Hard GCSE Transformations of graphs Questions

Challenging, exam-style GCSE Transformations of graphs questions with worked solutions. Stretch yourself on the hardest horizontal translation, expanding brackets, roots check, reflection in the x-axis problems.

horizontal translationexpanding bracketsroots checkreflection in the x-axisnegating every termreflection in the y-axis
GCSE Higher34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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
Question 2
6 markschallenging
The curve y=f(x−1)−4y = f(x - 1) - 4 has a minimum point at (5, −2)(5,\ -2). Find the coordinates of the minimum point of the curve y=f(x)y = f(x).
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Worked solution

  1. Write down what you are told

    min of y=f(x−1)−4 is (5, −2)\text{min of } y = f(x - 1) - 4 \text{ is } (5,\ -2)

    This is the turning point of the TRANSFORMED curve — you have to work backwards.

  2. Split the transformation

    y=f(x−1)−4y = f(x - 1) - 4

    Inside: x−1x - 1. Outside: −4-4.

  3. State the forward moves

    1 unit RIGHT,4 units DOWN1 \text{ unit RIGHT},\quad 4 \text{ units DOWN}

    These are the moves that take y=f(x)y = f(x) onto the given curve.

  4. Decide what to do

    reverse both moves\text{reverse both moves}

    To get back to y=f(x)y = f(x), undo them: 1 unit LEFT and 4 units UP.

  5. Undo the horizontal move

    5→5−1=45 \to 5 - 1 = 4

    Move the xx-coordinate back 1 unit.

  6. Undo the vertical move

    −2→−2+4=2-2 \to -2 + 4 = 2

    Move the yy-coordinate back up 4 units.

  7. State the turning point of ff

    (4, 2)(4,\ 2)

    The minimum of y=f(x)y = f(x) is at (4, 2)(4,\ 2).

  8. Check it forwards: horizontal

    4→4+1=54 \to 4 + 1 = 5

    Translating (4, 2)(4,\ 2) one unit right gives x=5x = 5.

  9. Check it forwards: vertical

    2→2−4=−22 \to 2 - 4 = -2

    Then four units down gives y=−2y = -2.

  10. Compare with the given point

    (5, −2)(5,\ -2)

    It matches, so the answer is right.

  11. Check with the function rule

    x=5: y=f(5−1)−4=f(4)−4x = 5:\ y = f(5 - 1) - 4 = f(4) - 4

    Substituting x=5x = 5 into the transformed equation.

  12. Use the turning point value

    f(4)=2f(4) = 2

    From the answer above.

  13. Finish the check

    y=2−4=−2y = 2 - 4 = -2

    Exactly the given minimum height.

  14. Note the nature of the point

    minimum→minimum\text{minimum} \to \text{minimum}

    Translations never turn a curve upside down, so (4, 2)(4,\ 2) is a minimum too.

  15. State the answer

    (4, 2)(4,\ 2)

    The minimum of y=f(x)y = f(x) is at (4, 2)(4,\ 2).

Answer
(4,2)(4, 2)
Question 3
5 markschallenging
Which points on the graph of y=f(x)y = f(x) are invariant (they stay exactly where they are) when the graph is transformed to y=f(−x)y = f(-x)?
Show worked solution

Worked solution

  1. Write down the rule for the transformation

    (x, y)→(−x, y)(x,\ y) \to (-x,\ y)

    y=f(−x)y = f(-x) is a reflection in the yy-axis.

  2. Say what "invariant" means

    (x, y) maps to itself(x,\ y) \text{ maps to itself}

    An invariant point ends up exactly where it started.

  3. Set up the condition

    (−x, y)=(x, y)(-x,\ y) = (x,\ y)

    The image must equal the original point.

  4. Compare the yy-coordinates

    y=yy = y

    The heights already match — the reflection never changes them, so this gives no information.

  5. Compare the xx-coordinates

    −x=x-x = x

    This is the condition that really matters.

  6. Solve the equation

    2x=02x = 0

    Add xx to both sides.

  7. Finish

    x=0x = 0

    The only xx-coordinate that survives the reflection unchanged.

  8. Interpret the answer

    x=0 is the y-axisx = 0 \text{ is the } y\text{-axis}

    So exactly the points of the curve on the yy-axis are invariant.

  9. Which point is that in practice?

    (0, f(0))(0,\ f(0))

    For a normal function there is just one: the yy-intercept.

  10. Check with an example

    f(x)=x2+3x+1, f(0)=1f(x) = x^2 + 3x + 1,\ f(0) = 1

    This curve passes through (0, 1)(0,\ 1).

  11. Reflect the example

    f(−x)=x2−3x+1f(-x) = x^2 - 3x + 1

    Substituting −x-x into the rule.

  12. Check the yy-intercept of the image

    x=0: y=1x = 0:\ y = 1

    The image still passes through (0, 1)(0,\ 1): invariant, as predicted.

  13. Check a point NOT on the yy-axis

    (1, 5)→(−1, 5)(1,\ 5) \to (-1,\ 5)

    For the same example f(1)=5f(1) = 5, and the image passes through (−1, 5)(-1,\ 5) instead — so it has moved.

  14. Contrast with y=−f(x)y = -f(x)

    there the x-axis points are invariant\text{there the } x\text{-axis points are invariant}

    Do not mix the two up: y=−f(x)y = -f(x) fixes the roots; y=f(−x)y = f(-x) fixes the yy-intercept.

  15. State the answer

    x=0x = 0

    The invariant points are the points on the yy-axis — for a function, its yy-intercept (0, f(0))(0,\ f(0)).

Answer
The points on the y-axis, i.e. where x=0x = 0 - for a function this is the y-intercept (0, f(0)).
Question 4
5 markschallenging
Tia says, 'Reflecting the graph of y=f(x)y = f(x) in the xx-axis and then reflecting the result in the yy-axis gives the graph of y=f(−x)y = f(-x).' Is Tia correct?
Show worked solution

Worked solution

  1. Write down the claim

    reflect in x-axis, then y-axis=?f(−x)\text{reflect in } x\text{-axis, then } y\text{-axis} \stackrel{?}{=} f(-x)

    Tia thinks the two reflections give y=f(−x)y = f(-x).

  2. Do the first reflection

    y=−f(x)y = -f(x)

    Reflecting y=f(x)y = f(x) in the xx-axis changes the sign of every output.

  3. Name the new function

    g(x)=−f(x)g(x) = -f(x)

    Call the reflected curve gg so the second step is clear.

  4. Do the second reflection

    y=g(−x)y = g(-x)

    Reflecting y=g(x)y = g(x) in the yy-axis replaces xx by −x-x.

  5. Substitute for gg

    g(−x)=−f(−x)g(-x) = -f(-x)

    The minus sign outside stays; the input becomes −x-x.

  6. Write the final equation

    y=−f(−x)y = -f(-x)

    BOTH signs have changed, not just one.

  7. Compare with Tia's answer

    −f(−x)≠f(−x)-f(-x) \neq f(-x)

    Tia has lost the minus sign that came from the reflection in the xx-axis.

  8. Test with a point

    Suppose f(3)=2, so (3, 2) is on the curve\text{Suppose } f(3) = 2 \text{, so } (3,\ 2) \text{ is on the curve}

    A concrete point makes the difference obvious.

  9. Reflect it in the xx-axis

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

    The height changes sign.

  10. Reflect that in the yy-axis

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

    Now the xx-coordinate changes sign.

  11. What does y=f(−x)y = f(-x) give?

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

    The height would be +2+2, not −2-2: a different point.

  12. What does y=−f(−x)y = -f(-x) give?

    x=−3: y=−f(3)=−2x = -3:\ y = -f(3) = -2

    This passes through (−3, −2)(-3,\ -2), matching the double reflection.

  13. Name the combined transformation

    rotation of 180∘ about the origin\text{rotation of } 180^\circ \text{ about the origin}

    Changing the sign of both coordinates is the same as a half-turn about the origin.

  14. Say what Tia should have written

    y=−f(−x)y = -f(-x)

    The correct equation.

  15. Give the verdict

    Tia is wrong\text{Tia is wrong}

    The result is y=−f(−x)y = -f(-x) — equivalent to rotating the graph 180∘180^\circ about the origin.

Answer
No. Both reflections change a sign, so the result is y=−f(−x)y = -f(-x) - the same as rotating the graph 180 degrees about the origin.
Question 5
5 markschallenging
f(x)=x2−1f(x) = x^2 - 1. The curve CC has equation y=−f(x−2)+1y = -f(x - 2) + 1. Find the equation of CC in the form y=ax2+bx+cy = ax^2 + bx + c.
Show worked solution

Worked solution

  1. Write down the function

    f(x)=x2−1f(x) = x^{2} - 1

    A parabola with minimum (0, −1)(0,\ -1).

  2. List the three ingredients

    y=−f(x−2)+1y = -f(x - 2) + 1

    Inside: x−2x - 2. Outside: the minus sign, then +1+1.

  3. Substitute (x−2)(x - 2) into ff

    f(x−2)=(x−2)2−1f(x - 2) = (x - 2)^2 - 1

    Replace xx by the bracket.

  4. Expand the square

    (x−2)2=x2−4x+4(x - 2)^2 = x^2 - 4x + 4

    Multiply out.

  5. Write f(x−2)f(x - 2) in expanded form

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

    x2−4x+4−1=x2−4x+3x^2 - 4x + 4 - 1 = x^2 - 4x + 3.

  6. Apply the minus sign

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

    Every term changes sign.

  7. Add the 1

    y=−x2+4x−2y = - x^{2} + 4 x - 2

    −3+1=−2-3 + 1 = -2.

  8. Track the vertex: start

    (0, −1)(0,\ -1)

    The minimum of ff.

  9. Track the vertex: 2 units right

    (2, −1)(2,\ -1)

    From the −2-2 inside the bracket.

  10. Track the vertex: reflect

    (2, 1)(2,\ 1)

    The height changes sign.

  11. Track the vertex: 1 unit up

    (2, 2)(2,\ 2)

    The +1+1 raises it.

  12. Check against the equation

    y=−(x−2)2+2y = -(x - 2)^2 + 2

    Completing the square on −x2+4x−2-x^2 + 4x - 2 gives a maximum at (2, 2)(2,\ 2) — it matches.

  13. Check a value of xx

    x=0: −f(−2)+1=−(4−1)+1=−2x = 0:\ -f(-2) + 1 = -(4 - 1) + 1 = -2

    Using the transformation route.

  14. Check the same value in the equation

    −0+0−2=−2-0 + 0 - 2 = -2

    Both give −2-2, so the algebra is right.

  15. State the equation of CC

    y=−x2+4x−2y = - x^{2} + 4 x - 2

    An n-shaped parabola with a maximum at (2, 2)(2,\ 2).

Answer
y=−x2+4x−2y = -x^2 + 4x - 2

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