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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- (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).
- (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.
- (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).
- (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) 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.
- (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.
- (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.
- (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) (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) 3 — y = f(x + 2) is f(x) translated 2 units to the LEFT (inside the bracket, adding moves the graph in the negative x-direction). The root moves with the whole graph: 5 − 2 = 3. Moving right instead of left gives 7; assuming a bracket shift leaves the root unchanged gives 5; writing down the shift amount 2 itself skips the translation altogether.
- (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.
- (a) x = 2, x = 7; y-intercept cannot be found here — Translating y = f(x) to y = f(x − 3) shifts the graph 3 units to the right, so each root increases by 3: x = −1 becomes x = 2, and x = 4 becomes x = 7. The y-intercept is the value at x = 0, which for this new graph is f(0 − 3) = f(−3) — and f(−3) is not one of the values given, so the new y-intercept cannot be worked out from the information given. Writing 'y-intercept stays at (0, −8)' wrongly assumes a horizontal translation leaves the y-intercept unchanged — it generally does not, since it moves the whole graph sideways, including the point that used to sit on the y-axis. Writing roots at x = −4 and x = 1 comes from translating 3 units to the LEFT instead of to the right — f(x − 3) shifts the graph in the positive x-direction, not the negative direction.
- (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.
- (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.
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