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Functions and transformations

One-to-one and many-to-one mappings, square-root and modulus functions, af(x), f(x)+a, f(x+a), f(ax) and compositions.

Spec MM1.7, MM8.2, MM8.3, MM8.4Section 1 Part 1: AS-level pure mathematics18 questions11 real past questions
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1Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = x^2\) is translated by \(3\) units in the positive \(x\)-direction and then stretched parallel to the \(y\)-axis with scale factor \(2\). What is the equation of the resulting graph?

  1. A\(y = 2(x-3)^2\)
  2. B\(y = (x-3)^2 + 2\)
  3. C\(y = 2(x+3)^2\)
  4. D\(y = 2x^2 - 3\)
  5. E\(y = (2x-3)^2\)
  6. F\(y = \frac12 (x-3)^2\)
Worked solution

Translation: \(y = (x-3)^2\). Stretch by factor \(2\) in the \(y\)-direction: \(y = 2(x-3)^2\).

Full method — how and why
Two transformations applied in a stated order. Apply them one at a time to the equation, in the order given.
  1. Translate \(3\) in the positive \(x\)-direction: replace \(x\) by \(x - 3\), giving \(y = (x - 3)^2\).
    A shift right by \(a\) replaces \(x\) with \(x - a\) - the sign is opposite to what it looks like, because the new graph reaches a given height \(3\) later.
  2. Stretch parallel to the \(y\)-axis with factor \(2\): multiply the whole function by \(2\), giving \(y = 2(x-3)^2\).
    A vertical stretch multiplies every \(y\)-value, so it acts on the outside of the function.
Where it goes wrong
  • Writing \((x+3)^2\) - shifting right uses \(x - 3\).
  • Putting the \(2\) inside as \((2x - 3)^2\), which is a horizontal stretch, not a vertical one.
Takeaway Changes inside the function affect \(x\) and act in the opposite sense; changes outside affect \(y\) and act as they read.
Answer A · Functions and transformations (MM8.2) · question FUN-01
2Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

\(f(x) = x^2 - 4x\). The graph of \(y = f(x + 2)\) has its minimum point at

  1. A\((0, -4)\)
  2. B\((2, -4)\)
  3. C\((4, -4)\)
  4. D\((2, 0)\)
  5. E\((-2, -4)\)
  6. F\((0, 0)\)
Worked solution

\(f(x) = (x-2)^2 - 4\) has minimum at \((2, -4)\). Replacing \(x\) by \(x + 2\) translates the graph \(2\) units to the left: \((0, -4)\).

Full method — how and why
Find the minimum of \(f\) first, then move it. Tracking the single turning point is quicker than re-completing the square.
  1. Complete the square: \(f(x) = x^2 - 4x = (x-2)^2 - 4\), so \(f\) has its minimum at \((2, -4)\).
    Half of \(-4\) is \(-2\); the \(-4\) outside is the minimum value.
  2. \(y = f(x + 2)\) translates the graph \(2\) units to the left.
    Replacing \(x\) by \(x + a\) shifts left by \(a\): the graph now reaches each value \(2\) earlier.
  3. Move the turning point: \((2, -4) \to (0, -4)\).
    Only the \(x\)-coordinate changes; a horizontal translation never alters \(y\).
Where it goes wrong
  • Shifting right and answering \((4, -4)\).
  • Changing the \(y\)-coordinate as well, giving \((0, -2)\) or similar.
Takeaway \(f(x + a)\) moves the graph \(a\) to the left; horizontal shifts leave \(y\)-values untouched.
Answer A · Functions and transformations (MM8.2) · question FUN-02
3Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

Let \(f(x) = 2x + 1\) and \(g(x) = x^2 - 3\). Find \(f(g(2)) - g(f(2))\).

  1. A\(-3\)
  2. B\(-17\)
  3. C\(-19\)
  4. D\(3\)
  5. E\(25\)
  6. F\(19\)
Worked solution

\(g(2) = 1\), \(f(1) = 3\). \(f(2) = 5\), \(g(5) = 22\). So \(3 - 22 = -19\).

Full method — how and why
Composite functions evaluated at a number. Work strictly from the inside out, and keep the two compositions separate.
  1. \(g(2) = 2^2 - 3 = 1\), then \(f(1) = 2(1) + 1 = 3\). So \(f(g(2)) = 3\).
    In \(f(g(2))\) the inner function acts first: feed \(2\) into \(g\), then its output into \(f\).
  2. \(f(2) = 2(2) + 1 = 5\), then \(g(5) = 25 - 3 = 22\). So \(g(f(2)) = 22\).
    The other order gives a completely different value - composition is not commutative.
  3. Subtract in the order asked: \(3 - 22 = -19\).
    The question is \(f(g(2)) - g(f(2))\), so the small one comes first.
Where it goes wrong
  • Computing \(g(f(2)) - f(g(2))\) and answering \(19\).
  • Applying the functions in the wrong order inside one composition.
Takeaway \(f(g(x))\) means \(g\) first; check the order twice, because the wrong order is always an offered option.
Answer C · Functions and transformations (MM8.2) · question FUN-03
4Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = f(x)\) passes through \((4, 6)\). Which point must lie on the graph of \(y = 3f(2x) - 1\)?

  1. A\((2, 11)\)
  2. B\((2, 17)\)
  3. C\((8, 17)\)
  4. D\((8, 18)\)
  5. E\((4, 17)\)
  6. F\((2, 18)\)
Worked solution

\(f(2x)\) halves \(x\)-coordinates: \((2, 6)\). Then \(3f - 1\): \(y = 18 - 1 = 17\). Point \((2, 17)\).

Full method — how and why
Track a single known point through the transformations, remembering that inside changes act inversely on \(x\).
  1. Start with the known point \((4, 6)\) on \(y = f(x)\).
    One point is all you are given, so the answer must be its image.
  2. \(f(2x)\) squashes horizontally by factor \(\frac12\): the point with \(x\)-coordinate \(4\) moves to \(x = 2\), since \(f(2 \times 2) = f(4) = 6\).
    To get the same output from \(f(2x)\) you need half the input - this is the inverse action of an inside change.
  3. Multiply by \(3\): \(y = 3 \times 6 = 18\). Then subtract \(1\): \(y = 17\).
    Outside changes act directly on the output, in the order written.
  4. The point is \((2, 17)\).
    Check: \(3f(2 \times 2) - 1 = 3(6) - 1 = 17\) ✓.
Where it goes wrong
  • Doubling the \(x\)-coordinate to \(8\) instead of halving it.
  • Forgetting the \(-1\) and answering \((2, 18)\).
Takeaway Substitute the point back into the new equation to check: \(f(2x)\) at \(x = 2\) really is \(f(4)\).
Answer B · Functions and transformations (MM8.2) · question FUN-05
5Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The curve \(y = x^2 - 6x + 5\) is reflected in the \(y\)-axis. Find the equation of the reflected curve.

  1. A\(y = x^2 + 6x + 5\)
  2. B\(y = x^2 + 6x - 5\)
  3. C\(y = -x^2 - 6x + 5\)
  4. D\(y = -x^2 + 6x - 5\)
  5. E\(y = x^2 - 6x - 5\)
Worked solution

Reflecting in the \(y\)-axis replaces \(x\) with \(-x\): \(y = x^2 + 6x + 5\).

Full method — how and why
Reflection in the \(y\)-axis replaces \(x\) by \(-x\). Only the odd powers change sign.
  1. Replace \(x\) with \(-x\): \(y = (-x)^2 - 6(-x) + 5\).
    Reflecting in the \(y\)-axis maps the point \((x, y)\) to \((-x, y)\), which is achieved by substituting \(-x\).
  2. Simplify: \((-x)^2 = x^2\) and \(-6(-x) = +6x\), so \(y = x^2 + 6x + 5\).
    Even powers are unaffected; odd powers flip sign. The constant is also unchanged.
Where it goes wrong
  • Changing the sign of the constant too, giving \(x^2 + 6x - 5\).
  • Confusing it with reflection in the \(x\)-axis, which would negate the whole right-hand side.
Takeaway \(y\)-axis reflection: \(x \to -x\), so only odd-degree terms change sign. \(x\)-axis reflection: \(y \to -y\), so everything does.
Answer A · Functions and transformations (MM8.2) · question FUN-06
6Functions and transformations · spec MM1.7 · Paper 1 style · ★☆☆

For which values of \(x\) is \(\sqrt{x^2} = -x\)?

  1. A\(x = 0\) only
  2. B\(x \geq 0\)
  3. C\(x \leq 0\)
  4. D\(x < 0\) only
  5. Eall real \(x\)
  6. Fno real \(x\)
Worked solution

\(\sqrt{x^2} = |x|\), and \(|x| = -x\) exactly when \(x \leq 0\).

Full method — how and why
The key fact is \(\sqrt{x^2} = |x|\), not \(x\). The question is then about when \(|x| = -x\).
  1. Rewrite the left-hand side: \(\sqrt{x^2} = |x|\).
    The square-root symbol always means the non-negative root, so the output cannot be negative even when \(x\) is.
  2. The equation becomes \(|x| = -x\).
    Both sides must therefore be \(\geq 0\), which already forces \(-x \geq 0\).
  3. \(|x| = -x\) exactly when \(x \leq 0\) (for \(x < 0\), \(|x| = -x\); at \(x = 0\) both sides are \(0\)).
    Split into cases \(x \geq 0\) and \(x < 0\), or recall the definition of \(|x|\) directly.
Where it goes wrong
  • Answering 'no real \(x\)' on the grounds that a square root cannot equal a negative - but \(-x\) is positive when \(x\) is negative.
  • Excluding \(x = 0\) and answering \(x < 0\) only.
Takeaway \(\sqrt{x^2} = |x|\); and \(-x\) is positive when \(x\) is negative, so expressions with a minus sign can still be non-negative.
Answer C · Functions and transformations (MM1.7) · question FUN-08
7Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = \sin x\) is transformed to \(y = 2\sin(3x) + 1\). What is the maximum value of the new function, and what is its period (in degrees)?

  1. Amaximum \(1\), period \(120^\circ\)
  2. Bmaximum \(3\), period \(360^\circ\)
  3. Cmaximum \(2\), period \(120^\circ\)
  4. Dmaximum \(3\), period \(120^\circ\)
  5. Emaximum \(3\), period \(1080^\circ\)
  6. Fmaximum \(2\), period \(180^\circ\)
Worked solution

\(\sin(3x)\) has period \(360^\circ/3 = 120^\circ\). \(2\sin(3x)\) ranges from \(-2\) to \(2\), so \(2\sin(3x) + 1\) has maximum \(3\).

Full method — how and why
For \(y = a\sin(bx) + c\), the amplitude, period and shift can be read straight off - as long as you know which letter does what.
  1. The \(3\) inside compresses horizontally: period \(= \dfrac{360^\circ}{3} = 120^\circ\).
    \(\sin(bx)\) completes a cycle when \(bx\) increases by \(360^\circ\), i.e. when \(x\) increases by \(360^\circ/b\).
  2. The \(2\) outside is the amplitude: \(2\sin(3x)\) ranges from \(-2\) to \(2\).
    Multiplying the function stretches the range by that factor, symmetrically about zero.
  3. The \(+1\) raises everything: the range becomes \(-1\) to \(3\), so the maximum is \(3\).
    A vertical translation moves the whole wave, including its maximum.
Where it goes wrong
  • Multiplying the period by \(3\) to get \(1080^\circ\) instead of dividing.
  • Giving the amplitude \(2\) as the maximum and ignoring the \(+1\).
Takeaway In \(a\sin(bx) + c\): amplitude \(|a|\), period \(360^\circ/b\), midline \(y = c\), maximum \(c + |a|\).
Answer D · Functions and transformations (MM8.2) · question FUN-09
8Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = f(x)\) is transformed by the sequence: reflect in the \(x\)-axis, then translate \(2\) units in the positive \(y\)-direction, then reflect in the \(y\)-axis. The resulting graph has equation

  1. A\(y = f(-x) - 2\)
  2. B\(y = 2 - f(x)\)
  3. C\(y = -f(-x + 2)\)
  4. D\(y = 2 - f(-x)\)
  5. E\(y = -f(x - 2)\)
  6. F\(y = -f(-x) - 2\)
Worked solution

Reflect in \(x\)-axis: \(y = -f(x)\). Translate up \(2\): \(y = -f(x) + 2\). Reflect in \(y\)-axis (replace \(x\) by \(-x\)): \(y = 2 - f(-x)\).

Full method — how and why
Three transformations in a stated order. Apply them to the equation one at a time, and be careful that the last one acts on everything built so far.
  1. Reflect in the \(x\)-axis: \(y = -f(x)\).
    Negating the output flips the graph vertically.
  2. Translate \(2\) up: \(y = -f(x) + 2\).
    Add to the whole expression, not just to \(f\).
  3. Reflect in the \(y\)-axis: replace every \(x\) by \(-x\), giving \(y = -f(-x) + 2 = 2 - f(-x)\).
    Only \(x\) is replaced; the constant \(2\) is unaffected because it does not involve \(x\).
Where it goes wrong
  • Writing \(-f(-x) - 2\) by also negating the constant.
  • Applying the transformations in the wrong order - with mixed inside/outside operations, order matters.
Takeaway Rewrite the equation after each step; a \(y\)-axis reflection substitutes \(-x\) for \(x\) everywhere it appears.
Answer D · Functions and transformations (MM8.2) · question FUN-11
9Functions and transformations · spec MM8.4 · Paper 1 style · ★☆☆

The graph of \(y = a(x + b)^2 + c\) has vertex \((-2, 5)\) and passes through \((0, -3)\). Find \(a\).

  1. A\(-2\)
  2. B\(2\)
  3. C\(-4\)
  4. D\(-1\)
  5. E\(1\)
  6. F\(-\dfrac12\)
Worked solution

\(b = 2\), \(c = 5\). At \(x = 0\): \(4a + 5 = -3\), so \(a = -2\).

Full method — how and why
The completed-square form is handed to you, so the vertex gives two constants and the extra point gives the third.
  1. Match the vertex: \(y = a(x + b)^2 + c\) has vertex \((-b, c)\), and we are told it is \((-2, 5)\), so \(b = 2\) and \(c = 5\).
    Note the sign: the form uses \(x + b\), so the vertex \(x\)-coordinate is \(-b\), not \(b\).
  2. Substitute the point \((0, -3)\): \(a(0 + 2)^2 + 5 = -3\).
    One point is enough to find the remaining unknown.
  3. Solve: \(4a = -8\), so \(a = -2\).
    Negative, which makes sense: the curve passes below its vertex, so it opens downwards.
Where it goes wrong
  • Taking \(b = -2\) and getting \(a = -\frac12\) from \(a(0-2)^2\)... (the same here, but the sign confusion bites when the point is not symmetric).
  • Answering \(+2\) and ignoring that the parabola must open downwards to reach \(y = -3\).
Takeaway Sanity-check the sign of \(a\) against the picture: if the curve drops below the vertex, \(a < 0\).
Answer A · Functions and transformations (MM8.4) · question FUN-13
10Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = f(x)\) has a maximum point at \((3, 8)\). The graph of \(y = f(x - a) + b\) has its maximum at \((1, 2)\). Find \(a + b\).

  1. A\(8\)
  2. B\(-2\)
  3. C\(-4\)
  4. D\(-8\)
  5. E\(4\)
  6. F\(0\)
Worked solution

The maximum moves from \((3, 8)\) to \((3 + a, 8 + b) = (1, 2)\), so \(a = -2\), \(b = -6\) and \(a + b = -8\).

Full method — how and why
Follow the turning point through the transformation and match coordinates - each coordinate gives one equation.
  1. \(y = f(x - a) + b\) translates the graph \(a\) right and \(b\) up, so the maximum moves from \((3, 8)\) to \((3 + a,\; 8 + b)\).
    A translation moves every point, including the turning point, by the same vector.
  2. Match the \(x\)-coordinates: \(3 + a = 1\), so \(a = -2\).
    A negative \(a\) simply means the shift is to the left.
  3. Match the \(y\)-coordinates: \(8 + b = 2\), so \(b = -6\).
    Same for the vertical shift.
  4. Add: \(a + b = -8\).
    The question asks for the sum, not for either value alone.
Where it goes wrong
  • Using \(3 - a = 1\) and getting \(a = 2\), which flips the sign of the answer.
  • Answering \(-2\) or \(-6\) (one of the two constants) instead of their sum.
Takeaway \(f(x - a) + b\) shifts by \((+a, +b)\); set up one equation per coordinate.
Answer D · Functions and transformations (MM8.2) · question FUN-14
11Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

Which of the following is the equation of the graph obtained by stretching \(y = x^3 - 3x\) parallel to the \(x\)-axis with scale factor \(\dfrac{1}{2}\)?

  1. A\(y = \dfrac{1}{2}x^3 - \dfrac{3}{2}x\)
  2. B\(y = 8x^3 - 3x\)
  3. C\(y = 2x^3 - 6x\)
  4. D\(y = \dfrac{x^3}{8} - \dfrac{3x}{2}\)
  5. E\(y = 8x^3 - 6x\)
Worked solution

A stretch parallel to the \(x\)-axis with factor \(\frac12\) gives \(y = f(2x) = (2x)^3 - 3(2x) = 8x^3 - 6x\).

Full method — how and why
A horizontal stretch of factor \(\frac12\) replaces \(x\) by \(2x\) - the reciprocal of the factor.
  1. A stretch parallel to the \(x\)-axis with scale factor \(k\) gives \(y = f\!\left(\frac{x}{k}\right)\); with \(k = \frac12\) that is \(y = f(2x)\).
    Inside changes act inversely: to squash the graph to half its width you must double the input.
  2. Substitute: \(y = (2x)^3 - 3(2x)\).
    Replace every \(x\) in the formula, including the one in the linear term.
  3. Simplify: \(8x^3 - 6x\).
    \((2x)^3 = 8x^3\) - the coefficient is cubed too.
Where it goes wrong
  • Replacing only the first \(x\), giving \(8x^3 - 3x\) - an offered option.
  • Using \(y = f(x/2)\) and getting \(\frac{x^3}{8} - \frac{3x}{2}\), which is a stretch of factor \(2\).
Takeaway Stretch factor \(k\) parallel to the \(x\)-axis \(\Rightarrow\) replace \(x\) by \(x/k\); substitute into every occurrence.
Answer E · Functions and transformations (MM8.2) · question FUN-15
12Functions and transformations · spec MM8.3 · Paper 1 style · ★☆☆

The line \(y = mx + c\) passes through \((2, 7)\). If \(m\) is increased by \(1\) and \(c\) is decreased by \(2\), the new line passes through \((2, k)\). Find \(k\).

  1. A\(6\)
  2. B\(7\)
  3. C\(5\)
  4. D\(9\)
  5. E\(8\)
  6. F\(10\)
Worked solution

New value at \(x = 2\): \((m+1)\cdot 2 + (c - 2) = 2m + c = 7\). So \(k = 7\).

Full method — how and why
The two changes are designed to cancel at \(x = 2\). Substitute and watch the terms collapse.
  1. From the original line: \(2m + c = 7\).
    The line passes through \((2, 7)\), so substituting gives one equation in \(m\) and \(c\).
  2. The new line is \(y = (m + 1)x + (c - 2)\); evaluate it at \(x = 2\): \(2(m+1) + (c - 2) = 2m + 2 + c - 2\).
    Substitute \(x = 2\) and expand - do not try to find \(m\) and \(c\), which are not determined.
  3. Simplify: \(2m + c = 7\), so \(k = 7\).
    The \(+2\) from the gradient change and the \(-2\) from the intercept change cancel exactly.
Where it goes wrong
  • Trying to solve for \(m\) and \(c\) individually - there is not enough information, and none is needed.
  • Forgetting that the gradient change is multiplied by \(x = 2\), and answering \(5\) or \(9\).
Takeaway When a question gives only one equation for two unknowns, look for a combination that is determined - here \(2m + c\).
Answer B · Functions and transformations (MM8.3) · question FUN-17
13Functions and transformations · spec MM1.7 · Paper 1 style · ★★☆

Which of the following functions, each with domain the set of all real numbers, is one-to-one?

  1. A\(f(x) = x^3 - x\)
  2. B\(f(x) = x^2 + 2x\)
  3. C\(f(x) = \cos x\)
  4. D\(f(x) = x^3 + x\)
  5. E\(f(x) = |x|\)
Worked solution

\(x^3 + x\) is strictly increasing (its derivative \(3x^2 + 1 > 0\)), so different inputs give different outputs. \(x^3 - x\) takes the value \(0\) at \(x = -1, 0, 1\); the others are even or periodic.

Full method — how and why
'One-to-one' means no horizontal line meets the graph twice. The quickest test is whether the function is always increasing (or always decreasing).
  1. Rule out the even and periodic functions: \(|x|\) and \(x^2 + 2x\) are symmetric (e.g. \(|1| = |-1|\)), and \(\cos x\) repeats every \(2\pi\).
    Any symmetry or repetition immediately produces two inputs with the same output.
  2. Test \(x^3 - x\): it equals \(0\) at \(x = -1, 0, 1\) - three inputs with the same output.
    A cubic with three distinct roots is certainly not one-to-one.
  3. Test \(x^3 + x\): its derivative \(3x^2 + 1\) is always positive, so the function is strictly increasing.
    A strictly increasing function can never take the same value twice, which is exactly one-to-one.
  4. So \(f(x) = x^3 + x\) is the one-to-one function.
    Alternatively, \(x^3\) and \(x\) both increase, so their sum does.
Where it goes wrong
  • Assuming every cubic is one-to-one - \(x^3 - x\) is a counterexample.
  • Thinking \(|x|\) is one-to-one because it has a 'simple' formula.
Takeaway One-to-one \(\Leftrightarrow\) strictly monotonic; check for symmetry or repeated values first, then look at the derivative's sign.
Answer D · Functions and transformations (MM1.7) · question FUN-04
14Functions and transformations · spec MM8.2 · Paper 1 style · ★★☆

The function \(f\) satisfies \(f(x) = x^2\) for all \(x\). Which of the following transformations maps \(y = f(x)\) onto \(y = 4x^2 - 8x + 4\)?

  1. Aa translation of \(1\) unit in the positive \(x\)-direction followed by a stretch parallel to the \(y\)-axis with factor \(4\)
  2. Ba translation of \(2\) units in the positive \(x\)-direction followed by a stretch parallel to the \(y\)-axis with factor \(2\)
  3. Ca translation of \(1\) unit in the negative \(x\)-direction followed by a stretch parallel to the \(y\)-axis with factor \(4\)
  4. Da stretch parallel to the \(y\)-axis with factor \(4\) followed by a translation of \(4\) units in the positive \(y\)-direction
  5. Ea stretch parallel to the \(x\)-axis with factor \(2\) followed by a translation of \(2\) units in the positive \(x\)-direction
Worked solution

\(4x^2 - 8x + 4 = 4(x-1)^2\), i.e. \(y = 4f(x - 1)\): translate \(1\) right, then stretch vertically by \(4\). (Equivalently \(y = f(2x - 2) = f(2(x-1))\): stretch parallel to the \(x\)-axis by factor \(\frac12\) then translate 1 right - not offered.)

Full method — how and why
Work backwards: put the target into a form that shows the transformations. Completing the square reveals them.
  1. Factorise the target: \(4x^2 - 8x + 4 = 4(x^2 - 2x + 1) = 4(x-1)^2\).
    Taking out the \(4\) exposes a perfect square - the shape of \(f\) with a translation.
  2. Read it as \(y = 4f(x - 1)\) where \(f(x) = x^2\).
    The inside \((x-1)\) is a translation, the outside \(4\) is a vertical stretch.
  3. Translate \(1\) unit in the positive \(x\)-direction, then stretch parallel to the \(y\)-axis by factor \(4\).
    With one inside and one outside change, the order between them does not matter - but the direction of each does.
  4. Confirm with a point: the vertex \((0,0)\) maps to \((1, 0)\), and the target \(4(x-1)^2\) does have its vertex at \((1, 0)\) ✓.
    A point check catches direction errors instantly.
Where it goes wrong
  • Translating left (\(x + 1\)), giving \(4(x+1)^2 = 4x^2 + 8x + 4\) - the wrong sign on the middle term.
  • Reading the \(4\) as a horizontal stretch; \(y = f(2x) = 4x^2\) has no translation at all.
Takeaway Factor the target into \(af(x - h) + k\) form; the numbers then name the transformations directly.
Answer A · Functions and transformations (MM8.2) · question FUN-07
15Functions and transformations · spec MM8.2 · Paper 2 style · ★★☆

\(f(x) = x^2 + 1\). Which of the following statements is/are true?

  1. I\(f(f(x)) = x^4 + 2x^2 + 2\) for all \(x\)
  2. II\(f(x + 1) = f(x) + 1\) for all \(x\)
  3. III\(f(2x) = 4f(x) - 3\) for all \(x\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: \(f(f(x)) = (x^2+1)^2 + 1 = x^4 + 2x^2 + 2\): true. II: \(f(x+1) = x^2 + 2x + 2 \neq x^2 + 2\): false. III: \(f(2x) = 4x^2 + 1\) and \(4f(x) - 3 = 4x^2 + 1\): true.

Full method — how and why
Three algebraic claims about one function. Each is settled by direct substitution - and the false one needs only a single value of \(x\).
  1. I: \(f(f(x)) = (x^2+1)^2 + 1 = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2\). True.
    Substitute the whole of \(f(x)\) into \(f\); expand \((x^2+1)^2\) carefully - the cross term \(2x^2\) is essential.
  2. II: \(f(x+1) = (x+1)^2 + 1 = x^2 + 2x + 2\), whereas \(f(x) + 1 = x^2 + 2\). These differ by \(2x\). False.
    A single value settles it: \(x = 1\) gives \(f(2) = 5\) but \(f(1) + 1 = 3\).
  3. III: \(f(2x) = 4x^2 + 1\) and \(4f(x) - 3 = 4x^2 + 4 - 3 = 4x^2 + 1\). Equal for all \(x\). True.
    Compute both sides independently and compare - do not try to 'see' it.
  4. I and III only.
    Map the verdicts onto the option list.
Where it goes wrong
  • Assuming II is true because functions 'usually' behave additively - almost none do.
  • Expanding \((x^2+1)^2\) as \(x^4 + 1\).
Takeaway Test each claim by substitution; one numerical counterexample is enough to destroy a 'for all \(x\)' statement.
Answer F · Functions and transformations (MM8.2) · question FUN-10
16Functions and transformations · spec MM1.7 · Paper 2 style · ★★☆

Let \(f(x) = \sqrt{x - 3}\) for \(x \geq 3\) and \(g(x) = x^2 + 3\). Which of the following statements is/are true?

  1. I\(g(f(x)) = x\) for all \(x \geq 3\)
  2. II\(f(g(x)) = x\) for all real \(x\)
  3. III\(f(g(x)) = |x|\) for all real \(x\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: \(g(f(x)) = (\sqrt{x-3})^2 + 3 = x\) for \(x \geq 3\): true. \(f(g(x)) = \sqrt{x^2} = |x|\), so II fails (e.g. \(x = -1\)) and III is true.

Full method — how and why
Two functions that almost invert each other. The asymmetry comes from \(\sqrt{x^2} = |x|\) and from the restricted domain.
  1. I: \(g(f(x)) = \left(\sqrt{x-3}\right)^2 + 3 = (x - 3) + 3 = x\) for \(x \geq 3\). True.
    Squaring a square root recovers the inside, provided the inside is non-negative - guaranteed by the stated domain \(x \geq 3\).
  2. Compute the other order: \(f(g(x)) = \sqrt{(x^2 + 3) - 3} = \sqrt{x^2} = |x|\).
    Do the algebra once and read both remaining statements off it.
  3. II claims this equals \(x\) for all real \(x\): false at \(x = -1\), where \(|x| = 1 \neq -1\). False.
    The square root returns the non-negative value, so it cannot reproduce a negative input.
  4. III says it equals \(|x|\), which is exactly what we found. True.
    Same computation, correct conclusion.
  5. I and III only.
    Match to the options.
Where it goes wrong
  • Assuming \(f(g(x)) = x\) because \(f\) and \(g\) 'look like' inverses - inverses require matching domains and ranges.
  • Writing \(\sqrt{x^2} = x\).
Takeaway \(f(g(x)) = x\) and \(g(f(x)) = x\) are different statements; \(\sqrt{x^2} = |x|\) is what breaks the symmetry here.
Answer F · Functions and transformations (MM1.7) · question FUN-12
17Functions and transformations · spec MM1.7 · Paper 1 style · ★★☆

The function \(f\) is defined by \(f(x) = |x - 2| + |x + 2|\) for all real \(x\). What is the minimum value of \(f(x)\), and for how many values of \(x\) is it attained?

  1. Aminimum \(0\), attained for exactly two \(x\)
  2. Bminimum \(2\), attained for exactly one \(x\)
  3. Cminimum \(4\), attained for exactly two \(x\)
  4. Dminimum \(4\), attained for infinitely many \(x\)
  5. Eminimum \(4\), attained for exactly one \(x\)
Worked solution

For \(-2 \leq x \leq 2\): \(f(x) = (2 - x) + (x + 2) = 4\). Outside this interval \(f(x) > 4\). So the minimum \(4\) is attained on the whole interval \([-2, 2]\).

Full method — how and why
A sum of two moduli. Splitting the line at the two 'corner' points shows the function is constant in the middle.
  1. The moduli change behaviour at \(x = 2\) and \(x = -2\); consider the three regions separately.
    \(|x - a|\) changes formula at \(x = a\); the critical points split the line into pieces where the function is linear.
  2. For \(-2 \leq x \leq 2\): \(|x-2| = 2 - x\) and \(|x+2| = x + 2\), so \(f(x) = (2 - x) + (x + 2) = 4\).
    The \(x\) terms cancel exactly - the function is constant on this whole interval.
  3. For \(x > 2\): \(f(x) = (x - 2) + (x + 2) = 2x > 4\). For \(x < -2\): \(f(x) = (2 - x) + (-x - 2) = -2x > 4\).
    Outside the middle the function grows, so nothing beats \(4\).
  4. Minimum \(4\), attained for every \(x\) in \([-2, 2]\): infinitely many values.
    The second half of the question is the real test - most students find the \(4\) and then guess.
Where it goes wrong
  • Answering 'attained for exactly one \(x\)' by assuming a minimum must occur at a single point.
  • Testing only \(x = 0\) and concluding the minimum is \(4\) without checking whether it is attained elsewhere.
Takeaway \(|x - a| + |x - b|\) is constant (and minimal) between \(a\) and \(b\) - it measures total distance to two fixed points.
Answer D · Functions and transformations (MM1.7) · question FUN-16
18Functions and transformations · spec MM8.2 · Paper 2 style · ★★☆

Let \(f(x) = x^2\) for all real \(x\), and let \(g\) be the function whose graph is obtained by translating the graph of \(y = f(x)\) by \(2\) units in the positive \(x\)-direction and then reflecting in the \(y\)-axis. Which of the following statements is/are true?

  1. I\(g(x) = (x + 2)^2\)
  2. II\(g(x) = (x - 2)^2\)
  3. IIIThe graph of \(g\) has its vertex at \((-2, 0)\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

Translation gives \((x - 2)^2\); reflecting in the \(y\)-axis replaces \(x\) by \(-x\): \((-x - 2)^2 = (x + 2)^2\), with vertex at \((-2, 0)\).

Full method — how and why
Composite transformations, applied in the stated order. Each transformation acts on the function you currently have.
  1. Translate \(y=x^2\) by \(2\) in the positive \(x\)-direction: replace \(x\) by \(x-2\), giving \(y = (x-2)^2\).
    A shift right subtracts inside the bracket - counter-intuitive but standard.
  2. Now reflect in the \(y\)-axis: replace \(x\) by \(-x\), giving \(y = (-x-2)^2\).
    Reflection in the \(y\)-axis is the substitution \(x \to -x\).
  3. Simplify: \((-x-2)^2 = \left(-(x+2)\right)^2 = (x+2)^2\). So \(g(x) = (x+2)^2\) - I is true, II is false.
    Squaring removes the overall minus sign.
  4. III: \((x+2)^2\) has its vertex where \(x+2=0\), i.e. at \((-2,0)\). True.
    Consistent: the vertex moved right to \((2,0)\), then reflected to \((-2,0)\).
  5. So I and III.
    Match to the list.
Where it goes wrong
  • Doing the transformations in the wrong order - reflecting first then translating gives \((x-2)^2\) instead.
  • Shifting right by adding \(2\) inside the bracket.
  • Forgetting to simplify \((-x-2)^2\), and so failing to recognise it as \((x+2)^2\).
Takeaway Right by \(a\): \(x \to x-a\). Reflect in the \(y\)-axis: \(x \to -x\). Apply them in the order given, tracking the vertex as a check.
Answer F · Functions and transformations (MM8.2) · question X2-11

How this topic has been examined

Real TMUA questions on Functions and transformations, newest first. Links open the official paper at the question and the online marking page.

2023 · Paper 1Question 11f(kx) and f(x - c) with the same minimumquestion papersit online
2023 · Paper 1Question 20graph of f with given max and minquestion papersit online
2022 · Paper 1Question 10translations of y = x^3question papersit online
2022 · Paper 2Question 20compositions of cos and sinquestion papersit online
2021 · Paper 1Question 5multiplicative function f(mn)question papersit online
2021 · Paper 1Question 15g built from f(2^r x)question papersit online
2021 · Paper 2Question 20f(n+1)(x) = |f(n)(x) + x|question papersit online
2020 · Paper 1Question 10sequence of transformations of 4x^2question papersit online
2019 · Paper 2Question 20reflect then translate the graph of fquestion papersit online
October 2018 · Paper 2Question 10symmetry conditions f(a - x) = f(a + x)question papersit online
Specimen papers · Paper 1Question 10cos x reflected then translatedquestion papersit online