Graph Transformations Described in Words

A question asks you to fully describe the single transformation that maps $y = \frac{6}{x}$ to $y = \frac{6}{x-2}$, and to give the equations of the new asymptotes. Which response would score full marks?

• Translation, 2 units in the positive x-direction; new asymptotes $x=2$, $y=0$

  Correct on every count that the mark scheme checks separately: the transformation TYPE is named correctly (translation), the magnitude and direction are both stated, and both new asymptotes are correct.

• BShift, 2 units in the positive x-direction; new asymptotes $x=2$, $y=0$

  The magnitude, direction and asymptotes are all correct — and this response still loses the type mark. A real examiner report on this exact pair of functions records that most candidates got everything right except this one word, using 'shift' or 'move' where the mark scheme required 'translation'.

• CTranslation, 2 units in the negative x-direction; new asymptotes $x=-2$, $y=0$

  The word is right and the direction is reversed. Tracking a point confirms it: (1,6) is on the original curve, and the same height is reached on the new curve at x−2=1, i.e. x=3 — the point moved right, to a LARGER x-value, not a smaller one.

• DStretch in the x-direction, scale factor 2; new asymptotes $x=2$, $y=0$

  The asymptotes happen to be right but the transformation type is wrong. $6/(x-2)$ has a constant ADDED to x, not multiplying x — that is the f(x+a) form (a translation), not the f(ax) form (a stretch).

Traps tested: Translation called a shift or move · X translation direction sign flipped · Translation called a stretch

Which single transformation maps $y = \cos x$ to $y = \cos 2x$?

• Stretch, scale factor $\frac{1}{2}$, in the x-direction (from the y-axis)

  Correct. $\cos(2x)$ has the input doubled before cosine acts on it — the f(ax) form with a = 2 — which moves every point to half its original distance from the y-axis: a stretch scale factor 1/a = 1/2, in the x-direction.

• BStretch, scale factor 2, in the y-direction (from the x-axis)

  This is the exact documented misconception on this exact pair of functions: an examiner report records candidates 'mistakenly thinking that the transformation mapping cos(x) to cos(2x) is a stretch in the y-axis direction' (Oct 2021, Q4(a)(i)). This description is actually correct for $y=2\cos x$, a genuinely different curve.

• CStretch, scale factor 2, in the x-direction (from the y-axis)

  The axis is right and the scale factor is not inverted. For y=f(ax), the scale factor is 1/a, the reciprocal of the number in the equation — using a itself instead is the mirror image of forgetting to invert.

• DTranslation, 2 units in the positive x-direction

  The transformation TYPE is wrong. Nothing is added to x here — x is multiplied by 2 before cosine acts on it, which is a stretch (f(ax) form), not a translation (f(x+a) form).

Traps tested: Af and fax stretch direction confused · Fax scale factor not inverted · Stretch called a translation

The curve $y=f(x)$ passes through $(1,-3)$. Which point must lie on the curve $y=-f(x)$?

• $(1, 3)$

  Correct. $y=-f(x)$ is $y=af(x)$ with $a=-1$: every OUTPUT has its sign flipped and every input is untouched, so the x-coordinate stays 1 and the y-coordinate flips from −3 to 3.

• B$(-1, -3)$

  This reflects in the y-axis instead of the x-axis — that describes $y=f(-x)$, not $y=-f(x)$. The minus sign here is on the OUTSIDE of $f$, so it acts on the output, not the input.

• C$(-1, 3)$

  This applies both reflections at once (through the origin), which is neither the transformation given here nor a single one of the four listed in spec 1.12 at all.

• D$(1, -3)$

  Unchanged. $y=-f(x)$ does move this point — it flips the sign of every output, and $-3 \neq 3$, so a point with a nonzero y-coordinate cannot stay where it was.

Traps tested: Reflected in wrong axis · Double reflection through origin · Reflection assumed not to move points

The curve $y = f(x)$, where $f(x) = x^3 - 4x$, passes through $(1, -3)$. What point does the curve $y = f(-x)$ pass through?

• $(-1, -3)$

  Correct. $y=f(-x)$ is $y=f(ax)$ with $a=-1$: the point that was at $x=1$ moves to $x = 1/a = -1$. Checking directly: $f(-(-1)) = f(1) = -3$ — the new curve does pass through $(-1,-3)$.

• B$(1, 3)$

  This reflects in the x-axis instead of the y-axis — that would be $y=-f(x)$, where the minus sign is OUTSIDE the function acting on the output. Here it is INSIDE, acting on the input.

• C$(-1, 3)$

  This applies both reflections at once. Reflecting only in the y-axis moves the x-coordinate and leaves the y-coordinate exactly as it was: still $-3$, not $3$.

• D$(1, -3)$

  Unchanged — but $f(-x)$ does move most points. Checking directly: at $x=1$, the new curve gives $f(-1) = -1+4 = 3 \neq -3$, so $(1,-3)$ is not on the new curve at all.

Traps tested: Reflected in wrong axis · Double reflection through origin · Reflection assumed not to move points

The curve $y=f(x)$ has a single minimum at $(2,-5)$, no other turning points, and $y \to \infty$ as $x \to \pm\infty$. For how many values of the constant $k$ does the curve $y = f(x) + k$ touch the x-axis at exactly one point (rather than crossing it twice, or missing it entirely)?

• Exactly one value of $k$

  Correct. $y=f(x)+k$ translates the whole curve vertically, so its minimum moves to $(2, -5+k)$. The curve touches the x-axis at exactly one point precisely when that minimum sits exactly ON the axis: $-5+k=0$, i.e. $k=5$ — one specific value, not a range.

• BInfinitely many values of $k$, since translating up or down always keeps the curve touching somewhere

  A vertical translation moves the minimum's HEIGHT; for almost every value of k the minimum sits either above the axis (curve misses it entirely) or below it (curve crosses twice). Touching happens only at the single boundary height where the minimum sits exactly on the axis.

• CNo values of $k$ — a curve translated vertically can only cross the axis twice or miss it, never touch it

  This dismisses the boundary case that separates 'crosses twice' from 'misses entirely'. Exactly at the value of k where the minimum's height is 0, the two intersection points merge into one — this is the same boundary logic as a discriminant equal to zero.

• DTwo values of $k$ — one for each direction the curve could be translated

  There is only one height at which the minimum sits exactly on the axis, because a minimum has one specific height that a vertical translation shifts by k — there is exactly one value of k that makes $-5+k=0$, not two.

Traps tested: Assumes translation always preserves tangency · Boundary case dismissed as impossible · Boundary case double counted

Which single transformation maps $y = \sin x$ to $y = 3\sin x$ — the specification's own worked example?

• Stretch, scale factor 3, in the y-direction (from the x-axis)

  Correct. $3\sin x$ is $y=af(x)$ with $a=3$: the OUTPUT of sine is multiplied by 3, so every height triples and no input changes — amplitude 3, period unchanged at 2π.

• BStretch, scale factor 3, in the x-direction (from the y-axis)

  This is the wrong axis for this transformation. The 3 multiplies the OUTPUT here (it sits outside sin, not inside it), so the effect is vertical, not horizontal — the reverse of the confusion documented for cos(2x), but the same underlying mix-up.

• CTranslation, 3 units in the positive y-direction

  The transformation type is wrong: 3 MULTIPLIES sin x here, it is not added to it. $y=af(x)$ (a stretch) and $y=f(x)+a$ (a translation) are different ones of the four listed transformations, and $3\sin x \neq \sin x + 3$.

• DStretch, scale factor $\frac{1}{3}$, in the y-direction (from the x-axis)

  The scale factor has been inverted when it should not have been. For $y=af(x)$, the OUTPUT is multiplied directly by a — the scale factor is a itself. Inverting to 1/a is the correct move for $y=f(ax)$, a different one of the four transformations, applied here to the wrong one.

Traps tested: Af x vs y stretch confused · Stretch confused with translation · Af scale factor wrongly inverted