OxfordAQA International AS and A-level Mathematics: Transformations and rational functions (9660) – Revision Notes

These revision notes condense the second half of section P2.1 (Algebra and functions) in the OxfordAQA International AS and A-level Mathematics specification, Version 5.2. The specification places this content in International A-level (A2) Unit P2 (Pure Maths); it is not examined at International AS. For full explanations and worked examples, read the study guide.

Related pages: the OxfordAQA Mathematics hub, the printable checklist, the practice questions and the earlier Unit P1 polynomials notes. Find gaps with the free 10-minute diagnostics.

Transformations at a glance

Equation	Description
y = af(x)	stretch, scale factor a, parallel to y-axis
y = f(x) + a	translation (0, a)
y = f(x + a)	translation (−a, 0)
y = f(ax)	stretch, scale factor 1/a, parallel to x-axis
y = −f(x)	reflection in the x-axis
y = f(−x)	reflection in the y-axis

Method in steps for a combined equation

1. Write the new equation as a function of the old one, for example 3f(2x + 1) − 4.
2. Treat the inside (x-direction) changes as the opposite of what they look like: add 1 means move left, multiply by 2 means halve the x-distances.
3. Treat the outside (y-direction) changes as they look: multiply by 3, subtract 4.
4. For an inside expression such as 2x − 1, you can either translate by 1 first and then halve the x-distances, or write 2(x − 1/2), halve first and then translate by 1/2.
5. On the y side, stretch first and then translate for 3f(x) + 2. Translate first only if the equation is 3(f(x) + 2).

Tracking a point. To find the image of (h, k) under y = a f(bx + c) + d, solve bx + c = h for x, then compute a k + d.

Spec examples worked as reminders

Start	Finish	Description
e^x	e^(2x − 1)	translate (1, 0), then stretch factor 1/2 in x. Asymptote y = 0.
ln x	2 ln(x − 1)	translate (1, 0), then stretch factor 2 in y. Asymptote x = 1, x-intercept (2, 0).
sec x	3 sec 2x	stretch factor 1/2 in x, stretch factor 3 in y. Period π. No values with −3 < y < 3.

Factor Theorem and Remainder Theorem

Result	Statement
Remainder Theorem	Remainder on dividing f(x) by (ax + b) is f(−b/a)
Factor Theorem	(ax + b) is a factor of f(x) if and only if f(−b/a) = 0
Rational function	p(x)/q(x), undefined where q(x) = 0

Method in steps for finding constants

1. Turn each condition into an equation: factor means f(−b/a) = 0, remainder r means f(−b/a) = r.
2. Solve the simultaneous equations.
3. Divide out the known factor to factorise the rest.
4. Check the discriminant of any remaining quadratic.

Small reminder. Is (3x + 1) a factor of 3x³ + 4x² − 17x − 6? f(−1/3) = −1/9 + 4/9 + 51/9 − 54/9 = 0, so yes, and f(x) = (3x + 1)(x − 2)(x + 3).

Simplifying rational expressions

1. Factorise numerator and denominator fully, using the Factor Theorem for cubics.
2. Cancel common factors, never common terms.
3. State excluded values if the question asks.
4. To add or subtract, use the lowest common denominator, then simplify the numerator.

Example: (x² − 9)/(x² + x − 6) = (x − 3)(x + 3)/((x − 2)(x + 3)) = (x − 3)/(x − 2).

Algebraic division

Method	Best used when
Inspection	divisor is linear and the numerator is linear or a simple quadratic
Equating coefficients	you want a quick, tidy quotient: write f(x) = (divisor)(quotient) + remainder and match powers
Formal division	cubic or quartic over a linear or quadratic divisor
Given identity A + (Bx + C)/(…)	same-degree fractions, found by matching coefficients

Always finish with a check: expand (divisor)(quotient) + remainder and compare.

Example by inspection: (7x − 2)/(x + 1) = 7 − 9/(x + 1), because 7x − 2 = 7(x + 1) − 9.

Partial fractions

Denominator	Form
(ax + b)(cx + d)	A/(ax + b) + B/(cx + d)
(ax + b)(cx + d)²	A/(ax + b) + B/(cx + d) + C/(cx + d)²
Numerator degree ≥ denominator degree	quotient + remainder form, then split

Method in steps

1. Compare degrees. If the numerator is not smaller, divide first.
2. Write the form with unknown constants.
3. Multiply both sides by the full denominator.
4. Substitute x-values that make each bracket zero, which gives the single-factor constants.
5. Equate a coefficient (usually x²) or use a spare value for the rest.
6. Check with a value of x not used before.

Irreducible quadratic factors are not tested.

Example: (4x + 1)/((x − 2)(x + 1)) = 3/(x − 2) + 1/(x + 1).

Worked reminders

Tracking a point. (3, 4) lies on y = f(x). On y = 2f(x − 1) + 5, solve x − 1 = 3 to get x = 4. Then y = 2 × 4 + 5 = 13. The image is (4, 13).

Dividing a cubic by a linear divisor. Divide 5x³ − 3x² + x + 6 by (x + 2). Write 5x³ − 3x² + x + 6 = (x + 2)(ax² + bx + c) + d. Matching x³: a = 5. Matching x²: b + 2a = −3, so b = −13. Matching x: c + 2b = 1, so c = 27. Matching constants: 2c + d = 6, so d = −48. The result is 5x² − 13x + 27 − 48/(x + 2). Check with the Remainder Theorem: the cubic at x = −2 is −40 − 12 − 2 + 6 = −48.

Repeated factor. Express (3x² − 14x − 1)/((x + 1)(x − 3)²) in partial fractions. Write the form A/(x + 1) + B/(x − 3) + C/(x − 3)². Then 3x² − 14x − 1 ≡ A(x − 3)² + B(x + 1)(x − 3) + C(x + 1). Put x = −1: 16 = 16A, so A = 1. Put x = 3: −16 = 4C, so C = −4. Compare x²: 3 = A + B, so B = 2. The result is 1/(x + 1) + 2/(x − 3) − 4/(x − 3)².

Must-know distinctions

• f(x + 2) moves the graph left 2; f(x) + 2 moves it up 2.
• f(2x) halves x-distances; 2f(x) doubles y-distances.
• Remainder uses f(−b/a); the divisor’s sign flips.
• Remainder (a number) is different from the quotient (a polynomial).
• A repeated factor needs one fraction for each power up to the repeat.

Wording to recognise

• “Describe” a transformation: name it (translation, stretch, reflection), give the vector or scale factor, and state the axis or direction.
• “Show that”: write every step, including the substitution and a final statement such as “so (3x − 2) is a factor”.
• “Hence”: use the result you have just found, such as a factor from the Factor Theorem, rather than starting again.
• “Exact”: leave surds, fractions and logarithms unevaluated.
• “Express in partial fractions”: give the final sum, not only the constants A, B and C.

Quick self-test

1. Find the remainder when 2x³ − x + 5 is divided by (2x + 1).
2. Show that (2x + 3) is not a factor of 2x³ + 3x² − 4x + 6 and give the remainder.
3. Describe the transformation from y = f(x) to y = f(x + 3) − 4.
4. The curve y = f(x) passes through (4, −3). Where is this point on y = 3f(x/2) + 1?
5. Describe, in order, how to obtain y = 3 − f(x) from y = f(x).
6. Simplify (x² − 9)/(x² + x − 6).
7. Write (7x − 2)/(x + 1) in the form A + B/(x + 1).
8. Divide x³ − 2x² + 3x + 4 by (x − 1), giving quotient and remainder.
9. Express (4x + 1)/((x − 2)(x + 1)) in partial fractions.
10. State, without evaluating any constants, the form for (x² + 1)/((x − 1)²(x + 3)).

Answers

1. f(−1/2) = −1/4 + 1/2 + 5 = 21/4.
2. f(−3/2) = −27/4 + 27/4 + 6 + 6 = 12, not 0, so it is not a factor. The remainder is 12.
3. Translation by (−3, −4).
4. Solve x/2 = 4 to get x = 8, and 3(−3) + 1 = −8. Image: (8, −8).
5. Reflect in the x-axis, then translate by (0, 3).
6. (x − 3)/(x − 2).
7. 7 − 9/(x + 1).
8. Quotient x² − x + 2, remainder 6.
9. 3/(x − 2) + 1/(x + 1).
10. The degree is below 3, so no division. The form is A/(x − 1) + B/(x − 1)² + C/(x + 3).

Where marks are usually lost

• Describing f(x + a) as a move to the right.
• Giving a stretch factor of 2 for f(2x), when the factor is 1/2.
• Applying transformations in the wrong order when two act on the same axis.
• Using f(b/a) in the Factor Theorem, and so missing a sign change.
• Cancelling a term instead of a factor in a rational expression.
• Leaving out the remainder when dividing, or writing the quotient only.
• Starting partial fractions on an improper fraction without dividing first.
• Writing a repeated factor as a single fraction.
• Not checking the final partial fractions by substituting a value of x.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. For International AS exams from May/June 2018 and International A-level exams from May/June 2019. Section P2.1: Algebra and functions.