AS & A Level · Practice questions · Transformations of graphs

Inside acts on x, backwards.

Six original Cambridge-style questions on transformations of graphs: describing them precisely, building new equations, tracking points and combining transformations in the right order.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Outside acts on \(y\); inside acts on \(x\), backwards.

Translation \(\begin{pmatrix}p\\q\end{pmatrix}\): \(y = f(x - p) + q\).
Reflection: \(-f(x)\) in the \(x\)-axis; \(f(-x)\) in the \(y\)-axis.
Stretch: \(af(x)\) parallel to the \(y\)-axis, factor \(a\); \(f(ax)\) parallel to the \(x\)-axis, factor \(\frac{1}{a}\).
Describe fully: name the transformation, the axis or direction, and the vector or factor.
Combine one step at a time, rewriting the equation after each, to check the order.

Recall

[2 marks]

Describe fully the single transformation that maps \(y = f(x)\) onto (a) \(y = f(x - 5)\) (b) \(y = \frac{1}{2}f(x)\).

(a) A translation by \(\begin{pmatrix}5\\0\end{pmatrix}\). ✓
(b) A stretch parallel to the \(y\)-axis with scale factor \(\frac{1}{2}\). ✓

Calculation

[3 marks]

The curve \(y = x^{2} - 3x\) is reflected in the \(y\)-axis and then stretched parallel to the \(y\)-axis with scale factor 2. Find the equation of the resulting curve.

Reflect in the \(y\)-axis: replace \(x\) by \(-x\): \(y = (-x)^{2} - 3(-x) = x^{2} + 3x\). ✓
Stretch parallel to the \(y\)-axis, factor 2: multiply by 2. ✓
\(y = 2x^{2} + 6x\). ✓

Calculation

[4 marks]

The point P\((3, -2)\) lies on the curve \(y = f(x)\). Find the image of P on each curve.

(a) \(y = f(x) + 5\) (b) \(y = f(x - 1)\) (c) \(y = -2f(x)\) (d) \(y = f(3x)\)

(a) Up 5: \((3,\ 3)\). ✓
(b) Right 1: \((4,\ -2)\). ✓
(c) Reflect in the \(x\)-axis and stretch by 2: \(y = -2 \times (-2) = 4\), so \((3,\ 4)\). ✓
(d) Stretch parallel to the \(x\)-axis, factor \(\frac{1}{3}\): \((1,\ -2)\). ✓

Analysis

[4 marks]

\(f(x) = x^{2} - 6x + 11\).

(a) Express \(f(x)\) in the form \((x - a)^{2} + b\) and hence describe the transformation that maps \(y = x^{2}\) onto \(y = f(x)\).

(b) Find the coordinates of the minimum point of \(y = 2f(x) + 1\).

(a) \(f(x) = (x - 3)^{2} - 9 + 11 = (x - 3)^{2} + 2\). ✓
A translation by \(\begin{pmatrix}3\\2\end{pmatrix}\). ✓
(b) The minimum of \(y = f(x)\) is \((3, 2)\). The stretch doubles \(y\) and the translation adds 1: \(y = 2(2) + 1 = 5\). ✓
Minimum point \((3,\ 5)\) (check: \(2f(x) + 1 = 2(x - 3)^{2} + 5\)). ✓

Proof

[3 marks]

Show that the curve \(y = x^{2} + 4x + 1\) is a translation of the curve \(y = x^{2}\), and state the translation vector.

Complete the square: \(x^{2} + 4x + 1 = (x + 2)^{2} - 4 + 1 = (x + 2)^{2} - 3\). ✓
This is \(f(x + 2) - 3\) with \(f(x) = x^{2}\): \(x\) replaced by \(x + 2\) and 3 subtracted. ✓
So it is a translation of \(y = x^{2}\) by \(\begin{pmatrix}-2\\-3\end{pmatrix}\). ✓

Analysis

[5 marks]

The curve \(y = f(x)\) has a maximum point at \((2, 5)\) and crosses the \(x\)-axis only at \((-1, 0)\) and \((4, 0)\).

For each curve, state the coordinates of its turning point and of the points where it crosses the \(x\)-axis, and say whether the turning point is a maximum or a minimum.

(a) \(y = f(x - 3)\) (b) \(y = 2f(x)\) (c) \(y = f(2x)\) (d) \(y = -f(x)\)

(a) Translation 3 right: maximum \((5, 5)\); crosses at \((2, 0)\) and \((7, 0)\). ✓
(b) \(y\)-values doubled: maximum \((2, 10)\); crosses at \((-1, 0)\) and \((4, 0)\) (zero doubled is still zero). ✓
(c) \(x\)-values halved: maximum \((1, 5)\); crosses at \(\left(-\frac{1}{2}, 0\right)\) and \((2, 0)\). ✓
(d) Reflection in the \(x\)-axis: the maximum becomes a MINIMUM at \((2, -5)\); crosses at \((-1, 0)\) and \((4, 0)\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.