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Transformations of graphsAQA A-Level Maths: Revision notes

Section 1

Translations: f(x)+af(x)+a and f(x+a)f(x+a)

A translation slides a graph without changing its shape.

  • y=f(x)+ay=f(x)+a moves the graph up by aa (down if a<0a<0): vector (0a)\begin{pmatrix}0\\ a\end{pmatrix}.
  • y=f(x+a)y=f(x+a) moves the graph left by aa (right if a<0a<0): vector (−a0)\begin{pmatrix}-a\\ 0\end{pmatrix}. If (3,−2)(3,-2) is a minimum of y=f(x)y=f(x), then y=f(x)+4y=f(x)+4 has its minimum at (3,2)(3,2) and y=f(x+4)y=f(x+4) at (−1,−2)(-1,-2). Changes outside the brackets act on yy in the obvious way; changes inside the brackets act on xx the opposite way.
Key termstranslationvector
Common mistake

Moving f(x+a)f(x+a) to the right. Adding aa inside moves the graph left, because xx must be smaller to give the same input.

Section 2

Stretches: af(x)af(x) and f(ax)f(ax)

A stretch changes size in one direction.

  • y=af(x)y=af(x): stretch parallel to the yy-axis with scale factor aa. Every yy-coordinate is multiplied by aa; xx-intercepts do not move.
  • y=f(ax)y=f(ax): stretch parallel to the xx-axis with scale factor 1a\frac1a. Every xx-coordinate is divided by aa; the yy-intercept does not move. For example, f(2x)f(2x) halves the xx-coordinates: roots −2-2 and 33 become −1-1 and 32\frac32. f(x2)f\left(\frac x2\right) doubles them.
Key termsstretchscale factor
Common mistake

Multiplying the xx-coordinates by 22 for f(2x)f(2x). The stretch factor is 12\frac12, so they are halved.

Section 3

Reflections: −f(x)-f(x)

Choosing a=−1a=-1 in af(x)af(x) gives y=−f(x)y=-f(x), a reflection in the xx-axis: each yy-coordinate changes sign, so maxima become minima. The xx-intercepts stay put. If y=f(x)y=f(x) has a maximum at (2,6)(2,6) and crosses the yy-axis at (0,4)(0,4), then y=−f(x)y=-f(x) has a minimum at (2,−6)(2,-6) and crosses the yy-axis at (0,−4)(0,-4). The sign of aa in af(x)af(x) decides whether the graph is flipped; its size decides the stretch.

Key termsreflection
Exam tip

For y=−f(x)y=-f(x), change the sign of every yy-value, and swap the labels maximum and minimum.

Section 4

Tracking points through a transformation

The safest method is to track key points. For each point (x,y)(x,y) on y=f(x)y=f(x):

  • y=f(x)+ay=f(x)+a: (x,y+a)(x,y+a)
  • y=f(x+a)y=f(x+a): (x−a,y)(x-a,y)
  • y=af(x)y=af(x): (x,ay)(x,ay)
  • y=f(ax)y=f(ax): (xa,y)\left(\frac xa,y\right) Example: if f(x)=x2−4x+7=(x−2)2+3f(x)=x^2-4x+7=(x-2)^2+3 has minimum (2,3)(2,3) and yy-intercept (0,7)(0,7), then y=f(2x)y=f(2x) has minimum (1,3)(1,3) and the same yy-intercept (0,7)(0,7). Check: f(2x)=4x2−8x+7f(2x)=4x^2-8x+7, with minimum at x=1x=1.
Key termsimage
Exam tip

Check one point by substitution: if (1,3)(1,3) is on y=f(2x)y=f(2x) then f(2)f(2) must equal 33.

Section 5

Describing a transformation and sketching the result

To describe a transformation fully, name the type and give its details: a translation needs a vector, a stretch needs a scale factor and a direction (parallel to which axis), a reflection needs a line. To sketch the transformed graph:

  1. Mark the key points of the original (intercepts, turning points, asymptotes).
  2. Move each key point using the rules above.
  3. Redraw the same shape through the new points, labelling coordinates. Translations and stretches by themselves do not change the shape's turning-point nature (a minimum stays a minimum), unless a<0a<0.
Key termsdescribe fully
Common mistake

Saying only 'a translation'. The vector, or the left, right, up or down amount, is needed for the marks.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Transformations of graphs

  1. The curve y=f(x)y=f(x) has a minimum point at (3,−2)(3,-2).
    Write down the coordinates of the minimum point of the curve y=2f(x−1)y=2f(x-1).2 marks
  2. The curve y=f(x)y=f(x) crosses the xx-axis at (−2,0)(-2,0) and (3,0)(3,0) only, and crosses the yy-axis at (0,6)(0,6).
    Describe fully the single transformation that maps y=f(x)y=f(x) onto y=f(x+2)y=f(x+2), and write down the xx-intercepts of y=f(x+2)y=f(x+2).2 marks
  3. The function ff is defined by f(x)=x2−4x+7f(x)=x^2-4x+7.
    Show that the curve y=f(x+1)y=f(x+1) has equation y=x2−2x+4y=x^2-2x+4.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).