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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Answer key: Translations and reflections of graphs worksheet — GCSE Higher
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- (a) y = 2ˣ + 3 — A translation of 3 units in the positive y-direction shifts the whole graph up, which means adding to the output: y = f(x) + k with k = 3, so the image is y = 2ˣ + 3. Adding the 3 inside the power instead of outside it, which translates the graph horizontally instead of vertically, gives y = 2ˣ⁺³. Using a negative 3, which moves the graph down instead of up, gives y = 2ˣ − 3. Mistaking 2ˣ for the linear expression 2x and adding 3 inside brackets gives y = 2(x + 3).
- (a) Positive x-direction, 90°; image is y = sin x. — Writing cos(x − 90°) as cos(x − a) with a = 90 shows this is a horizontal translation, y = f(x − a), which moves the graph 90° in the positive x-direction; the identity cos(x − 90°) = sin x confirms the image is y = sin x. Choosing the negative x-direction reverses the sign inside the bracket — subtracting inside the bracket always translates in the positive x-direction, not the negative one, so that statement is wrong on direction. Getting the direction right but conflating the subtraction inside the bracket with an extra reflection of the output flips the sign of the resulting graph, wrongly giving y = −sin x. Treating the subtraction as if it changed the output directly, rather than the input, wrongly calls this a vertical translation even while still correctly recalling that the image simplifies to y = sin x.
- (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.
- (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.
- (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.
- (b) a = 7; turning point (4, 2) — A vertical translation y = f(x) + a moves every point on the graph up or down by a, so the x-coordinate of the turning point stays at 4 and the minimum value becomes −5 + a. Setting −5 + a = 2 and solving gives a = 7, so the new turning point is (4, 2). Rearranging −5 + a = 2 with a sign error, treating it as a = −5 − 2, gives a = −7 while still landing on the correct turning-point coordinates. Correctly finding a = 7 but then writing down the original turning point instead of the shifted one gives (4, −5). Assuming a is simply equal to the new minimum value itself, ignoring the original −5 entirely, gives a = 2.
- (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) y = −f(x + 2) has a maximum turning point at (1, 5). — y = f(x + 2) translates the graph 2 units in the negative x-direction, so the turning point's x-coordinate moves from 3 to 1; reflecting in the x-axis then negates the y-coordinate, turning −5 into 5, and also turns a minimum into a maximum — giving a true maximum turning point at (1, 5). Translating the root at x = 1 in the wrong direction, adding 2 instead of subtracting it, gives x = 3, but the true image root is at 1 − 2 = −1, so that statement is false. Keeping the coordinates (1, 5) correct but forgetting that reflecting in the x-axis turns a minimum into a maximum gives a false "minimum turning point" statement. Getting the reflection's effect on the turning-point type right (a maximum) but translating the x-coordinate in the wrong direction, using 3 + 2 = 5 instead of 3 − 2 = 1, gives a false statement at (5, 5).
- (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.
- (c) Translate +1 in y; new range 0 ≤ y ≤ 2 — Adding 1 to sin x translates the graph 1 unit in the positive y-direction, and every y-value on the range increases by 1: −1 + 1 = 0 and 1 + 1 = 2, giving a new range of 0 ≤ y ≤ 2. A translation in the x-direction would not add anything to the y-values, so 'Translate +1 in x; range unchanged' correctly leaves the range at −1 ≤ y ≤ 1 for that (wrong) transformation, but the transformation itself is not what y = sin x + 1 shows. 'Translate +1 in y; new range −2 ≤ y ≤ 0' correctly spots the translation but subtracts 1 from each bound instead of adding it. 'Reflect in the x-axis' mistakes the transformation for a reflection rather than a translation; reflecting sin x in the x-axis does leave the range at −1 ≤ y ≤ 1 unchanged in size, but that is not the transformation y = sin x + 1 actually applies.
- (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.
- (d) x = −2, x = 6; y-intercept (0, 12) — Reflecting y = f(x) in the x-axis, to get y = −f(x), negates every y-value but leaves every x-value fixed. The x-intercepts happen where y = 0, and −0 = 0, so they are unaffected: y = −f(x) still crosses the x-axis at x = −2 and x = 6. The y-intercept is the value at x = 0: f(0) = −12, so −f(0) = 12, giving the point (0, 12) — the sign flips because the y-intercept is a nonzero y-value, unlike the roots. Writing 'x = 2, x = −6; y-intercept (0, −12)' comes from confusing −f(x) with f(−x) — reflecting in the y-axis instead of the x-axis, which negates the x-values of the intercepts (turning −2 into 2 and 6 into −6) but leaves f(0) unchanged, since f(−0) = f(0) = −12. Writing 'x = −2, x = 6; y-intercept (0, −12)' comes from forgetting that −f(x) is a reflection at all, and assumes both intercepts stay exactly as they were. Writing 'x = 2, x = −6; y-intercept (0, 12)' correctly negates the y-intercept but wrongly negates the x-intercepts too, as if a reflection in the x-axis also flipped the sign of every x-value.
- (a) (4, −8), a minimum point — The transformation x → x + 1 inside f translates the graph 1 unit to the LEFT, so the x-coordinate becomes 5 − 1 = 4. The minus sign in front of f reflects the graph in the x-axis, so the y-coordinate becomes −8, and a reflection in the x-axis turns every maximum into a minimum, so (4, −8) is a minimum point. Writing '(4, −8), a maximum point' gets the coordinates right but forgets that a reflection in the x-axis swaps maximum and minimum points. Writing '(6, −8), a minimum point' comes from translating 1 unit to the RIGHT instead of the left — f(x + 1) always moves the graph in the negative x-direction. Writing '(4, 8), a minimum point' gets the x-coordinate and the type right, but forgets to actually negate the y-coordinate, even though it does correctly reclassify the point as a minimum.
- (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.
- (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.
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