Revision Notes
AQA A-level Mathematics: Transformations of Graphs (7357) – Revision Notes
Revision notes for AQA A-level Maths B9: the four transformations of y = f(x), order of combinations, asymptotes, a self-test and common lost marks.
- Subject
- Mathematics
- Level
- A LEVELS
- Topic
- Transformations of Graphs
- Author
- Marlbridge Academic Team
- Updated
Aligned to AQA A Level Mathematics (7357), For first teaching 2017. Official specification .
Syllabus page (what it covers and how it is assessed): AQA A Level Mathematics.
Syllabus points this page covers
7357
- B9 Transformations of graphs
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These revision notes condense content reference B9 of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018): the effect of simple transformations on the graph of y = f(x), including y = af(x), y = f(x) + a, y = f(x + a), y = f(ax) and combinations. For the explanations and fully worked examples, read the Transformations of Graphs study guide first.
When you are ready to test yourself, use the practice questions. The full course is at the AQA A-level Mathematics hub, with a printable checklist. The free 10-minute diagnostics show which topics need work.
The four transformations
| Graph | Describe it as | Point (x, y) becomes |
|---|---|---|
| y = f(x) + a | translation by (0, a) | (x, y + a) |
| y = f(x + a) | translation by (−a, 0) | (x − a, y) |
| y = af(x) | stretch, scale factor a, parallel to the y-axis | (x, ay) |
| y = f(ax) | stretch, scale factor 1/a, parallel to the x-axis | (x/a, y) |
| y = −f(x) | reflection in the x-axis | (x, −y) |
| y = f(−x) | reflection in the y-axis | (−x, y) |
Inside or outside?
- Outside the brackets: acts on y, does the obvious thing.
- Inside the brackets: acts on x, does the opposite.
A full description needs the type, the direction or axis, and the number. “Stretch” alone, or “move”, is not a full description.
What stays fixed
| Transformation | Unchanged |
|---|---|
| Stretch parallel to the y-axis | x-intercepts |
| Stretch parallel to the x-axis | y-intercept |
| Reflection in the x-axis | x-intercepts |
| Reflection in the y-axis | y-intercept |
| Any translation | nothing (recalculate the intercepts) |
Under a reflection or a negative scale factor, maxima and minima swap when the reflection is in the x-axis. A reflection in the y-axis does not swap them.
Method for a sketch
- Sketch or read y = f(x). List the key points: intercepts, turning points, endpoints, and the equations of any asymptotes.
- Write the new equation in the form y = af(b(x + c)) + d if it is not already.
- Write the point map: (x, y) → (x/b − c, ay + d).
- Apply the map to every key point and to every asymptote.
- Recalculate any intercept that is not a mapped key point. For the y-intercept, substitute x = 0 in the new equation.
- Label every mapped point and every asymptote on the sketch.
Order of combined transformations
For y = af(b(x + c)) + d, write the bracket as b(x + c) and apply the changes in this order.
y = a f( b (x + c) ) + d
1. stretch parallel to the x-axis, scale factor 1/b
2. translate by (−c, 0)
3. stretch parallel to the y-axis, scale factor a
4. translate by (0, d)
- The x pair (1, 2) and the y pair (3, 4) do not interfere, so either pair can come first.
- Within a pair the order matters. Stretch then translate in the same direction gives a different translation from translate then stretch. For example, f(2x − 1) = f(2(x − ½)) is “stretch, scale factor ½, then translate by (½, 0)” or “translate by (1, 0), then stretch, scale factor ½”. Check one point.
Asymptotes
| Graph | Horizontal asymptote y = k becomes | Vertical asymptote x = h becomes |
|---|---|---|
| y = af(x) + d | y = ak + d | x = h |
| y = f(b(x + c)) | y = k | x = h/b − c |
Example: if y = f(x) has asymptotes x = 3 and y = 1, then y = 2f(x) − 5 has asymptotes x = 3 and y = −3.
Reading and writing equations
- Completing the square gives a translation of y = x². For y = x² − 10x + 31 = (x − 5)² + 6, the vertex is (5, 6).
- For y = a(x + p)² + q, stretch first, then translate by (−p, q).
- To write the new equation, replace x with the expression that reverses the change. A stretch with scale factor 2 parallel to the x-axis replaces x by x/2. A translation by (3, 0) replaces x by x − 3.
Worked reminder
The curve y = x² − 4x + 1 has its minimum at (2, −3). Find the equation of the image under a translation by (5, 2) and the new minimum.
y = (x − 5)² − 4(x − 5) + 1 + 2 = x² − 14x + 48
minimum: (2 + 5, −3 + 2) = (7, −1)
check: 49 − 98 + 48 = −1
Must-know distinctions
| Compare | Key difference |
|---|---|
| f(x) + 3 and f(x + 3) | Up 3, and left 3. One acts on y, the other on x. |
| 2f(x) and f(2x) | Twice as tall, and half as wide. |
| −f(x) and f(−x) | Reflection in the x-axis, and in the y-axis. |
| f(x) + a and af(x) | A translation moves every point by the same amount. A stretch moves points by amounts that depend on their distance from the axis. |
| Translation and stretch | A translation keeps the shape. A stretch changes it, and points on the invariant axis do not move. |
Worked reminder: a curve with asymptotes
Sketch y = 2 − 6/(x − 4), starting from y = 1/x.
Write it as −6f(x − 4) + 2 with f(x) = 1/x. Translate by (4, 0), stretch with scale factor 6 parallel to the y-axis, reflect in the x-axis, translate by (0, 2).
asymptotes: x = 0 → x = 4 y = 0 → y = 2
y-intercept: x = 0 gives 2 − 6/(−4) = 2 + 3/2 = 7/2
x-intercept: 2 − 6/(x − 4) = 0, so x − 4 = 3 and x = 7
Because of the reflection, the branches lie in the top-left and bottom-right regions made by the asymptotes, the opposite way round from y = 1/x. The point (1, 1) goes to (5, −4).
Worked reminder: a trigonometric graph
Describe y = 4 − 3cos(2x) as a transformation of y = cos x, and give its greatest and least values.
- Stretch, scale factor ½, parallel to the x-axis.
- Stretch, scale factor 3, parallel to the y-axis.
- Reflect in the x-axis.
- Translate by (0, 4).
The greatest value is 4 + 3 = 7 and the least is 4 − 3 = 1. In degrees, the period is 180°, so y = 1 at x = 0, 180 and 360, and y = 7 at x = 90 and 270.
Checking your sketch
- Does each mapped point lie where the map says it should? Test one by substituting its new x-value into the new equation.
- Has every maximum, minimum and asymptote been moved the same way as the rest of the curve?
- If the curve goes through the origin, did the transformation keep it there? Only a stretch or a reflection does.
- Does the sketch show the correct behaviour at the ends, going up or down for large positive and large negative x?
Quick self-test
- The point (−2, 5) lies on y = f(x). Give its image on y = f(x) − 3.
- Give its image on y = f(x + 4).
- Give its image on y = ½f(x).
- Give its image on y = f(3x).
- Give its image on y = f(−x).
- Give its image on y = −f(x − 1).
- Describe the graph of y = (x − 2)² − 9 as a transformation of y = x², and give its x-intercepts.
- State the asymptotes of y = 3/(x − 1) + 2.
- The curve y = √x is stretched with scale factor ¼ parallel to the x-axis. Write its new equation.
- Describe fully the transformations that map y = f(x) to y = 3f(x/2).
- If y = g(x) has the asymptote y = −3, state the horizontal asymptote of y = 2g(x) + 1.
- State the period of y = cos(4x), in degrees.
Answers
- (−2, 2).
- (−6, 5). The translation is 4 to the left.
- (−2, 2.5).
- (−2/3, 5).
- (2, 5).
- Translate right by 1: (−1, 5). Then reflect in the x-axis: (−1, −5).
- Translation by (2, −9). The x-intercepts come from (x − 2)² = 9, so x = 5 and x = −1.
- x = 1 and y = 2.
- y = √(4x), which equals 2√x.
- Stretch, scale factor 2, parallel to the x-axis, and stretch, scale factor 3, parallel to the y-axis. The order of these two does not matter.
- y = 2(−3) + 1 = −5.
- 360°/4 = 90°.
Where marks are usually lost
- Describing a transformation with a single word such as “move” or “stretch”, without the direction, axis, vector or scale factor.
- Treating f(x + a) as a translation to the right.
- Writing scale factor 2 for f(2x), when the scale factor is ½.
- Giving two transformations in the wrong order when the order changes the numbers, for example f(2x − 1).
- Translating an asymptote and leaving the intercepts as they were. After a translation, recalculate the intercepts from the new equation.
- Forgetting that a negative multiplier swaps maxima and minima.
- Labelling the images of key points but not the new asymptotes, or sketching a curve that crosses an asymptote.
- Combining a vertical stretch with a vertical translation in the wrong order, so that 2f(x) + 1 becomes 2f(x) + 2.
- Reading the sketch’s y-intercept from the old curve instead of substituting x = 0 into the new equation.
Official syllabus
AQA, A-level Mathematics (7357) specification, for teaching from September 2017 onwards, version 1.3 (31 January 2018), section 3.3 (B: Algebra and functions), content reference B9.
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