Transform a graph

New equation, image points, sketch, range

In 7 of 20 recent exams
Paper 1 or 2
1 to 4 marks

How it is set in the exam

Paper 1 (no calculator) mostly (5 of 7 exam versions), Paper 2 (calculator) in 2; 1–4 marks per part: a short Section A question on a given graph, or the later parts of a Section B question on a rational or exponential function. Common (7 of 20 exam versions). The original function is given as a graph (on a grid, or as labelled graphs to match) or as a formula. One version stretches a sine curve both ways and asks for a sketch.

Spot it in the exam

Exam phrases to spot (5)

“the graph of $f$ is reflected in the $x$ -axis and then translated by $\binom{0}{3}$ to give the graph of $g$ ; find $g(x)$ ”
“sketch the graph of $y = 2f(x) - 1$ on the grid”
“write down the domain of $g$ ”
“find the range of $g$ ”
“match each function to its graph”

In disguise (4)

a composite with a linear outer function: $g(x) = h(f(x))$ with $h(x) = x - 5$ , then "find the range of $g$ ": the asymptote of $f$ moves down $5$
a sinusoid stretched both ways and then sketched (even when the new curve is given by its own formula, to be drawn on the same axes as the given graph): the new amplitude, period and domain are the images of the old ones
the range of $-f$ on a restricted domain, after the $y$ -intercept and the horizontal asymptote of $f$ have been written down from its formula or its graph
"show that the graph of $g$ is the graph of $f^{-1}$ ", where $g$ comes from reflections in both axes

Variants

given = a graph or a formula; transformation = a vertical stretch and a translation, a reflection in the $x$ -axis and a translation, reflections in both axes, two stretches, or four single ones to match; asked = the new formula, its domain, its range, a sketch, a show-that.

Looks like it, but isn't (5)

both graphs (or both formulas) are given, and the transformation or a constant in it is asked for: Name the transformation
a quadratic written in vertex form (completing the square), with its vertex, axis or range read off: Find the vertex
a sinusoid sketched from its own equation, with no earlier graph to transform: Read a sinusoid's features from its equation (Topic 3)
a reflection in $y = x$ (the inverse, not an SL 2.11 transformation): its sketch, or where it meets the graph of $f$ , is The graphs of f and f⁻¹; its equation is Find the inverse function, with its domain
the features of the function itself, before any transformation (the asymptotes and intercepts of $\frac{ax + b}{cx + d}$ ): Rational functions: asymptotes, intercepts, sketch, range

Worked example

A graph made of segments: map the corners

Paper 1 (no calculator)

Question

The graph of $y = f(x)$ , for $-4 \le x \le 4$ , is made of three straight line segments. They join the points $A(-4, 2)$ , $B(-2, -2)$ , $C(1, -2)$ and $D(4, 4)$ , in that order. (On the exam paper, the graph is drawn on a grid.) Let $g(x) = \frac{1}{2}f(x) + 2$ .
(a) Sketch the graph of $y = g(x)$ . [3]
(b) Write down the range of $g$ . [1]
(c) Let $h(x) = f\left(\frac{1}{2}x\right)$ . Write down the domain and the range of $h$ . [2]

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Steps

Write each transformation as a change to the formula
( $f(x) + b$ , $f(x - a)$ , $-f(x)$ , $f(-x)$ , $pf(x)$ , $f(qx)$ ) and apply them in the order given, each to the whole previous expression. A horizontal change replaces every $x$ .
Map the key points:
vertical changes act on $y$ , horizontal changes on $x$ (the table in Images of points). Asymptotes and endpoints move in the same way.
Domain and range:
only horizontal changes alter the domain, and only vertical changes alter the range. A reflection in the $x$ -axis swaps the ends of the range, and each end keeps its type ( $\le$ or $<$ ).
Sketch
through the image points with the same shape: a straight piece stays straight, a maximum goes to a maximum (or, after a reflection in the $x$ -axis, a minimum). Stop at the ends of the domain, with closed dots at included ends, and draw asymptotes dashed, with their equations. On a grid, the image points need not be labelled unless the question asks for their coordinates, but each must sit on the right grid point: each placed feature earns its mark.
A "show that" about a transformation
works with the general expression $f(x)$ , never with one particular point.

Common mistakes

The inside going the wrong way
$f(x - 3)$ is $3$ to the right, and $f(2x)$ halves the $x$ -coordinates. When unsure, ask which $x$ makes the inside equal to the old $x$ .
Translating before reflecting (or stretching)
when the question says the opposite: reflecting in the $x$ -axis and then translating $7$ up is $-f(x) + 7$ ; the other order gives $-f(x) - 7$ .
Replacing only one $x$
a horizontal change replaces every $x$ in the formula, top and bottom of a fraction included.

More mistakes (6)

Domain and range on the wrong side
a vertical change never alters the domain, and a horizontal one never alters the range. But a horizontal change does move the ends of a restricted domain: $f(x + 1)$ , where $f$ is defined for $x \le 1$ , is defined for $x \le 0$ , and $f\left(\frac{1}{2}x\right)$ doubles each end.
Range signs after a reflection
$-3 < f(x) \le 5$ gives $-5 \le -f(x) < 3$ , not $-5 < -f(x) \le 3$ . The value that is reached stays reached.
One particular point in a "show that"
it scores . Work with $f(x)$ .
A sketch that moves the curve but not its features
the asymptotes, the endpoints (closed dots where included) and the turning points all move, and each one placed correctly earns its mark.
An asymptote that is not an equation
write $y = 5$ (or $x = -2$ ). A bare number, or " $y \ne 5$ ", scores even when the value is right. (A range may be written $y \ne 5$ : markschemes accept it there. It is the asymptote that must be an equation.)
A transformed sine curve drawn with straight lines or sharp corners
the sketch marks (amplitude, domain, maximum and minimum points, zeros) are only awarded when the curve looks like a sine curve.

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