THE MODULUS FUNCTION

The graph of y = | x | can be found by starting with y = x: The modulus of x, written | x | means the size of x ( ignoring the sign). Hence, | – 3 | = 3, | 7.1 | = 7.1 The graph of y = | x | can be found by starting with y = x: x y y = x The modulus of any negative values of y, will be positive x y y = | x | For the graph of y = | f(x) |, sketch y = f(x) and reflect the negative values of f(x) in the x-axis.

Also consider the graph of y = f ( | x | ). If x is negative, then | x | will be positive i.e. f ( | – a | ) = f ( | a | ) So a graph of y = f ( | x | ) will be symmetrical about the y-axis. For the graph of y = f ( | x | ), sketch y = f(x) for x ≥ 0 and reflect the graph in the y-axis.

Example 1: Draw a sketch of y = | 2x + 4 |, and hence solve the equation Firstly, sketch y = 2x + 4 x = 0, y = 4 y = 0, x = –2 For the modulus graph, reflect the negative part in the x-axis The line segment for x ≥ –2 has equation: y = 2x + 4 The line segment for x ≤ –2 has equation: y = –2x – 4 Add the line y = 6: y x 4 –2 For A: –2x – 4 = 6 – 2x = 10 A B y = 6 x = –5 For B: 2x + 4 = 6 2x = 2 x = 1

Example 2: Shown here is a graph of y = f(x). The graph crosses the y-axis at the point ( 0, 7 ) and has maximum points at the points ( 3, 10 ) and ( –4, 5 ). Sketch the graph of y = f ( | x | ). ( 3, 10 ) 7 x y ( –4, 5 ) y = f(x) The part of the graph for negative values of x disappears: ( 3, 10 ) 7 x y ( –3, 10 ) The part of the graph for positive values of x is reflected in the y axis: Hence, there is a maximum point at ( –3, 10 ).

Hence solve the equation | 2x – 8 | = x + 2. Example 3: Draw a sketch of y = | 2x – 8 |, and on the same axes sketch the graph of y = x + 2. Hence solve the equation | 2x – 8 | = x + 2. Firstly, sketch y = 2x – 8 –8 x = 0, y = For the modulus graph, reflect the negative part in the x-axis y = 0, x = 4 The line segment for x ≥ 4 has equation: y = 2x – 8 y = 2x – 8 x y The line segment for x ≤ 4 has equation: y = –2x + 8 Add the line y = x + 2: x y 4 8 2 B A y = x + 2 For A: – 2x + 8 = x + 2 6 = 3x x = 2 For B: 2x – 8 = x + 2 x = 10

a) Find the co-ordinates of the points P and Q and R. Example 4: Shown here is a sketch of y = f(x), where f(x) = | x + 1 | – 2 a) Find the co-ordinates of the points P and Q and R. b) Sketch the graph of y = | f(x) | c) Solve | f(x) | = 2x. x y P R Q a) For P and R, y = 0 | x + 1 | – 2 = 0 | x + 1 | = 2 x = – 3 or 1 i.e. P is ( –3, 0 ), and R is ( 1, 0 ) For Q, the smallest value of | x + 1 | is x = –1 i.e. Q is ( –1, –2 ) x y –3 1 b) For y = | f(x) |, we have: ( –1, 2 ) Q will move to the point:

For the point of intersection: 1 – x = 2x 1 = 3x y ( 1, 0 ) ( –3, 0 ) ( –1, 2 ) c) We have, y = | f(x) |: Where f(x) = | x + 1 | – 2 A y = 2x Add the line y = 2x For A, we need the equation of the line through ( –1, 2 ) and ( 1, 0 ). 2 – 0 –1 – 1 The gradient, m = = –1 y – 0 = –1( x – 1 ) y = 1 – x x = 1 3 For the point of intersection: 1 – x = 2x 1 = 3x

For the graph of y = | f(x) |, reflect the Summary of key points: For the graph of y = | f(x) |, reflect the negative values of f(x) in the x-axis. The part of the graph of y = f(x) which is positive and has therefore not been reflected in the x-axis, has equation y = f(x). The part of the graph of y = f(x) which is negative and has been reflected in the x-axis has equation y = – f(x). For the graph of y = f ( | x | ), sketch y = f(x) for x ≥ 0 and reflect the graph in the y-axis. This PowerPoint produced by R.Collins ; Updated Feb. 2009