Pearson Edexcel International Advanced Level Mathematics: Functions, modulus and graph transformations (YMA01)

This guide covers section 1 (Algebra and functions) of Unit P3 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification, Issue 3 (April 2019): topics 1.1 to 1.4. The specification makes P3 a compulsory unit for IAL Mathematics and IAL Pure Mathematics, and lists its IAS weighting as N/A. It also says that knowledge of P1 and P2 is assumed, so keep the Pure 1 revision notes and Pure 2 revision notes close.

Next, use the revision notes for recall and the practice questions for timed work. The course pages are the Edexcel A-level Mathematics hub and the printable checklist. To find weak spots first, try the free 10-minute diagnostics.

Outcomes in this section

Topic	Skills this guide builds
1.1	Simplify rational expressions by factorising, cancelling and algebraic division. Denominators are linear or quadratic.
1.2	Define a function, and give its domain and range. Compose functions (fg means do g first). Find inverse functions and sketch their graphs. Use f⁻¹f(x) = ff⁻¹(x) = x.
1.3	Sketch y = |ax + b|, y = |f(x)| and y = f(|x|). Solve equations and inequalities such as |2x − 1| = x + 5.
1.4	Sketch combined transformations such as y = 2f(3x), y = f(−x) + 1, y = 3 + sin 2x and y = −cos(x + π/4). The graph of y = f(ax + b) is not required.

1.1 Rational expressions

A rational expression is a fraction whose top and bottom are both polynomials. To simplify it, factorise the top and bottom completely, then cancel any bracket that appears on both. Brackets cancel; separate terms do not.

Example 1. Simplify (3x² − 5x − 12)/(x² − 9).

3x² − 5x − 12 = (3x + 4)(x − 3)
x² − 9 = (x − 3)(x + 3)
(3x² − 5x − 12)/(x² − 9) = (3x + 4)/(x + 3),   x ≠ 3

The original expression is undefined at x = 3, so the simplified form agrees with it everywhere except x = 3.

When the numerator has degree equal to or greater than the denominator and nothing cancels, use algebraic division. Divide to get a quotient and a remainder, then write numerator/denominator = quotient + remainder/denominator.

Example 2. Write (3x³ − 2x² + x − 5)/(x² + 2) as a quotient plus a remainder term.

3x³ − 2x² + x − 5 ÷ (x² + 2):
   3x³ ÷ x² = 3x     3x(x² + 2) = 3x³ + 6x       remainder −2x² − 5x − 5
  −2x² ÷ x² = −2    −2(x² + 2) = −2x² − 4       remainder −5x − 1

So (3x³ − 2x² + x − 5)/(x² + 2) = 3x − 2 + (−5x − 1)/(x² + 2).

A linear denominator works the same way. Here x³ − 8 = (x − 2)(x² + 2x + 4), so (x³ − 8)/(x − 2) = x² + 2x + 4 with no remainder.

1.2 Functions: domain, range, composites and inverses

A function maps each input in its domain to exactly one output. The set of outputs is the range. The specification uses the notation f: x ↦ … and f(x). A function can be one-one (each output comes from one input) or many-one (two or more inputs share an output). Always state the domain when it is not all real numbers. Division by zero and square roots of negatives are the usual restrictions.

Composite functions. fg(x) means f(g(x)): do g first, then f. In general fg ≠ gf.

Example 3. f(x) = x² + 1 for all real x, and g(x) = 3/(x − 2) for x ≠ 2. Find fg(x) and gf(x), solve fg(x) = 10 and state the range of fg.

fg(x) = f(3/(x − 2)) = 9/(x − 2)² + 1
gf(x) = g(x² + 1)    = 3/(x² − 1),   x ≠ ±1
fg(x) = 10:  9/(x − 2)² = 9,  (x − 2)² = 1,  x = 1 or x = 3

The term 9/(x − 2)² is positive for every x ≠ 2 and can be as large as you like, so the range of fg is fg(x) > 1.

Inverse functions. A function has an inverse f⁻¹ only if it is one-one. Then:

• inputs and outputs swap roles: f⁻¹ accepts the outputs of f as its inputs and returns the inputs of f, so a restricted domain on f shows up as a restricted range on f⁻¹
• f⁻¹f(x) = ff⁻¹(x) = x
• reflecting y = f(x) in the line y = x produces the graph of y = f⁻¹(x)

To find f⁻¹: set y equal to f(x), rearrange to give x in terms of y, then rename the variable so that f⁻¹ is written in terms of x. A many-one function such as x² has no inverse unless you restrict its domain.

Example 4. f(x) = 3 − √(x + 1) for x ≥ −1. Find the range of f, find f⁻¹(x), and state its domain.

x + 1 ≥ 0, so √(x + 1) ≥ 0 and f(x) ≤ 3.  Range of f: f(x) ≤ 3.
y = 3 − √(x + 1)  →  √(x + 1) = 3 − y  →  x = (3 − y)² − 1
f⁻¹(x) = (3 − x)² − 1,   domain x ≤ 3 (the range of f)

Check with a point: f(8) = 3 − 3 = 0, so (8, 0) is on y = f(x) and (0, 8) is on y = f⁻¹(x). Indeed f⁻¹(0) = 9 − 1 = 8. The range of f⁻¹ is f⁻¹(x) ≥ −1.

To confirm the algebra, check that f⁻¹f(x) = x. Here f⁻¹(f(x)) = (3 − (3 − √(x + 1)))² − 1 = (x + 1) − 1 = x for x ≥ −1.

To sketch the inverse, reflect key points: swap the x- and y-coordinates, swap any horizontal and vertical asymptotes, and keep the curve symmetrical about y = x.

1.3 The modulus function

|x| is the non-negative value of x: |x| = x for x ≥ 0, and |x| = −x for x < 0.

• y = |ax + b|: draw the line y = ax + b, then reflect the part below the x-axis in the x-axis. The graph is a V with its vertex on the x-axis at x = −b/a.
• y = |f(x)|: reflect every part of y = f(x) that lies below the x-axis in the x-axis. Nothing changes where f(x) ≥ 0, and the graph never goes below the axis.
• y = f(|x|): keep the part of y = f(x) with x ≥ 0, delete the part with x < 0, and replace it by the reflection of the kept part in the y-axis. The graph is symmetrical about the y-axis.

Example 5. Sketch y = |4x − 3| and solve |4x − 3| = x + 6 and |4x − 3| > x + 6.

The V has vertex (3/4, 0) and meets the y-axis at (0, 3). The line y = x + 6 has intercept (0, 6) and gradient 1, so it meets the V twice, once on each arm. Solve each branch:

4x − 3 = x + 6          3x = 9           x = 3
−(4x − 3) = x + 6       3 − 4x = x + 6   x = −3/5
Check: x = 3:   |9| = 9 = 3 + 6.   x = −3/5:   |−27/5| = 27/5 = −3/5 + 6.

The V is above the line to the left of the first intersection and to the right of the second. So |4x − 3| > x + 6 when x < −3/5 or x > 3.

Always check each solution in the original equation. When you solve by branches, an answer that does not belong to its branch’s range of x must be rejected.

Example 6. Given f(x) = x² − 3x − 10, sketch y = |f(x)| and y = f(|x|).

f(x) = (x − 5)(x + 2) has roots −2 and 5, y-intercept −10 and minimum point (3/2, −49/4).

y = |f(x)|:  the dip below the axis is reflected.
  x-intercepts (−2, 0) and (5, 0); y-intercept (0, 10);
  local maximum (3/2, 49/4).
y = f(|x|) = x² − 3|x| − 10:  keep x ≥ 0, mirror it in the y-axis.
  x-intercepts (−5, 0) and (5, 0); y-intercept (0, −10);
  minimum points (3/2, −49/4) and (−3/2, −49/4).

The root x = −2 disappears from y = f(|x|), because the part of the curve with x < 0 is discarded. The y-intercept of y = f(|x|) is −10, but that of y = |f(x)| is 10. Label both intercepts with their correct signs.

1.4 Combining transformations

Start from y = f(x). Each change has a standard effect.

Equation	What happens to y = f(x)
y = f(x) + a	translation up by a, vector (0, a)
y = f(x + a)	translation left by a, vector (−a, 0)
y = af(x)	stretch in the y-direction, scale factor a
y = f(ax)	stretch in the x-direction, scale factor 1/a
y = −f(x)	mirror image in the x-axis
y = f(−x)	mirror image in the y-axis

Changes that act on x (inside the brackets) do the opposite of what you expect and change x-coordinates. Changes outside the brackets change y-coordinates as you expect. Asymptotes move in the same way as the curve.

For a combination, work through the order you would use to evaluate the expression for a given x. For y = 2f(x) + 1, stretch first, then translate. For y = 2(f(x) + 1), translate first, then stretch. These two are different graphs. A stretch parallel to the x-axis and one parallel to the y-axis can be done in either order.

Example 7. The curve y = f(x) has a minimum at (−2, −5) and crosses the x-axis at (1, 0) and (4, 0). Find the images of these three points on y = 2f(3x) and y = f(−x) + 1.

y = 2f(3x): x-coordinate ÷ 3, y-coordinate × 2
  (−2, −5) → (−2/3, −10)   (1, 0) → (1/3, 0)   (4, 0) → (4/3, 0)
y = f(−x) + 1: x-coordinate × (−1), then y-coordinate + 1
  (−2, −5) → (2, −4)       (1, 0) → (−1, 1)     (4, 0) → (−4, 1)

Example 8. Sketch y = 3 + sin 2x and y = −cos(x + π/4) for 0 ≤ x ≤ 2π.

• y = 3 + sin 2x: the period is 2π/2 = π, so there are two full waves. The graph oscillates about y = 3, with minimum 2 and maximum 4. It passes through (0, 3).
• y = −cos(x + π/4): translate y = cos x left by π/4, then reflect in the x-axis. The maximum value 1 occurs at x = 3π/4, the minimum value −1 at x = 7π/4, and the zeros are at x = π/4 and x = 5π/4. The y-intercept is −√2/2 and the end value at x = 2π is also −√2/2.

Common errors

• Cancelling a term such as the x² in (x² + 3)/x², which is not a factor of the whole top line.
• Reading fg as “f first”. Work from the inside out: g is applied first.
• Giving the domain of f⁻¹ as the domain of f. It is the range of f.
• Finding an inverse of a many-one function without restricting the domain first.
• Forgetting to test solutions of a modulus equation in the original equation.
• Drawing y = f(|x|) by reflecting the part with x < 0 instead of discarding it.
• Translating y = f(x + 2) to the right. It moves left.
• Applying a stretch and a translation in the wrong order.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019, Unit P3 content section 1 (Algebra and functions), topics 1.1 to 1.4. Pearson Education Limited.