A-Level Transformations of graphs Practice Questions

Free A-Level Transformations of graphs practice questions with full step-by-step worked solutions. Covers vertical translation, f(x)+a, image of a point, f(x)-a. Practise exam-style problems and check your method.

vertical translationf(x)+aimage of a pointf(x)-ahorizontal translationf(x-a)
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The point P(2, 5)P(2,\ 5) lies on the curve y=f(x)y=f(x). Write down the coordinates of the image of PP on the curve y=f(x)+3y=f(x)+3.
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Worked solution

  1. Recall the transformation rule

    y=f(x)+3y=f(x)+3

    Adding 3 to the whole function moves every point up by 3, so the y-coordinate increases by 3.

  2. Write down the original point

    P(2, 5)P(2,\ 5)

    P has coordinates (2,\ 5) and lies on the curve y=f(x).

  3. Apply the rule to the coordinates

    (2, 5) ⟶ (2, 8)(2,\ 5)\ \longrightarrow\ (2,\ 8)

    Only change the coordinate that this transformation affects; the other coordinate stays exactly the same. Remember from plotting graphs that 'up' means the y-value grows.

  4. State the image

    P′(2, 8)P'(2,\ 8)

    So this is the position of the point after the transformation.

Answer
(2, 8)(2,\ 8)
Question 2
2 markseasy
The blue curve is y=x2y=x^{2}. Select the graph of y=(x−3)2y=(x-3)^{2}.
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Worked solution

  1. Identify the base curve

    y=x2y=x^{2}

    The blue curve is the standard parabola, vertex at the origin.

  2. Apply the (x-3)

    vertex (0,0)→(3,0)\text{vertex } (0,0)\to(3,0)

    Replacing x with (x-3) slides the parabola 3 units to the right.

  3. Choose the sketch

    y=(x−3)2y=(x-3)^{2}

    The correct graph keeps the same shape with vertex at (3,0).

Answer
The parabola with vertex at (3,0)(3,0).
Question 3
3 marksintermediate
Describe fully the single transformation that maps y=cos⁡xy=\cos x onto y=cos⁡2xy=\cos 2x.
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Worked solution

  1. Locate the change

    y=cos⁡2xy=\cos 2x

    The 2 multiplies x inside the function, so it affects the input (horizontal direction).

  2. Decide stretch factor

    factor=12\text{factor}=\tfrac{1}{2}

    For f(ax) the horizontal stretch factor is 1/a, so here it is 1/2.

  3. Check the effect on period

    2π→π2\pi\to\pi

    The period halves, so the wave repeats twice as often.

  4. State the transformation

    horizontal stretch, factor 12\text{horizontal stretch, factor }\tfrac12

    So the graph is squashed towards the y-axis by scale factor 1/2.

  5. Name the transformation

    translation / stretch / reflection\text{translation / stretch / reflection}

    State clearly what type of transformation it is, because exam marks are often awarded for the correct name and full description, not just the final numbers.

  6. Is the answer reasonable?

    sense-check the size and sign\text{sense-check the size and sign}

    Step back and ask whether the size and sign of the answer make sense for the transformation described. Unreasonable values usually mean an arithmetic error.

  7. Compare with the original

    how has each feature moved?\text{how has each feature moved?}

    Compare the transformed curve feature by feature with the original: has the vertex moved, has the width changed, has it flipped? Each should match your working.

Answer
A horizontal stretch of scale factor 12\tfrac{1}{2} (parallel to the xx-axis).
Question 4
6 markshard
Describe fully a sequence of transformations that maps y=f(x)y=f(x) onto y=3−2f(x+1)y=3-2f(x+1).
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Worked solution

  1. Rewrite in standard order

    y=−2f(x+1)+3y=-2f(x+1)+3

    Rearranging makes the outside operations clearer: stretch, reflect, then shift.

  2. Inside the bracket

    f(x+1) : left 1f(x+1)\ \text{: left }1

    The +1 inside translates the graph 1 unit to the left.

  3. The factor 2

    vertical stretch, factor 2\text{vertical stretch, factor }2

    Multiplying by 2 outside doubles every height.

  4. The minus sign

    reflect in the x-axis\text{reflect in the }x\text{-axis}

    The negative sign flips the stretched graph in the x-axis.

  5. The +3

    translation (03)\text{translation }\binom{0}{3}

    Finally the +3 moves everything up 3 units.

  6. State the full sequence

    left 1, ×2, reflect x, up 3\text{left }1,\ \times2,\ \text{reflect }x,\ \text{up }3

    Do the inside shift, then stretch, then reflect, then the vertical shift.

  7. Mind the order

    inside and outside act independently\text{inside and outside act independently}

    When several transformations are combined, a change to x and a change to y can be done in either order, but two changes to the same direction may not commute. Keep track carefully.

  8. Picture the graph

    original (blue) vs image (pink)\text{original (blue) vs image (pink)}

    Sketching the original curve and its image on the same axes turns the algebra into a picture, which makes the shift, stretch or reflection obvious.

  9. Check by substituting

    put a value back in\text{put a value back in}

    Substitute a simple value or a known point back into the transformed equation to confirm it works. A quick numerical check catches slips.

  10. Link to earlier work

    completing the square / plotting\text{completing the square / plotting}

    This uses skills from earlier topics such as plotting graphs and completing the square, so the same coordinate and algebra methods apply here.

  11. Avoid the classic slip

    change the correct coordinate only\text{change the correct coordinate only}

    A very common error is to change the wrong coordinate or reflect in the wrong axis. Re-read which variable the transformation acts on before writing the answer.

Answer
Translate left 1, stretch vertically factor 2, reflect in the xx-axis, then translate up 3.
Question 5
9 markschallenging
The function is f(x)=x3−4xf(x)=x^{3}-4x. (a) Find the roots of ff. (b) Find the equation of y=f(x−1)y=f(x-1) in expanded form. (c) State the roots of y=f(x−1)y=f(x-1).
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Worked solution

  1. Set f(x)=0 for the roots

    x3−4x=0x^{3}-4x=0

    For part (a), the roots are where the curve meets the x-axis.

  2. Factor out x

    x(x2−4)=0x(x^{2}-4)=0

    Take out the common factor x.

  3. Factor the difference of squares

    x(x−2)(x+2)=0x(x-2)(x+2)=0

    x^2-4 factorises using a^2-b^2=(a-b)(a+b).

  4. Read off the roots

    x=0, 2, −2x=0,\ 2,\ -2

    A product is zero when a factor is zero; this answers part (a).

  5. Set up part (b)

    y=f(x−1)=(x−1)3−4(x−1)y=f(x-1)=(x-1)^{3}-4(x-1)

    Replace every x in f with (x-1) for the right-shift by 1.

  6. Expand the cube

    (x−1)3=x3−3x2+3x−1(x-1)^{3}=x^{3}-3x^{2}+3x-1

    Use the binomial expansion of (x-1)^3.

  7. Expand the linear part

    −4(x−1)=−4x+4-4(x-1)=-4x+4

    Multiply out the second term.

  8. Collect all terms

    y=x3−3x2+3x−1−4x+4y=x^{3}-3x^{2}+3x-1-4x+4

    Add the two expansions together.

  9. Simplify

    y=x3−3x2−x+3y=x^{3}-3x^{2}-x+3

    Combine 3x-4x=-x and -1+4=3.

  10. Set up part (c)

    shift each root right by 1\text{shift each root right by }1

    A translation right by 1 moves every root by +1.

  11. Shift the roots

    0→1, 2→3, −2→−10\to1,\ 2\to3,\ -2\to-1

    Add 1 to each original root.

  12. State the new roots

    x=1, 3, −1x=1,\ 3,\ -1

    So y=f(x-1) crosses the x-axis at these values.

  13. Give the required equation

    y=x3−3x2−x+3y=x^{3}-3x^{2}-x+3

    This expanded cubic is the main answer for part (b).

  14. Inside or outside?

    inside→x-direction, outside→y-direction\text{inside}\to x\text{-direction},\ \text{outside}\to y\text{-direction}

    Decide whether each change is inside f(...) or outside it. Inside changes move the graph horizontally (and act the opposite way to the sign); outside changes move it vertically.

  15. Read the sign for direction

    +⇒up/right, −⇒down/left+\Rightarrow\text{up/right},\ -\Rightarrow\text{down/left}

    Use the sign to fix the direction. Outside: plus is up, minus is down. Inside the bracket the movement is the opposite way round, which is easy to get wrong.

  16. Follow one key point

    track vertex / intercept\text{track vertex / intercept}

    Pick one important point, such as a turning point or an axis crossing, and follow it through the transformation. If that point lands correctly the whole curve is right.

Answer
y=x3−3x2−x+3y=x^{3}-3x^{2}-x+3

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