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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- (b) y = f(−x) — y = f(−x) reflects the graph of y = f(x) in the y-axis: every x-coordinate changes sign. The x-intercepts −3 and 7 become 3 and −7, matching the second graph's intercepts −7 and 3. A point already on the y-axis is unaffected, since −0 = 0, so the y-intercept (0, 21) stays exactly where it is — matching the second graph as well. y = −f(x) leaves the x-intercepts unchanged at −3 and 7, since f(x) = 0 exactly where −f(x) = 0, which does not match; it also sends the y-intercept to (0, −21), a second mismatch. y = −f(−x) does send the x-intercepts to the right places, −7 and 3, but it sends the y-intercept to (0, −21) instead of (0, 21), so it fails the second clue. y = f(x) − 4 moves every point down 4, sending the y-intercept to (0, 17) instead of (0, 21), so it fails the y-axis clue. Test each option against BOTH clues — the pair of x-intercepts and the point on the y-axis — because more than one option gets only one of the two right.
- (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) (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) 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) 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.
- (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.
- (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.
- (a) −1 — Reflecting y = sin x in the x-axis gives y = −sin x, so g(x) = −sin x. Since sin 90° = 1, g(90) = −1. Reading sin 90° = 1 and forgetting to apply the reflection gives 1. Misreading the angle as 0° instead of 90° gives sin 0° = 0, so 0. Confusing sin 90° with sin 30° = 0.5, then reflecting it, gives −0.5.
- (d) y = x² − 10x + 21 — A translation by the vector (3, 0) moves the graph 3 units in the positive x-direction, which means replacing every x in the equation with (x − 3). Substitute into x² − 4x: (x − 3)² − 4(x − 3). Expand (x − 3)² to x² − 6x + 9, and expand −4(x − 3) to −4x + 12. Collecting like terms, x² − 6x + 9 − 4x + 12 = x² − 10x + 21, so the image is y = x² − 10x + 21. Substituting (x + 3) instead of (x − 3) — translating in the wrong direction — gives y = x² + 2x − 3. Adding 3 straight onto the original equation, treating the translation as vertical instead of horizontal, gives y = x² − 4x + 3. Expanding (x − 3)² as x² − 3x + 9, using −3x instead of −6x for the middle term, and then combining with −4(x − 3) gives y = x² − 7x + 21.
- (a) (3, 7) — y = f(x) + 5 is a vertical translation of y = f(x) by 5 units up — the translation vector is (0, 5) — so only the y-coordinate of any point changes. Turning point (3, 2) → (3, 2 + 5) = (3, 7). Adding the 5 to the x-coordinate, or treating it as a horizontal shift like y = f(x + 5), moves the wrong coordinate — check first whether the number sits inside or outside the brackets.
- (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) y = −x³ + 5x — Reflecting a graph in the y-axis replaces every x in the equation with −x: y = (−x)³ − 5(−x) = −x³ + 5x. Writing y = −x³ − 5x comes from substituting −x into the x³ term only and leaving the −5x term as it was. Writing y = x³ + 5x comes from substituting −x into the −5x term only and leaving the x³ term as it was. Writing y = x³ − 5x is the original equation with no reflection applied at all — every term needs the substitution, not just one of them.
- (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.
- (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).
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