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.
Higher only
Translations and reflections of graphs worksheet — GCSE Higher
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- 1.The graph of y = cos x is transformed onto the graph of y = cos(x − 90°). State the direction of the translation and which standard graph the image is.y = cos(x)
- 2.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
- 3.The graph of y = x³ − 5x is reflected in the y-axis. Work out the equation of the image.y = x
- 4.The graph of y = f(x) has a root (an x-intercept) at x = 5. Work out the x-coordinate of the corresponding root on the graph of y = f(x + 2).
- 5.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.
- 6.The graph of y = f(x) has x-intercepts at x = −2 and x = 6 and crosses the y-axis at (0, −12). Work out the x-intercepts and the y-intercept of y = −f(x).
- 7.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)?
- 8.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.
- 9.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.
- 10.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)
- 11.The graph of y = f(x) has a minimum turning point at (4, −5). The graph of y = f(x) + a has a minimum turning point whose minimum VALUE is 2. Work out the value of a, and state the coordinates of the minimum turning point of y = f(x) + a.
- 12.The graph of y = sin x is transformed onto the graph of y = sin x + 1. Which statement correctly describes the transformation and the new range of the graph?y = sin(x)
- 13.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.
- 14.Describe a sequence of two transformations that maps the graph of y = x² onto the graph of y = −(x − 5)².y = x²
- 15.The graph of y = x² − 4x is translated by the vector (3, 0). Work out the equation of the image, giving your answer in the form y = x² + bx + c.y = x² − 4xy = x²
Answer key
- (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.
- (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.
- (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.
- (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.
- (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.
- (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.
- (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) 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) (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).
- (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) 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.
- (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) −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.
- (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) 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.
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