GCSE Transformations of graphs Practice Questions

Free GCSE Transformations of graphs practice questions with full step-by-step worked solutions. Covers vertical translation, image of a point, horizontal translation, reflection in the x-axis. Practise exam-style problems and check your method.

vertical translationimage of a pointhorizontal translationreflection in the x-axisreflection in the y-axisdescribing a transformation
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
The point (2, 5)(2,\ 5) lies on the curve y=f(x)y = f(x). Write down the coordinates of the corresponding point on the curve y=f(x)+3y = f(x) + 3.
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Worked solution

  1. Identify the transformation

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

    The +3 is written outside the function, so it changes only the yy-coordinates: the whole curve slides 3 units up. The xx-coordinate of every point stays where it is.

  2. Apply the rule to the point

    (2, 5)→(2, 5+3)(2,\ 5) \to (2,\ 5 + 3)

    Keep the xx-coordinate as 2 and add 3 to the yy-coordinate.

  3. State the image

    (2, 8)(2,\ 8)

    The point (2, 5) on y=f(x)y = f(x) maps to (2, 8). The arrow on the diagram shows the move.

Answer
(2,8)(2, 8)
Question 2
2 markseasy
The graph of y=f(x)y = f(x) has a maximum point at (1, 6)(1,\ 6). Write down the coordinates of the maximum point of the graph of y=f(x)+2y = f(x) + 2.
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Worked solution

  1. Identify the transformation

    y=f(x)+2y = f(x) + 2

    The +2+2 is outside the function: a translation 2 units up.

  2. Apply it to the maximum point

    (1, 6)→(1, 6+2)(1,\ 6) \to (1,\ 6 + 2)

    The xx-coordinate is unchanged; the yy-coordinate increases by 2. A maximum stays a maximum under a translation.

  3. State the image

    (1, 8)(1,\ 8)

    The maximum of y=f(x)+2y = f(x) + 2 is at (1, 8)(1,\ 8).

Answer
(1,8)(1, 8)
Question 3
2 marksintermediate
Jordan says, 'The graph of y=f(x+5)y = f(x + 5) is the graph of y=f(x)y = f(x) translated 55 units to the right.' Is Jordan correct?
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Worked solution

  1. Write down the claim

    y=f(x+5)=?5 units righty = f(x + 5) \stackrel{?}{=} 5 \text{ units right}

    Jordan has read the +5+5 as a move in the positive xx-direction.

  2. Test it with a point

    Suppose f(0)=7\text{Suppose } f(0) = 7

    Then the original curve passes through (0, 7)(0,\ 7).

  3. Find where that output now appears

    x+5=0⇒x=−5x + 5 = 0 \Rightarrow x = -5

    On y=f(x+5)y = f(x + 5) the output 7 appears when the input to ff is 0, i.e. at x=−5x = -5.

  4. Compare the two points

    (0, 7)→(−5, 7)(0,\ 7) \to (-5,\ 7)

    The point has moved 5 units to the LEFT, not right.

  5. State the rule properly

    f(x+a)⇒a units leftf(x + a) \Rightarrow a \text{ units left}

    Inside the bracket the direction is the opposite of the sign.

  6. Give the verdict

    translation by (−50)\text{translation by } \begin{pmatrix} -5 \\ 0 \end{pmatrix}

    Jordan is wrong: y=f(x+5)y = f(x + 5) is a translation 5 units to the left.

Answer
No. y=f(x+5)y = f(x + 5) is a translation 5 units to the LEFT, by the vector (-5, 0).
Question 4
3 markshard
The point (7, k)(7,\ k) lies on the curve y=f(x−4)y = f(x - 4). Given that f(3)=−2f(3) = -2, find the value of kk.
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Worked solution

  1. Write down the equation of the curve

    y=f(x−4)y = f(x - 4)

    The input to ff is x−4x - 4, not xx.

  2. Substitute the xx-coordinate of the point

    k=f(7−4)k = f(7 - 4)

    The point (7, k)(7,\ k) lies on the curve, so put x=7x = 7.

  3. Simplify the input

    k=f(3)k = f(3)

    The bracket gives 7−4=37 - 4 = 3.

  4. Use the given value

    f(3)=−2f(3) = -2

    This is the fact you are given.

  5. State kk

    k=−2k = -2

    So the point is (7, −2)(7,\ -2).

  6. Interpret it as a transformation

    f(x−4): 4 units RIGHTf(x - 4):\ 4 \text{ units RIGHT}

    The curve has been translated 4 units to the right.

  7. Check with the point mapping

    (3, −2)→(7, −2)(3,\ -2) \to (7,\ -2)

    The original curve passes through (3, −2)(3,\ -2); translating 4 right gives (7, −2)(7,\ -2).

  8. Note the trap

    k≠f(11)k \neq f(11)

    It is tempting to add 4 to the 7; but the rule replaces xx by x−4x - 4, so you subtract.

  9. Check the yy-coordinate is untouched

    no change outside f\text{no change outside } f

    Nothing is added outside the function, so heights are unchanged.

  10. Final answer

    k=−2k = -2

    The value of kk is −2-2.

Answer
k=−2k = -2
Question 5
6 markschallenging
f(x)=x2+4x+3f(x) = x^2 + 4x + 3. The graph of y=f(x)y = f(x) is reflected in the yy-axis and the image is then translated by the vector (0−2)\begin{pmatrix} 0 \\ -2 \end{pmatrix}. Find the equation of the resulting curve in the form y=x2+bx+cy = x^2 + bx + c.
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Worked solution

  1. Write down the function

    f(x)=x2+4x+3f(x) = x^{2} + 4 x + 3

    The given quadratic.

  2. Note the order of the transformations

    reflect first, then translate\text{reflect first, then translate}

    The question says the reflection happens first, and the IMAGE is then translated.

  3. Write the reflection in function form

    y=f(−x)y = f(-x)

    A reflection in the yy-axis replaces the input xx by −x-x.

  4. Substitute −x-x into ff

    f(−x)=(−x)2+4(−x)+3f(-x) = (-x)^2 + 4(-x) + 3

    Brackets around −x-x everywhere.

  5. Simplify the squared term

    (−x)2=x2(-x)^2 = x^2

    A negative squared is positive.

  6. Simplify the linear term

    4×(−x)=−4x4 \times (-x) = -4x

    The sign of the xx term flips.

  7. Write the reflected equation

    y=x2−4x+3y = x^{2} - 4 x + 3

    So the reflected curve is y=x2−4x+3y = x^2 - 4x + 3.

  8. Read the translation vector

    (0−2)\begin{pmatrix} 0 \\ -2 \end{pmatrix}

    No horizontal movement; 2 units down.

  9. Apply the translation

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

    Subtract 2 from the whole expression.

  10. Simplify the constant

    y=x2−4x+1y = x^{2} - 4 x + 1

    3−2=13 - 2 = 1, giving y=x2−4x+1y = x^2 - 4x + 1.

  11. Find the turning point of ff

    f(x)=(x+2)2−1⇒(−2, −1)f(x) = (x + 2)^2 - 1 \Rightarrow (-2,\ -1)

    Completing the square.

  12. Reflect the turning point

    (−2, −1)→(2, −1)(-2,\ -1) \to (2,\ -1)

    The xx-coordinate changes sign.

  13. Translate the turning point

    (2, −1)→(2, −3)(2,\ -1) \to (2,\ -3)

    Two units down.

  14. Check against the answer

    y=x2−4x+1=(x−2)2−3y = x^2 - 4x + 1 = (x - 2)^2 - 3

    Completing the square on the final equation gives the minimum (2, −3)(2,\ -3) — it matches.

  15. State the equation

    y=x2−4x+1y = x^{2} - 4 x + 1

    Reflecting ff in the yy-axis and lowering it 2 gives y=x2−4x+1y = x^2 - 4x + 1.

Answer
y=x2−4x+1y = x^2 - 4x + 1

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