axis are unchanged.
Mistake 6 — Confusing |f(x)| with f(|x|)
✗ Treating y = f(|x|) as "reflect all negative y-values upward" (which is y = |f(x)|).
✓ y = f(|x|): keep the right half (x ≥ 0) unchanged, then mirror it to form the left half. The left half of the original curve is discarded entirely.
Mistake 7 — Incomplete Description of a Transformation
✗ "Translation of 3 units." (No direction given.)
✓ Always specify: type (translation/stretch/reflection), axis/direction, and amount. Example: "Translation of 3 units in the positive x-direction (to the right)."
Mistake 8 — Applying −f(x) as a Horizontal Reflection
✗ "−f(x) reflects in the y-axis."
✓ −f(x) reflects in the x-axis (all y-values change sign). f(−x) reflects in the y-axis (all x-values change sign).
Mistake 9 — Not Checking for Fixed Points
✗ Moving all points when applying a y-stretch, including those on the x-axis.
✓ Under kf(x): any point with y = 0 is a fixed point (k×0 = 0). x-intercepts do not move under a y-stretch or a reflection in the x-axis.
Key Formulas — Graph Transformations
Transformation
Equation
Coordinate Effect
Notes
Vertical translation up a
y = f(x) + a
(x, y) → (x, y+a)
Horizontal asymptotes shift by +a
Horizontal translation left a
y = f(x + a)
(x, y) → (x−a, y)
Vertical asymptotes shift by −a; left when a>0
Vertical stretch scale k
y = kf(x)
(x, y) → (x, ky)
x-intercepts fixed; asymptote y=c → y=kc
Horizontal stretch scale 1/k
y = f(kx)
(x, y) → (x/k, y)
y-intercept fixed; vertical asymptote x=c → x=c/k
Reflection in x-axis
y = −f(x)
(x, y) → (x, −y)
x-intercepts fixed; maxima become minima
Reflection in y-axis
y = f(−x)
(x, y) → (−x, y)
y-intercept fixed; left/right swapped
Modulus of function
y = |f(x)|
y < 0 parts reflected up
Graph always ≥ 0; fold at x-intercepts of f
Function of modulus
y = f(|x|)
x ≥ 0 half mirrored
Even function; left half = mirror of right half
Combined: translate then stretch
y = af(x+b)+c
(x,y) → (x−b, ay+c)
Read inside out; b shifts before a scales for x
Period of f(kx)
T_new = T/k
—
For trig: period of sin(kx) = 2π/k
Amplitude of kf(x)
A_new = k·A
—
For trig: amplitude of k·sin(x) = k
Inverse transformation
Undo f(x+a) with f(x−a)
—
Undo kf(x) with (1/k)f(x)
Proof Bank
These proofs and justifications appear in exam questions and underpin the key results.
Proof 1 — Why f(x + a) Moves the Graph LEFT when a > 0
Let y = f(x) and consider y = f(x + a).
Suppose (p, q) is a point on y = f(x), so f(p) = q.
On y = f(x + a), we want to find where the output is also q.
f(x + a) = q when x + a = p, i.e. x = p − a.
So the point that had x-coordinate p now has x-coordinate p − a.
Since a > 0, p − a < p: the point has moved LEFT by a. Conclusion: f(x + a) with a > 0 translates the graph a units to the left. Q.E.D.
Proof 2 — Order Matters: A Counterexample
Let f(x) = x². Compare: (A) stretch by factor 2 in y, then translate right 1; (B) translate right 1, then stretch by factor 2 in y.
Order A: stretch gives y = 2x². Translate right 1 gives y = 2(x−1)² = 2x²−4x+2.
Order B: translate gives y = (x−1)². Stretch gives y = 2(x−1)² = 2x²−4x+2.
In this case they agree! Now try: (A) x-stretch by 1/2, then translate right 1; (B) translate right 1, then x-stretch by 1/2.
Order A: f(2x) = (2x)² = 4x². Translate: 4(x−1)² = 4x²−8x+4.
Order B: f(x−1) = (x−1)². Then f(2x−1) = (2x−1)² = 4x²−4x+1.
Different results! 4x²−8x+4 ≠ 4x²−4x+1. Horizontal stretches and horizontal translations do NOT commute. Q.E.D.
Proof 3 — f(kx) gives a Stretch of Scale Factor 1/k
Let (p, q) be on y = f(x), so f(p) = q.
On y = f(kx), the output is q when kx = p, i.e. x = p/k.
The new x-coordinate is p/k, which is the original p multiplied by 1/k.
Since every x-coordinate is scaled by 1/k, this is a stretch parallel to the x-axis with scale factor 1/k. Note: when k > 1, 1/k < 1, so the graph is compressed (narrower). When 0 < k < 1, 1/k > 1, so the graph is stretched (wider). Q.E.D.
Proof 4 — y = |f(x)| has no Negative y-values
By definition, |t| ≥ 0 for all real t (since |t| = t if t ≥ 0, and |t| = −t ≥ 0 if t < 0).
Therefore |f(x)| ≥ 0 for all x in the domain of f. Conclusion: the graph of y = |f(x)| lies entirely on or above the x-axis. Q.E.D.
Proof 5 — y = f(|x|) is an Even Function
An even function satisfies g(−x) = g(x) for all x in its domain.
Let g(x) = f(|x|). Then g(−x) = f(|−x|) = f(|x|) = g(x).
(Using the fact that |−x| = |x| for all real x.) Therefore y = f(|x|) is an even function, and its graph is symmetric about the y-axis. Q.E.D.
Interactive Graph Transformation Visualiser
Choose a base function and apply transformations. The original (grey dashed) and transformed (pink) curves are shown side by side.
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Equation: y = x²
Exercise 1 — Identify the Transformation (10 Questions)
Exercise 2 — Write the Equation After Transformation (10 Questions)
8 Cambridge-style questions with mark schemes. Attempt each before revealing the solution.
Q1 [3 marks]
Describe fully the single transformation that maps the graph of y = x³ onto the graph of y = (x + 4)³ − 7.
This is a translation. [B1 type]
The graph moves 4 units to the left and 7 units downward. [B1 direction]
Translation by vector (−4, −7). [B1 complete description]
Q2 [4 marks]
The graph of y = f(x) passes through the point (6, 8). Find the coordinates of the corresponding point on the graph of y = 3f(2x).
Solve 2x = 6 → x = 3 [M1 correct x]
New x-coordinate: 3 [A1]
New y-coordinate: 3 × 8 = 24 [M1 A1] New point: (3, 24)
Q3 [5 marks]
The curve y = f(x) has a vertical asymptote at x = −2 and a horizontal asymptote at y = 3. Find the equations of the asymptotes of y = 2f(x + 1) − 5.
Identify transformations: f(x+1) shifts left 1, ×2 stretches y, −5 shifts down 5. [M1]
Vertical asymptote: x = −2 shifts left 1 → x = −3. [M1 A1]
Horizontal asymptote: y = 3 → 2(3) − 5 = 1. [M1 A1] Asymptotes: x = −3 and y = 1
Q4 [4 marks]
Sketch the graph of y = |2x − 6|, labelling all intercepts with the coordinate axes.
y = |2x − 6| = 0 when x = 3 → x-intercept (3, 0). [B1]
At x = 0: y = |−6| = 6 → y-intercept (0, 6). [B1]
V-shape with vertex at (3, 0). [B1 shape]
Left arm: slope −2 (for x < 3, y = −(2x−6) = 6−2x). Right arm: slope +2 (for x > 3, y = 2x−6). [B1 gradients or correct arms]
Q5 [5 marks]
The function f(x) = sin x is transformed to g(x) = 3 sin(2x) + 1. Describe all transformations applied, stating the correct order and axes affected.
sin x → sin(2x): stretch parallel to x-axis, scale factor 1/2. [B1]
sin(2x) → 3 sin(2x): stretch parallel to y-axis, scale factor 3. [B1]
3 sin(2x) → 3 sin(2x) + 1: translation 1 unit in positive y-direction. [B1]
New amplitude = 3. [B1]
New period = π. [B1]
Q6 [5 marks]
The diagram shows the graph of y = f(x) with key points A(−3, 0), B(0, 2), C(2, −4). Find the images of A, B, C under y = −f(x − 1).
Solve the equation |3x − 2| = |x + 4|. Show all cases clearly.
Case 1: 3x − 2 = x + 4 [M1]
2x = 6 → x = 3. Check: |7| = |7| ✓ [A1]
Case 2: 3x − 2 = −(x + 4) [M1]
3x − 2 = −x − 4 → 4x = −2 → x = −1/2. Check: |−3/2−2| = |7/2|, |−1/2+4| = |7/2| ✓ [A1 M1] x = 3 or x = −1/2 [A1 both]
Q8 [6 marks]
The graph of y = f(x) has a maximum at (2, 5) and passes through (0, 1). Write down the coordinates of the maximum and the y-intercept of: (i) y = f(2x), (ii) y = 2f(x) + 3, (iii) y = −f(x − 1).
(i) f(2x): rule (x,y) → (x/2, y). Max: (1, 5). y-intercept: (0, 1). [A1 A1]
(ii) 2f(x)+3: rule (x,y) → (x, 2y+3). Max: (2, 13). y-intercept: (0, 5). [A1 A1]
(iii) −f(x−1): rule (x,y) → (x+1, −y). Max becomes minimum (2+1, −5) = (3, −5). y-intercept: (0+1, −1) → point (0,1) maps to (1, −1); at x=0 we need f(0−1) = f(−1) which is not given. Note: y-intercept of (iii) requires f(−1) which is not stated — only the image of the given points can be found. Image of (0,1) under −f(x−1): (1, −1). Image of max (2,5): (3, −5) — minimum. [A1 A1]
Past Paper Questions
5 questions drawn from Cambridge A-Level 9709 Pure 1 past papers on Graph Transformations.
Past Paper Q1 — 9709/11/O/N/18 Q3 [5 marks]
The graph of y = f(x) is transformed to the graph of y = 2f(x − 3). Describe fully the two transformations involved, stating the correct order.
First transformation: translation of 3 units in the positive x-direction (right). [B1 type B1 description]
Second transformation: stretch parallel to the y-axis with scale factor 2. [B1 type B1 axis B1 scale factor]
Note: the two horizontal/vertical transformations commute here, so either order is acceptable as long as both are stated correctly. [B1]
Past Paper Q2 — 9709/12/M/J/19 Q2 [4 marks]
The curve y = x² − 2x + 3 is translated by the vector (2, −5). Find the equation of the new curve, giving your answer in the form y = x² + bx + c.
Translation (2, −5): replace x with (x−2) and subtract 5. [M1]
y = (x−2)² − 2(x−2) + 3 − 5 [M1]
= x² − 4x + 4 − 2x + 4 + 3 − 5
= x² − 6x + 6 [A1 A1] y = x² − 6x + 6
Past Paper Q3 — 9709/11/M/J/21 Q5 [6 marks]
The function f is defined by f(x) = 3/(x − 2) + 1 for x > 2. (i) State the equations of the asymptotes of y = f(x). (ii) The graph of y = f(x) is reflected in the x-axis. State the equation of the resulting curve.
(i) Vertical asymptote: x = 2. [B1]
Horizontal asymptote: as x → ∞, 3/(x−2) → 0, so y → 1. Asymptote: y = 1. [B1]
(ii) Reflect in x-axis: replace y with −y. So −y = 3/(x−2) + 1, giving y = −3/(x−2) − 1. [M1 A1]
New asymptotes: x = 2 (unchanged) and y = −1. [A1 A1]
Past Paper Q4 — 9709/13/O/N/20 Q4 [5 marks]
On a single diagram, sketch the graphs of y = |2x − 4| and y = x, stating the coordinates of their intersection(s).
y = |2x − 4|: V-shape, vertex at (2, 0), y-intercept (0, 4). [B1 B1]
y = x: straight line through origin, gradient 1. [B1]
Case 1 (x ≥ 2): 2x − 4 = x → x = 4. Point (4, 4). [M1 A1]
Case 2 (x < 2): −(2x−4) = x → 4−2x = x → x = 4/3. Point (4/3, 4/3). [A1] Intersections: (4, 4) and (4/3, 4/3)
Past Paper Q5 — 9709/12/O/N/22 Q3 [6 marks]
The graph of y = f(x) passes through the points (0, 4), (2, 0) and (5, −3). (i) Write down the coordinates of the corresponding points on y = f(x + 2) − 1. (ii) Write down the coordinates of the corresponding points on y = f(2x).