A-level Maths revision · Pure, year 1

Graph transformations

The idea

Changes outside the function act on yy as you expect; changes inside act on xx the opposite way. y=f(x)+ay = f(x) + a moves up aa; y=f(x+a)y = f(x + a) moves left aa; y=af(x)y = af(x) stretches vertically by aa; y=f(ax)y = f(ax) stretches horizontally by 1a\frac{1}{a}.

  1. Decide whether the change is inside or outside the brackets.
  2. Outside: change the yy-coordinates. Inside: change the xx-coordinates, inverted.
  3. Move each key point, asymptote and intercept.

Watch out f(2x)f(2x) halves the xx-coordinates; it does not double them.

See it worked

Worked example

y=f(x)y = f(x) has a maximum at (2,5)(2, 5). Find the maximum of y=f(x−3)y = f(x - 3), y=2f(x)y = 2f(x) and y=f(2x)y = f(2x).

  1. f(x−3)f(x - 3): move right 33, giving (5,5)(5, 5).
  2. 2f(x)2f(x): double the yy-coordinate, giving (2,10)(2, 10).
  3. f(2x)f(2x): halve the xx-coordinate, giving (1,5)(1, 5).

Answer (5,5)(5, 5), (2,10)(2, 10), (1,5)(1, 5)

Your turn

Question 1Non-calculator3 marks

The curve y=f(x)y = f(x) has a minimum at (−1,3)(-1, 3). Find the minimum of y=f(x+2)y = f(x + 2).

Question 2Non-calculator3 marks

The curve y=f(x)y = f(x) has a minimum at (−1,3)(-1, 3). Find the minimum of y=3f(x)y = 3f(x).

Question 3Non-calculator3 marks

The curve y=f(x)y = f(x) has a minimum at (−1,3)(-1, 3). Find the minimum of y=f(12x)y = f\left(\frac{1}{2}x\right).

In past papers

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