Printable · GCSE Higher · ages 14-16
Translations and reflections of graphs worksheet — GCSE Higher
Fifteen questions on "translations and reflections of graphs" — DfE statement A13. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Translations and reflections of graphs worksheet — GCSE Higher
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- 1.The graph of y = f(x) passes through the point (2, 5). Write down the coordinates of the corresponding point on the graph of y = f(x − 4) + 1.
- 2.The graph of y = f(x) passes through the point (0, 4). Work out the y-coordinate of the point where the graph of y = −f(x) + 5 crosses the y-axis.
- 3.A company's weekly profit, in £, is modelled by y = f(x), where x is the number of weeks since launch. The graph of y = f(x) has a maximum at (10, 45000). A rival company uses the same marketing strategy but starts trading 6 weeks later and has fixed costs £8000 higher every week, so its profit is modelled by y = f(x − 6) − 8000. In which week does the rival's maximum weekly profit occur, and what is it?
- 4.The graph of y = f(x) has a maximum turning point at (−1, 6). Write down the coordinates of the maximum turning point of the graph of y = f(x − 3).
- 5.The graph of y = f(x) has a minimum point at (4, −1). Write down the coordinates of the corresponding turning point on the graph of y = −f(x).
- 6.Using the table of values of f(x) (x = 0, 1, 2, 3 gives f(x) = 5, 8, 4, 1), work out the value of −f(x) when x = 1.
- 7.The graph of y = f(x) passes through the point (2, 7). The graph of y = f(x) + a passes through the point (2, 3). Work out the value of a.
- 8.The table shows some values of f(x): when x = 0, f(x) = 5; when x = 1, f(x) = 8; when x = 2, f(x) = 4; when x = 3, f(x) = 1. Work out the value of f(x − 1) when x = 2.
- 9.Describe the single transformation that maps the graph of y = f(x) onto the graph of y = f(−x).
- 10.The graph of y = f(x) has a minimum turning point at (2, −3). The graph of y = −f(x) + a has a maximum turning point at (2, 9). Work out the value of a.
- 11.The graph of y = f(x) has x-intercepts at x = −1 and x = 4. Which statement correctly describes the x-intercepts of y = f(2 − x)?
- 12.Describe a sequence of two transformations that maps the graph of y = cos x onto the graph of y = cos(x + 90°) − 2.y = cos(x)
- 13.Describe a sequence of two transformations that maps the graph of y = x² onto the graph of y = −(x − 5)².y = x²
- 14.The graph of y = f(x) passes through the point (0, 4). Work out the y-coordinate of the point where the graph of y = f(x) − 6 crosses the y-axis.
- 15.The graph of y = 2x² + 6x − 1 is translated by the vector (−2, 0). Work out the equation of the image, giving your answer in the form y = 2x² + bx + c.y = 2x² + 6x − 1y = 2x²
Answer key
- (d) (6, 6) — y = f(x − 4) + 1 combines two translations: the −4 inside the brackets moves the graph 4 units to the RIGHT, and the +1 outside the brackets moves it 1 unit UP. Applying both to (2, 5): add 4 to the x-coordinate 2, and add 1 to the y-coordinate 5, giving (6, 6). Moving left instead of right, as the minus sign inside the bracket might suggest, gives (−2, 6) or (−2, 4); forgetting the horizontal shift altogether and only applying the vertical one gives (2, 6).
- (c) 1 — At x = 0, f(0) = 4. Applying the transformations in order — reflect in the x-axis first, then translate up by 5 — gives −f(0) + 5 = −4 + 5 = 1. Applying the translation but forgetting the reflection gives f(0) + 5 = 9. Applying the reflection to the whole expression, including the +5, gives −f(0) − 5 = −9. Applying the reflection but forgetting the translation gives −f(0) = −4.
- (c) Week 16, £37,000 — y = f(x − 6) − 8000 combines a horizontal translation of 6 units RIGHT (subtracting 6 inside the brackets) with a vertical translation of £8000 DOWN (subtracting 8000 outside). Applying both to the maximum (10, 45000): 10 + 6 = 16, so the new maximum is in week 16. And 45000 − 8000 = 37000, so the maximum weekly profit is £37,000.
- (b) (2, 6) — y = f(x − 3) translates y = f(x) horizontally by 3 units to the RIGHT — inside the brackets, subtracting moves the graph in the positive x-direction. Turning point (−1, 6) → (−1 + 3, 6) = (2, 6). The common slip is to move LEFT instead, since the sign inside the bracket is negative — that gives (−4, 6). Changing the y-coordinate instead of the x-coordinate, as in (−1, 3) or (−1, 9), treats this as a vertical shift, which y = f(x − 3) is not.
- (a) (4, 1) — y = −f(x) reflects the graph of y = f(x) in the x-axis: every point (x, y) maps to (x, −y). Applying this to (4, −1): the x-coordinate stays 4, and the y-coordinate −1 becomes its negative, 1. Leaving the y-coordinate unchanged skips the reflection entirely, giving (4, −1); reflecting the x-coordinate instead, or reflecting both, mixes this up with a reflection in the y-axis or a rotation, giving (−4, −1) or (−4, 1).
- (a) −8 — −f(x) means take the output value from the table and change its sign, without changing which x-value is looked up. From the table, f(1) = 8, so −f(1) = −8. Reading f(1) = 8 from the table but forgetting to apply the negative sign gives 8. Misreading the row and using f(0) = 5 instead of f(1) = 8, then negating it, gives −5. Confusing −f(x) with f(x) − 1 — taking f(1) = 8 and subtracting 1 instead of negating — gives 7.
- (c) −4 — f(2) = 7, and y = f(x) + a passing through (2, 3) means f(2) + a = 3, so 7 + a = 3, giving a = 3 − 7 = −4. Writing a = 4 comes from subtracting the wrong way round, 7 − 3 instead of 3 − 7. Writing a = 10 comes from adding the two y-values instead of subtracting one from the other. Writing a = 3 comes from taking the new y-coordinate as the value of a directly, without accounting for the original y-value of 7 at all.
- (a) 8 — f(x − 1) means every input to f is reduced by 1 before it is looked up. At x = 2, the input to f becomes 2 − 1 = 1, so f(x − 1) at x = 2 is f(1) = 8. Reading f(2) = 4 directly from the table, without applying the shift, gives 4. Shifting in the wrong direction, using x + 1 = 3 instead of x − 1 = 1, gives f(3) = 1. Confusing f(x − 1) with f(x) − 1 — taking f(2) = 4 and then subtracting 1 — gives 3.
- (c) A reflection in the y-axis — Replacing x with −x reflects the graph in the y-axis: each point (x, y) maps to (−x, y). Reflecting the OUTPUT instead, y = −f(x), gives a reflection in the x-axis — that is a different function. Combining both reflections gives a rotation of 180° about the origin, and reflecting in the line y = x swaps the x- and y-values, which is what produces the inverse function, not f(−x). Check first which side of f the minus sign sits on.
- (b) 6 — Reflecting y = f(x) in the x-axis turns the minimum point (2, −3) into a maximum point at (2, 3), since −f(x) negates every y-value: −(−3) = 3. Adding a then gives 3 + a = 9, so a = 9 − 3 = 6. Forgetting the reflection and using the original y-value of −3 gives −3 + a = 9, so a = 12 — this ignores that −f(x) changes the sign of the y-coordinate before a is added. Writing a = −12 comes from subtracting in the wrong order, working out 9 − (−3) as −3 − 9 instead. Writing a = −6 comes from taking the negative of the correct answer, as if the final value of a needed to be reflected too, on top of the turning point.
- (d) Reflect in the y-axis, +2 in x; roots 3, −2 — f(2 − x) is zero exactly when 2 − x equals one of f's roots: 2 − x = −1 or 2 − x = 4. Solving each correctly (x = 2 − (−1) = 3, and x = 2 − 4 = −2) gives the new roots x = 3 and x = −2. This is the same as reflecting y = f(x) in the y-axis to get f(−x), then translating 2 units in the positive x-direction to get f(−(x − 2)) = f(2 − x). Solving 2 − x = k as x = k − 2 instead of x = 2 − k is a sign slip when rearranging, and gives x = −3 and x = 2. Translating +2 in x with no reflection at all uses f(x − 2), whose roots are the original roots plus 2: x = 1 and x = 6 — this misses the reflection completely. Assuming 'no overall change' wrongly treats a reflection-and-translation pair as always cancelling out, when here the roots genuinely move, from x = −1 and x = 4 to x = 3 and x = −2.
- (b) Translate −90° in x, then translate −2 in y. — cos(x + 90°) translates the graph 90° in the NEGATIVE x-direction, since a positive shift inside the bracket moves a graph left, not right, and subtracting 2 afterwards translates it 2 units in the negative y-direction (down). So the sequence is: translate −90° in x, then translate −2 in y. Using +90° in x reverses the direction of the horizontal shift — the sign inside the bracket moves the graph the opposite way to what it looks like. Using +2 in y reverses the direction of the vertical shift; subtracting 2 outside the function moves the graph down, not up. Describing the −2 as a reflection in the x-axis is wrong because a reflection turns positive y-values negative and vice versa, whereas here every y-value is simply reduced by the fixed amount 2, which is what a translation does, not a reflection.
- (b) Translate +5 in x, then reflect in the x-axis. — Translating y = x² by 5 units in the positive x-direction gives y = (x − 5)². Reflecting this in the x-axis, which replaces y with −y, gives y = −(x − 5)², matching the target. Using a translation of −5 in x instead gives y = (x + 5)², and reflecting that in the x-axis gives y = −(x + 5)² — the sign inside the bracket is wrong. Reflecting in the y-axis first does nothing to y = x², since (−x)² = x², so translating afterwards only reaches y = (x − 5)² with no negative sign at all. Translating by 5 units in y instead of x gives y = x² + 5, and reflecting that in the x-axis gives y = −x² − 5, a different curve altogether — a vertical shift does not create the (x − 5)² term the target equation needs.
- (d) −2 — y = f(x) − 6 is f(x) shifted down by 6, so every y-value on the graph decreases by 6. At x = 0, f(0) = 4, so the new y-value is 4 − 6 = −2. Adding 6 instead of subtracting gives 10; writing down the shift itself, −6, or leaving the original value 4 unchanged both skip the translation altogether.
- (b) y = 2x² + 14x + 19 — A translation by the vector (−2, 0) moves the graph 2 units in the negative x-direction, which means replacing x with (x + 2): 2(x + 2)² + 6(x + 2) − 1. Expanding 2(x + 2)² gives 2x² + 8x + 8, and 6(x + 2) gives 6x + 12; collecting terms 2x² + 8x + 8 + 6x + 12 − 1 simplifies to 2x² + 14x + 19. Substituting (x − 2) instead of (x + 2) — translating in the wrong direction — gives 2(x−2)² + 6(x−2) − 1, which simplifies to 2x² − 2x − 5. Distributing the outer 2 only onto the x² term of the expansion, instead of onto every term of (x + 2)², gives 2x² + 10x + 15. Dropping the −1 when collecting the constant terms, treating 8 + 12 as 20 rather than 8 + 12 − 1 as 19, gives 2x² + 14x + 20.
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