01 · Inside and outside
Evaluate (a) |3 − 11|, (b) −|3 − 11|, (c) |3| − |11| and (d) ||−8| − 5|.
Hint
For nested bars, begin with the innermost pair.
Worked solution
(a) |−8| = 8. (b) −8. (c) 3 − 11 = −8. (d) |8 − 5| = 3.
Understand · explore · practise
The modulus function
Understand the modulus (absolute value) function, sketch |ax + b| and solve basic modulus equations. Move the graph yourself and practise with worked solutions.
Before you startNegative numbers, straight-line graphs and solving linear equations
01 / What modulus means
Distance has no negative direction.
The modulus of a real number is its distance from zero on a number line. We write it using vertical bars. Both −6 and 6 are six units from zero, so |−6| = |6| = 6. Also |0| = 0: modulus is nonnegative, not necessarily positive.
|u| = u when u ≥ 0
|u| = −u when u < 0
For a negative u, the expression −u is positive. For example, if u = −6 then −u = 6.
The bars enclose the whole input to the modulus operation, called its argument. Work out that argument first. A minus sign outside the bars is a separate operation.
The placement of the bars mattersWorked example
|4 − 9| = |−5| = 5
Subtract first, then find the distance from zero.
|4| − |9| = 4 − 9 = −5
This is a different expression: two moduli followed by subtraction.
−|4 − 9| = −5
The outer minus remains after evaluating the modulus.
02 / Evaluate a function
Substitute, simplify inside, then take the modulus.
Let f(x) = |2x − 2|. The domain is every real number. Every output is at least zero, and every nonnegative output is possible, so its range is [0, ∞).
Three inputs, two outputsWorked example
f(−1) = |2(−1) − 2|
= |−4| = 4
Keep the negative input in brackets when substituting.
f(1) = |2 − 2| = 0
The argument is zero at x = 1.
f(3) = |6 − 2| = 4
Two different inputs can produce the same output. This is still a function.
03 / Reflect the graph
Negative outputs move above the axis; inputs stay put.
Start with y = 2x − 2. Where the line is already on or above the x-axis, taking modulus leaves its output unchanged. Where it is below the axis, replace the negative output with its positive opposite.
(x, y) → (x, |y|)
The vertex is (1, 0), where the original line crosses the x-axis. The y-intercept is (0, 2). Both arms continue upward without a maximum. The model only shows a finite window of the graph.
Move the input through x = 1. Below 1, the original negative output is reflected; above 1, the two points coincide. This is an output change, not a reflection of the entire original line.
Same input, nonnegative outputExplore
Dashed gold: y = 2x − 2. Solid green: y = |2x − 2|. The vertical guide joins the two outputs at the same input.
x = −1: 2x − 2 = −4, so |2x − 2| = 4.
The original output is negative, so reflect it across the x-axis. The x-coordinate stays the same.
Watch the negative line segment reflect
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Two straight pieces
Use the sign of the argument to choose the rule.
The expression inside the bars, 2x − 2, changes sign at x = 1. Therefore the same graph can be described without modulus bars:
|2x − 2| = 2 − 2x for x < 1
|2x − 2| = 2x − 2 for x ≥ 1
The left arm has slope −2 and the right arm has slope 2. At x = 1 both formulas give zero, so the pieces meet. We include the boundary in just one piece to make the rule easy to read.
Why not replace every x by |x|?
That would form |2x| − 2 or 2|x| − 2, a different function. For example, at x = 0 our original function gives 2, but 2|0| − 2 gives −2. The bars apply to the entire argument 2x − 2.
05 / Any linear argument
Find the zero before drawing the V.
For f(x) = |ax + b| with a ≠ 0, the argument is zero at x = −b/a. The vertex is (−b/a, 0), the y-intercept is (0, |b|), and the range is [0, ∞).
|ax + b| = |a| |x + b/a|
The left arm has slope −|a| and the right arm has slope |a|, even if a is negative. If a = 0, the function is instead the horizontal line y = |b|, with a single-value range; there is no V-shaped vertex.
Sketch y = |5 − x|Worked example
5 − x = 0 ⇒ x = 5
Vertex: (5, 0). The y-intercept is (0, 5).
For x ≤ 5: y = 5 − x
For x > 5: y = x − 5
The descending original line is reflected only after it goes below the x-axis.
Domain: all real x
Range: y ≥ 0
Use the slopes −1 and +1 to complete the sketch.
06 / Basic equations
A distance can point in either direction.
For c > 0, the equation |u| = c means u = c or u = −c. For c = 0, only u = 0 works. For c < 0, there is no real solution. These facts do not require guessing from a drawing.
Solve |4x + 1| = 7Worked example
4x + 1 = 7
or 4x + 1 = −7
The two allowed argument values are 7 and −7.
x = 3/2 or x = −2
Check: |4(3/2) + 1| = 7 and |4(−2) + 1| = 7.
|4x + 1| = 0 ⇒ x = −1/4
A horizontal line at height zero meets the vertex once.
|4x + 1| = −7: no real solution
A modulus cannot equal a negative number.
07 / Your turn
Explain the sign choice as well as the answer.
Attempt these without opening the solution. Label vertices and intercepts on sketches; give both roots when a positive distance permits them.
02 · Evaluate a shifted function
For g(x) = |3x + 2| − 4, find g(−2), g(0) and g(2).
Hint
The subtraction of 4 happens after the modulus.
Worked solution
g(−2) = |−4| − 4 = 0. g(0) = |2| − 4 = −2. g(2) = |8| − 4 = 4. A negative output is possible because of the outer subtraction.
03 · Negative coefficient
Sketch y = |6 − 2x|, giving its vertex, y-intercept, range and two linear rules.
Hint
The argument changes sign at x = 3.
Worked solution
Vertex (3, 0), y-intercept (0, 6), range y ≥ 0. For x ≤ 3 use 6 − 2x; for x > 3 use 2x − 6. The arms have slopes −2 and +2.
04 · Two roots
Solve |3x − 4| = 8.
Hint
Solve 3x − 4 = 8 and 3x − 4 = −8.
Worked solution
The first equation gives x = 4; the second gives x = −4/3. Both give modulus 8 in the original equation.
05 · Count before calculating
How many real solutions does |5x + 2| = c have when (a) c = −1, (b) c = 0 and (c) c = 3? Find any solutions.
Hint
The graph is a V with minimum height zero.
Worked solution
(a) None. (b) One: x = −2/5. (c) Two: 5x + 2 = ±3 gives x = 1/5 or x = −1.
06 · A false identity
Is |a + b| = |a| + |b| always true? Give a counterexample and one case in which it is true.
Hint
Try numbers with opposite signs first.
Worked solution
With a = −4 and b = 7, the left side is 3 and the right side is 11. With a = 2 and b = 7, both sides are 9. One counterexample disproves the universal claim.
07 · Distance from a point
Explain |x − 4| = 3 using a number line, then solve it.
Hint
|x − 4| is the distance between x and 4.
Worked solution
The input is three units from 4, either to its left or right. Thus x = 1 or x = 7. Algebraically, x − 4 = −3 or 3.
08 · Constant special case
Describe f(x) = |0x − 9| and solve f(x) = 9.
Hint
The argument no longer depends on x.
Worked solution
f(x) = 9 for every real x. The graph is horizontal, its range is {9}, and every real x solves f(x) = 9. The usual two-root claim for |ax + b| = c > 0 assumes a ≠ 0.
08 / Recap
Track what the bars enclose.
- Modulus is distance from zero: nonnegative, including zero.
- Evaluate the entire argument before taking its modulus.
- Reflect negative outputs across the x-axis and leave other outputs unchanged.
- For a nonconstant linear argument, its zero locates the vertex.
- For a positive target, solve both signed equations and check the results.
Section 1 of 8 · What modulus means