The modulus function |x| and the V-shaped graph of y = |ax + b|

Sketching y = |f(x)| by reflecting the negative parts in the x-axis

Sketching y = f(|x|) by reflecting the right-hand half in the y-axis

Solving |ax+b| = c, |ax+b| = cx + d and |ax+b| = |cx+d|, with a check for false roots

Solving modulus inequalities with a sketch and with algebra

Combined transformations such as y = 2f(3x), y = f(-x) + 1 and y = 3 - f(x)

What a combined transformation does to points, intercepts, turning points and asymptotes

1. The modulus |x| is the size of x without its sign, so y = |f(x)| is y = f(x) with every negative part reflected upwards, while y = f(|x|) keeps the part with x ≥0 and mirrors it in the y-axis.

2. |x| = x when x ≥0 and |x| = -x when x < 0; a modulus is never negative.

3. |A| = B means A = B or A = -B: solve both and check every root. |A| = |B| can be squared safely: A² = B².

4. |A| < c ⇔-c < A < c and |A| > c ⇔A < -c or A > c. Find critical values, then read the answer from a sketch.

5. Under y = af(bx) + c, the point (p, q) moves to (p/b,  aq + c), and asymptotes move by the same rule.

Many quantities in mathematics have a size but no direction: a distance, an error, a length. The modulus function captures exactly this idea. It takes any real number and throws away its sign, so that positive numbers are unchanged and negative numbers become positive. Geometrically, |x| is the distance of x from 0 on the number line, and so it can never be negative.

The distance idea gives a quick way into simple inequalities. The statement |x - 2| < 5 says "x is less than 5 units away from 2", so x lies between 2 - 5 = -3 and 2 + 5 = 7. No algebra is needed once you read it as a distance.

Now apply the modulus to a linear expression. The graph of y = |ax + b| is the straight line y = ax + b with every point below the x-axis reflected upwards. The line crosses the axis where ax + b = 0, so the result is a V shape whose vertex sits on the x-axis at (-b/a, 0). The y-intercept is (0, |b|). One arm has gradient a and the other has gradient -a, so the V is symmetrical about the vertical line through its vertex.

WORKED EXAMPLE · QUESTION

(a) Sketch the graph of y = |2x - 3|, stating the coordinates of the points where it meets the coordinate axes. (b) Solve |2x - 3| = 5.

SOLUTION

(a) The vertex is where 2x - 3 = 0, so x = (3/2). When x = 0, y = |-3| = 3. The graph is a V with vertex ((3/2), 0) and y-intercept (0, 3).

(b) Split into two cases. 2x - 3 = 5 ⇒x = 4, or 2x - 3 = -5 ⇒x = -1. Check: |2(4) - 3| = 5 and |2(-1) - 3| = |-5| = 5. x = -1 or x = 4. The sketch confirms two solutions, one on each arm.

The modulus here is wrapped around the whole function, so it acts on the output y. Whatever height the graph of y = f(x) reaches, y = |f(x)| reaches the same height without the sign. Points that were already on or above the x-axis are unchanged; points below it are reflected in the x-axis. That gives a reliable three-step method.

Sketch y = f(x) first, lightly.

Keep every part that is on or above the x-axis.

Reflect every part that is below the x-axis in the x-axis.

Some consequences follow straight away. The x-intercepts stay where they are, because 0 is its own modulus. A minimum below the axis at (p, -q) becomes a maximum at (p, q). Where the original curve crossed the axis, the new graph has a sharp corner (a cusp), because a descending piece meets its own reflection. Where the original only touched the axis, nothing is reflected and the curve stays smooth. Finally, y = |f(x)| is never below the x-axis, so its range contains no negative values.

A horizontal line y = k is a powerful tool with these graphs. The number of times it meets y = |f(x)| is the number of solutions of |f(x)| = k, so questions that ask "for which values of k are there exactly four solutions?" are answered by sliding a horizontal line up and down the sketch.

WORKED EXAMPLE · QUESTION

(a) Sketch y = |x² - 2x - 3|, giving the coordinates of the points where it meets the axes and of its maximum point. (b) Solve |x² - 2x - 3| = 3. (c) Find the values of k for which |x² - 2x - 3| = k has exactly four solutions.

SOLUTION

(a) x² - 2x - 3 = (x - 3)(x + 1) = (x - 1)² - 4. The graph meets the axes at (-1, 0), (3, 0), (0, 3), and the minimum (1, -4) of the parabola reflects to the maximum (1, 4).

(b) Case 1: x² - 2x - 3 = 3 ⇒x² - 2x - 6 = 0 ⇒x = 1 ±√7. Case 2: x² - 2x - 3 = -3 ⇒x(x - 2) = 0 ⇒x = 0 or 2. x = 1 - √7, 0, 2, 1 + √7.

(c) A horizontal line y = k cuts the graph four times only when it lies strictly between the axis and the maximum: 0 < k < 4. At k = 4 the line touches the maximum (three solutions); for k > 4 there are two, and at k = 0 two (the roots).

Now the modulus is around the x, so it acts on the input. For x ≥0 we have |x| = x, so f(|x|) = f(x): the right-hand half of the graph is unchanged. For x < 0 we have f(|x|) = f(-x), which takes the value the function had at the mirror-image point on the right. So the left half becomes a copy of the right half, reflected in the y-axis.

Delete everything to the left of the y-axis.

Reflect the part with x ≥0 in the y-axis.

The result is always symmetrical about the y-axis (an even function), and unlike |f(x)| it may go below the x-axis. Anything that happened only for x < 0 on the original graph, such as a root, a turning point or an asymptote, disappears completely, and anything on the right gains a mirror image.

The three figures below use the cubic f(x) = (x + 1)(x - 2)². Compare them carefully: the same original curve produces two quite different modulus graphs.

WORKED EXAMPLE · QUESTION

The function f is defined by f(x) = (x + 1)(x - 2)², x ∈ℝ. (a) Show that the curve y = f(x) has turning points at (0, 4) and (2, 0). (b) Sketch y = |f(x)| and y = f(|x|), labelling where each meets the axes. (c) Solve f(|x|) = 2, giving exact answers.

SOLUTION

(a) By the product rule, f′(x) = (x - 2)² + 2(x + 1)(x - 2) = (x - 2)[(x - 2) + 2(x + 1)] = 3x(x - 2). Setting f′(x) = 0 gives x = 0 or x = 2. Then f(0) = 1 ×4 = 4 and f(2) = 0, so the turning points are (0, 4) and (2, 0).

(b) f(x) < 0 only for x < -1, so y = |f(x)| flips just that part: it meets the axes at (-1, 0), (2, 0), (0, 4), with a cusp at (-1, 0). y = f(|x|) is a W shape meeting the axes at (-2, 0), (2, 0), (0, 4).

(c) Solve f(x) = 2 for x ≥0, then add the mirror images. (x + 1)(x - 2)² = 2 ⇒x³ - 3x² + 2 = 0. Since x = 1 is a root, (x - 1)(x² - 2x - 2) = 0, giving x = 1 or x = 1 ±√3. Reject 1 - √3 ≈-0.732, which is negative and so not on the kept half. Reflecting gives x = ±1, ±(1 + √3).

Every modulus equation is really two equations in disguise. If |A| = B, then A is either B or -B, so you solve both. The subtle point is that a modulus can never equal a negative number. When the right-hand side B contains x, a root from one of the cases might make B negative, and that root does not satisfy the original equation. It is called a false root, and it must be rejected.

|ax + b| = c with a number c > 0: two solutions, from ax + b = ±c.

|ax + b| = cx + d: solve both cases, then substitute each root back and reject any that make cx + d negative. A sketch of the V and the line shows how many solutions to expect.

|ax + b| = |cx + d|: square both sides and solve the quadratic, or solve ax + b = ±(cx + d).

The figure explains a false root visually. Solving the case x - 4 = 2x + 1 is the same as intersecting the line y = 2x + 1 with the full line y = x - 4. But the V only uses y = x - 4 for x ≥4; to the left it has been reflected away. The intersection at x = -5 lies on the dashed part, which is not on the graph of y = |x - 4|.

WORKED EXAMPLE · QUESTION

(a) Solve |x - 4| = 2x + 1. (b) Solve |3x - 1| = |x + 5|.

SOLUTION

(a) Case x - 4 = 2x + 1: x = -5, but then 2x + 1 = -9 < 0, so reject. Case -(x - 4) = 2x + 1: 4 - x = 2x + 1 ⇒3x = 3 ⇒x = 1. Check: |1 - 4| = 3 = 2(1) + 1. x = 1 only.

(b) Square both sides: 9x² - 6x + 1 = x² + 10x + 25 ⇒8x² - 16x - 24 = 0 ⇒x² - 2x - 3 = 0 ⇒(x - 3)(x + 1) = 0. x = 3 or x = -1. Check: |8| = |8| and |-4| = |4|. Squaring cannot create false roots here, because both sides were moduli.

The distance picture from the first section gives the two basic rules. If |A| < c, then A is within c of zero, which is one interval. If |A| > c, then A is further than c from zero, on either side, which is two separate intervals.

When both sides involve x, use a three-step method:

sketch both sides on the same axes;

find the critical values by solving the equation, with the false-root check;

read off the intervals where one graph is above the other.

For |A| < |B| or |A| > |B| you may square, because both sides are non-negative and squaring non-negative numbers preserves the order. The problem then becomes a quadratic inequality. Strict signs (<, >) give open intervals; ≤ and ≥ include the critical values.

WORKED EXAMPLE · QUESTION

(a) Solve |2x - 3| > x + 1. (b) Solve |x + 2| ≤|2x - 1|.

SOLUTION

(a) Critical values from |2x - 3| = x + 1: 2x - 3 = x + 1 ⇒x = 4 (right-hand side 5, valid); -(2x - 3) = x + 1 ⇒x = ⅔ (right-hand side (5/3), valid). From the sketch, the V is above the line outside the critical values: x < ⅔ or x > 4.

(b) Both sides are non-negative, so square: (x + 2)² ≤(2x - 1)² ⇒0 ≤3x² - 8x - 3 = (3x + 1)(x - 3). The critical values are -⅓ and 3, and a positive quadratic is ≥0 outside its roots: x ≤-⅓ or x ≥3. Test x = 0: |2| ≤|-1| is false, so the middle interval is correctly excluded.

These come from P1. Every combined transformation is built from them, so they need to be automatic. The pattern is easy to remember: a change outside the function acts on y and does what it looks like; a change inside the bracket acts on x and does the opposite of what it looks like. Adding c inside moves the graph left, and multiplying x by b squeezes by 1/b, because the input reaches each value sooner.

A quick way to check any rule is substitution. On y = f(x + 3), put x = p - 3: you get y = f(p) = q, so (p - 3, q) lies on the new curve, and the graph has moved 3 to the left.

A combined transformation such as y = af(bx) + c changes x-coordinates and y-coordinates independently, so you can handle the two directions separately. Horizontally, the only change is the factor b: each x-coordinate is divided by b. Vertically, the output is first multiplied by a and then has c added, so in that direction the stretch (or reflection) comes before the translation.

y = 2f(3x): stretch parallel to the x-axis, scale factor ⅓, and stretch parallel to the y-axis, scale factor 2. (p, q) ↦(p/3, 2q).

y = f(-x) + 1: reflect in the y-axis, then translate 1 up. (p, q) ↦(-p, q + 1).

y = 3 - f(x) = -f(x) + 3: reflect in the x-axis, then translate 3 up. (p, q) ↦(p, 3 - q), so a maximum becomes a minimum.

A stationary point stays stationary under any of these, but a reflection in the x-axis swaps maximum and minimum. Points on the x-axis stay on the axis only if there is no vertical translation. To describe a combined transformation, name each single one fully (type, direction, scale factor or vector) and give the order where it matters.

The figures use f(x) = 4 - (x - 1)², which passes through A(-1, 0), B(1, 4), C(3, 0) and D(0, 3).

WORKED EXAMPLE · QUESTION

The curve y = f(x) has a maximum at B(1, 4), meets the x-axis at A(-1, 0) and C(3, 0), and meets the y-axis at D(0, 3). (a) Write down the coordinates of the images of A, B, C and D on y = 2f(3x). (b) Write down the coordinates of the maximum point of y = f(-x) + 1 and of the point where it meets the y-axis. (c) State the coordinates of the turning point of y = 3 - f(x) and its nature.

SOLUTION

(a) (p, q) ↦(p/3, 2q): A′(-⅓, 0), B′(⅓, 8), C′(1, 0), D′(0, 6).

(b) (p, q) ↦(-p, q + 1): maximum (-1, 5); y-axis (0, 4).

(c) (p, q) ↦(p, 3 - q): B(1, 4) ↦ (1, -1), a minimum, because the reflection in the x-axis turns the maximum upside down.

An asymptote is just a line the curve approaches, so it moves exactly as the points near it do. Treat a vertical asymptote x = h with the x-rule and a horizontal asymptote y = v with the y-rule. Under y = af(bx) + c, x = h becomes x = h/b and y = v becomes y = av + c. Vertical changes never move a vertical asymptote, and horizontal changes never move a horizontal one.

The modulus operations have their own effects. Under y = |f(x)| a horizontal asymptote y = v becomes y = |v|. Under y = f(|x|) a vertical asymptote with h < 0 disappears, and one with h > 0 gains a mirror image at x = -h. In every case, find the new intercepts from the new equation: an intercept of the old graph need not map to an intercept of the new one.

WORKED EXAMPLE · QUESTION

The function f is defined by f(x) = 2 + 3/(x - 1), x ≠1. (a) State the equations of the asymptotes of y = f(x). (b) The curve y = f(x) is transformed to y = 2f(½x) - 1. Find the equations of the asymptotes of the new curve and the coordinates of the points where it meets the axes.

SOLUTION

(a) The denominator is zero at x = 1, and the fraction tends to 0 for large |x|: x = 1 and y = 2.

(b) The x-rule multiplies by 2; the y-rule doubles, then subtracts 1. So x = 2 ×1 = 2 and y = 2 ×2 - 1 = 3: x = 2, y = 3.

Intercepts from the new equation. When x = 0: y = 2f(0) - 1 = 2(-1) - 1 = -3, giving (0, -3). When y = 0: f(½x) = ½ ⇒2 + 3/(½x - 1) = ½ ⇒½x - 1 = -2 ⇒x = -2, giving (-2, 0).

Check by simplifying: 2f(½x) - 1 = 3 + 12/(x - 2), which has asymptotes x = 2 and y = 3.

Longer questions usually start with a modulus graph, ask for the solutions of an equation (where the sketch tells you which cases are real), and finish with a transformation of the same graph. The function f(x) = 2|x + 1| - 4 is a typical example: it is itself a transformed modulus graph, the V of |x| moved left 1, stretched vertically by 2 and moved down 4.

WORKED EXAMPLE · QUESTION

The function f is defined by f(x) = 2|x + 1| - 4, x ∈ℝ. (a) Sketch the graph of y = f(x), giving the coordinates of its vertex and of the points where it meets the coordinate axes. (b) Solve f(x) = ½x. (c) Sketch y = |f(x)| and find the set of values of k for which |f(x)| = k has exactly four solutions. (d) Find the coordinates of the vertex of y = -½f(x) + 1, state whether it is a maximum or a minimum, and find where this graph meets the y-axis.

SOLUTION

(a) The V of 2|x + 1| has its vertex at x = -1, then moves down 4: vertex (-1, -4). 2|x + 1| = 4 ⇒x + 1 = ±2, giving (-3, 0) and (1, 0); f(0) = 2 - 4 = -2, giving (0, -2).

(b) Right arm (x ≥-1): 2(x + 1) - 4 = ½x ⇒2x - 2 = ½x ⇒(3/2)x = 2 ⇒x = (4/3), valid since (4/3) ≥-1. Left arm (x < -1): -2(x + 1) - 4 = ½x ⇒-2x - 6 = ½x ⇒-(5/2)x = 6 ⇒x = -(12/5), valid since -(12/5) < -1. x = -(12/5) or x = (4/3).

(c) Reflect the part between x = -3 and x = 1 upwards: a W shape through (-3, 0) and (1, 0), with local maximum (-1, 4) and y-intercept (0, 2). Four solutions when 0 < k < 4.

(d) (p, q) ↦(p, -½q + 1): (-1, -4) ↦ (-1, 3), a maximum (reflected in the x-axis); (0, -2) ↦ (0, 2).

In (b) each case is valid only on its own arm, so each root is checked against its condition. The sketch in (a) shows the line through O crossing both arms, so two roots are expected.

P3.1 Algebraic fractions and functions

Range, composite and inverse functions use the same graphs. The range of |f(x)| never contains negative values, and the graph of an inverse is a reflection in y = x: another transformation, but one that swaps x and y.

P3.4 R-form and P3.5 Exponentials and logarithms

Rsin(x + α), y = eˣ + c and y = ln(bx) are transformations of standard graphs, and their maxima, intercepts and asymptotes move by the rules on this page.

P3.8 Numerical methods and earlier units

P1 introduced single transformations and quadratic inequalities, which reappear when you square a modulus inequality. Locating roots and iteration in P3.8 often start from a sketch of two graphs, just like the modulus equations here. In mechanics and statistics a distance is the modulus of a displacement, and a tolerance such as |L - 120| ≤0.5 is a modulus inequality.

1. Solve |x - 2| < 5.

2. Describe the difference between the graphs of y = |x - 3| and y = |x| - 3.

3. The point (4, -2) lies on y = f(x). Find its image on y = 3f(½x) + 1.

4. The curve y = f(x) has asymptotes x = 6 and y = -2. Find the asymptotes of y = ½f(3x) + 4.

5. Solve |x + 1| = 3x - 5.

6. Find the values of k for which |x² - 2x - 3| = k has exactly three solutions.

1. -5 < x - 2 < 5, so -3 < x < 7.

2. y = |x - 3| is a V with vertex (3, 0); it is |f(x)| for f(x) = x - 3. y = |x| - 3 is a V with vertex (0, -3); it is f(|x|) and crosses the x-axis at x = ±3.

3. x: 4 ÷½ = 8. y: 3(-2) + 1 = -5. The image is (8, -5).

4. x = 6 ÷3 = 2 and y = ½(-2) + 4 = 3.

5. x + 1 = 3x - 5 gives x = 3 (right-hand side 4, valid). -(x + 1) = 3x - 5 gives x = 1, but then 3x - 5 = -2 < 0: reject. So x = 3 only.

6. The graph has a local maximum at (1, 4). A horizontal line meets it exactly three times only when it touches that maximum, so k = 4. (Then x = 1 and x² - 2x - 7 = 0, i.e. x = 1 ±2√2.)