AQA A-level Mathematics: Transformations of Graphs (7357) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover content reference B9 of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018): transformations of the graph of y = f(x), including y = af(x), y = f(x) + a, y = f(x + a), y = f(ax) and combinations. Read the study guide first, or use the revision notes for a quick refresher. The course pages are the AQA A-level Mathematics hub and the printable checklist. Use the free 10-minute diagnostics to find weak topics.

Sketches should be described clearly in your own working: label the coordinates of key points and the equations of asymptotes.

Questions

1. The point P(6, −3) lies on the curve y = f(x). Write down the coordinates of the image of P on each curve.

(a) y = f(x) + 4 [1] (b) y = f(x + 1) [1] (c) y = f(3x) [1] (d) y = −f(x) [1]

2. A curve has equation y = x² + 8x + 9.

(a) Write the equation in the form y = (x + p)² + q. [2] (b) Describe the transformation that maps y = x² to this curve. [1]

3. The curve y = 3ˣ is translated by the vector (−2, 0) and then stretched with scale factor 4 parallel to the y-axis.

(a) Write down the equation of the image. [2] (b) Find the coordinates of the point where the image meets the y-axis. [1] (c) State the equation of the asymptote of the image. [1]

4. The curve y = f(x), where f(x) = (x + 1)(x − 2)², crosses the x-axis at (−1, 0) and touches it at (2, 0). It has a maximum at (0, 4) and a minimum at (2, 0).

(a) Sketch y = f(x + 2). Give the coordinates of the turning points and of the points where the curve meets the axes. [3] (b) Write down the coordinates of the turning points of y = −½f(x). [2]

5. The curve y = f(x) has asymptotes x = 1 and y = 2, and passes through (3, 0).

(a) For y = f(x + 3), state the equations of the asymptotes and the image of (3, 0). [3] (b) For y = 2f(x) − 5, state the equations of the asymptotes. [2]

6. Let f(x) = x² − 2x. Find the equation of the image of y = f(x) in each case, in the form y = ax² + bx + c.

(a) A translation by (3, 0). [2] (b) A stretch with scale factor 2 parallel to the y-axis, followed by a translation by (0, −1). [2] (c) A stretch with scale factor ½ parallel to the x-axis. [1]

7. (a) Describe a sequence of two transformations that maps y = f(x) to y = f(2x − 6). [3] (b) The point (4, −1) lies on y = f(x). Find its image on y = f(2x − 6). [1]

8. The curve C has equation y = 4 − 3cos(2x), for 0° ≤ x ≤ 360°.

(a) Describe the sequence of transformations that maps y = cos x to C. [3] (b) Find the greatest and least values of y, and the values of x at which each occurs. [3]

9. The curve y = 1/x is stretched with scale factor 5 parallel to the y-axis, then translated by (−2, 0), then translated by (0, 1). The result is the curve C.

(a) Write down the equation of C. [2] (b) State the equations of the asymptotes of C. [2] (c) Find the coordinates of the points where C meets the axes. [3]

10. The curve y = f(x) has x-intercepts (−3, 0), (1, 0) and (5, 0), a y-intercept (0, 3), a maximum at (−1, 4) and a minimum at (3, −2). The curve G has equation y = 2 − f(x/2).

(a) Describe the sequence of transformations that maps y = f(x) to G. [3] (b) Find the coordinates of the turning points of G, and say which is a maximum and which is a minimum. [2] (c) Find the y-intercept of G. [1] (d) Explain why the information given does not let you write down the x-intercepts of G. [2]

11. The curve y = x² − 4x + 1 is translated by the vector (p, q). The image has equation y = x² + 2x + 8.

(a) Find p and q. [4] (b) Find the coordinates of the minimum point of the image. [1]

Answers

1. (a) (6, −3 + 4) = (6, 1) [1] (b) The translation is 1 to the left, so (5, −3) [1] (c) x is divided by 3, so (2, −3) [1] (d) y changes sign, so (6, 3) [1] Examiner insight: Write each image next to the letter it answers. A common slip is moving f(x + 1) to the right, which gives (7, −3), a wrong x-coordinate.

2. (a) x² + 8x = (x + 4)² − 16 [1], so y = (x + 4)² − 7 [1] (b) Translation by the vector (−4, −7) [1] Examiner insight: For a “describe” part, a bare vector with the word translation is complete. Giving (4, 7) from the signs in the brackets is the usual way to lose this mark.

3. (a) After the translation, y = 3^(x+2) [1]. After the stretch, y = 4 × 3^(x+2) [1] (b) At x = 0, y = 4 × 9 = 36, so the point is (0, 36) [1] (c) The stretch does not move y = 0, so the asymptote is y = 0 [1] Examiner insight: The order is stated in the question, so the 4 multiplies the whole of 3^(x+2). Writing 3^(x+2) + 4 would be a translation, not a stretch.

4. (a) Replace x by x + 2: each point moves 2 to the left [1]. Turning points: maximum (−2, 4) and minimum (0, 0) [1]. The curve crosses the x-axis at (−3, 0) and touches it at the origin, and the y-intercept is (0, f(2)) = (0, 0), which is the minimum [1]. (b) The map is (x, y) → (x, −½y) [1]. The maximum becomes a minimum at (0, −2), and the minimum becomes a maximum at (2, 0) [1] Examiner insight: In part (a), a clear sketch has the y-intercept and the minimum at the same point. Recalculating the y-intercept from the new equation avoids placing it at (0, 4), the old maximum.

5. (a) x = 1 becomes x = −2 [1]; y = 2 is unchanged [1]; (3, 0) goes to (0, 0) [1] (b) The vertical asymptote does not change: x = 1 [1]. For the horizontal one, 2(2) − 5 = −1, so y = −1 [1] Examiner insight: Apply the multiplication first and then the subtraction to y = 2. Subtracting before doubling gives −2, which is a different asymptote.

6. (a) Replace x by x − 3: (x − 3)² − 2(x − 3) [1] = x² − 8x + 15 [1] (b) 2(x² − 2x) − 1 [1] = 2x² − 4x − 1 [1] (c) f(2x) = (2x)² − 2(2x) = 4x² − 4x [1] Examiner insight: In (c) the scale factor is ½, so x is replaced by 2x. Using x/2 gives x²/4 − x, the equation for scale factor 2.

7. (a) Write 2x − 6 = 2(x − 3). Stretch with scale factor ½ parallel to the x-axis [1], then translate by the vector (3, 0) [1]. (Alternatively, translate by (6, 0) and then stretch with scale factor ½ parallel to the x-axis.) The numbers must match the order given [1] (b) 2x − 6 = 4 gives x = 5, so the image is (5, −1) [1] Examiner insight: If you state the order, the translation must match it: 3 after the stretch, 6 before it. Using the part (b) point to test both versions is a quick way to catch a mismatch.

8. (a) Stretch with scale factor ½ parallel to the x-axis [1], stretch with scale factor 3 parallel to the y-axis and reflect in the x-axis [1], translate by the vector (0, 4) [1] (b) The greatest value is 4 + 3 = 7 [1], when cos 2x = −1, so 2x = 180° or 540°, giving x = 90° and 270° [1]. The least value is 4 − 3 = 1, when cos 2x = 1, so 2x = 0°, 360°, 720°, giving x = 0°, 180°, 360° [1] Examiner insight: The reflection is what turns the cosine peaks into minima. Without it, the greatest value would appear at x = 0°, and the sketch would be upside down.

9. (a) After the three stages, y = 5/(x + 2) + 1 [1], which equals (x + 7)/(x + 2) [1] (b) The asymptotes are x = −2 [1] and y = 1 [1] (c) At x = 0, y = 5/2 + 1 = 7/2, so the point is (0, 7/2) [1]. At y = 0, 5/(x + 2) = −1, so x + 2 = −5 [1] and the point is (−7, 0) [1] Examiner insight: A graph with both asymptotes and both axis crossings labelled is much easier to credit than one with only the shape drawn. Keep the exact fraction 7/2 unless the question asks for a decimal.

10. (a) Stretch with scale factor 2 parallel to the x-axis [1], reflect in the x-axis [1], translate by the vector (0, 2) [1] (b) The map is (x, y) → (2x, 2 − y) [1]. The maximum (−1, 4) becomes the minimum (−2, −2), and the minimum (3, −2) becomes the maximum (6, 4) [1] (c) The y-intercept (0, 3) goes to (0, −1) [1] (d) The x-intercepts of G are where f(x/2) = 2 [1]. The given points have f equal to 0, 4 or −2, which is never 2, so none of them is an x-intercept of G, and the x-values where f takes the value 2 are not given [1] Examiner insight: Students often map the old x-intercepts and call them the new x-intercepts. Those points are now at y = 2, and a note saying so shows you understand the transformation.

11. (a) The image is y = (x − p)² − 4(x − p) + 1 + q [1] = x² − (2p + 4)x + p² + 4p + 1 + q [1]. Comparing the x-coefficients, −(2p + 4) = 2, so p = −3 [1]. Then 9 − 12 + 1 + q = 8, so q = 10 [1] (b) The original minimum is at (2, −3) (from (x − 2)² − 3), so the image minimum is (−1, 7) [1] Examiner insight: A shorter route is to use the vertex form of both curves. Checking that (−1, 7) satisfies y = x² + 2x + 8 gives a quick test of both p and q.

Where marks are usually lost

• Moving f(x + a) to the right, or reading the vector from the sign inside the bracket.
• Using scale factor a for f(ax), when the scale factor is 1/a.
• Describing a stretch without saying which axis it is parallel to.
• Applying two transformations in an order that does not match the numbers given.
• Leaving the y-intercept unchanged after a horizontal translation.
• Forgetting to move or label the asymptotes, or applying the stretch to the vertical asymptote of a reciprocal curve.
• Not swapping a maximum and a minimum after a reflection in the x-axis.
• Mapping x-intercepts as if they were still x-intercepts after a vertical translation.

Next steps

• Review the Transformations of Graphs revision notes.
• Re-read the study guide for any question you got wrong.
• Visit the AQA A-level Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics to find your next topic.
• Book a free trial class.

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.