Form 4 · Relationship and Algebra
Quadratic Functions and Equations in One Variable
Quadratic functions are the first big idea of Form 4, curves, roots and the shape of a parabola. Master them and a lot of later maths gets easier.
What is Quadratic Functions and Equations in One Variable?
This chapter is about quadratic functions of the form y = ax² + bx + c and the equations you get when you set them to zero. You learn to recognise a quadratic, find its roots (the x-values where the curve meets the x-axis), and sketch the parabola, including which way it opens and where its turning point sits.
Content standards (DSKP)
The DSKP KSSM sets these content standards for this chapter:
- 1.1 Quadratic Functions and Equations in One Variable
The key ideas
The shape of a parabola
If a is positive the curve opens upward (a smile); if a is negative it opens downward. The sign of a is the first thing to read.
Roots of a quadratic equation
The roots are where y = 0. You can find them by factorisation, by completing the square, or with the quadratic formula, each is a tool for a different question.
The discriminant b² − 4ac
Its sign tells you how many real roots there are: positive means two, zero means one repeated root, negative means none.
The turning point
Completing the square rewrites the function so the maximum or minimum point can be read straight off, a favourite of Paper 2.
How this chapter is examined
Paper 1 tends to test roots, the discriminant and reading a graph quickly. Paper 2 goes deeper: completing the square to find a turning point, sketching with intercepts labelled, or solving a real-context problem that hides a quadratic inside it.
The chapter reappears indirectly all year, so a shaky start here is felt in later topics too.
How to study this chapter
Common mistakes to avoid
- Forgetting the ± when taking a square root, so one root goes missing
- Sign errors when the coefficient a is negative
- Reading the discriminant’s sign wrong and miscounting the roots
Three routes to the roots, and how to choose
There are three standard ways to solve ax² + bx + c = 0, and picking the right one saves time. Factorising is fastest when the roots are whole numbers or simple fractions, look for two numbers that multiply to ac and add to b.
If the factors do not appear within a few seconds, switch to the quadratic formula x = (−b ± √(b² − 4ac)) / 2a, which is printed on the exam formula sheet and always works, so it is your safe fallback whenever a question asks for the answer to two decimal places (that phrasing is a strong hint the roots are not whole numbers). Completing the square is the third route; it is slower for just finding roots but it is the only method that also hands you the turning point, so reach for it when a question wants both.
Completing the square, step by step
Completing the square rewrites ax² + bx + c in the form a(x − h)² + k, called vertex form, because it shows the vertex (turning point) directly at (h, k). The routine is fixed: first factor a out of the x² and x terms only, leaving c outside; inside the bracket, halve the coefficient of x and square it, adding and subtracting that square so the bracket becomes a perfect square; then tidy the constants.
For example 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3. Once it is in this form you can read the minimum or maximum value straight off (here the smallest y is −3, since a squared bracket is never negative), which is exactly what many Paper 2 turning-point questions are really asking.
Building a quadratic from what you are told
Some questions run the process backwards: instead of a function, you are given clues and must write the quadratic. If you know the roots are p and q, start from y = a(x − p)(x − q), because setting each bracket to zero recovers those roots; the a is a stretch factor you find from one extra point on the curve.
If you know the turning point is (h, k), start from vertex form y = a(x − h)² + k for the same reason. If you are only given that the graph passes through certain points, substitute each point to make equations in a, b and c and solve them together.
The habit to build is to ask what form makes the given information easiest to plug in, roots suggest factor form, a turning point suggests vertex form, and scattered points suggest the general form.
A worked exam-style example
This example ties completing the square, the turning point and the roots into one typical Paper 2 question.
- (a) Factor 2 out of the x-terms only: y = 2(x² − 4x) + 5.
- Half of −4 is −2, and (−2)² = 4, so add and subtract 4 inside the bracket: y = 2[(x² − 4x + 4) − 4] + 5.
- The first three terms are a perfect square: y = 2[(x − 2)² − 4] + 5.
- Expand the 2 and combine constants: y = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3.
- (b) In vertex form h = 2 and k = −3, so the turning point is (2, −3). Since a = 2 is positive, the parabola opens upward, so it is a minimum.
- (c) Set the vertex form to zero: 2(x − 2)² − 3 = 0, so (x − 2)² = 3/2 = 1.5.
- Take the square root of both sides, keeping the ±: x − 2 = ±√1.5 = ±1.2247.
- So x = 2 + 1.2247 = 3.2247 or x = 2 − 1.2247 = 0.7753.
- Check with the quadratic formula: b² − 4ac = (−8)² − 4(2)(5) = 64 − 40 = 24, and x = (8 ± √24) / 4 = (8 ± 4.899) / 4, giving 3.22 and 0.78, the same values.
Study Quadratic Functions and Equations in One Variable
Methods
Key terms
Frequently asked questions
How this chapter is examined
SPM Mathematics assesses this chapter across Mathematics Paper 1 (Objective) and Mathematics Paper 2 (Subjective), drawing on the DSKP content standards above. Paper 2 gives marks for working, so showing every step matters.
Common mistakes to avoid
Forgetting the ± when taking a square root, so one root goes missing; Sign errors when the coefficient a is negative; Reading the discriminant’s sign wrong and miscounting the roots.
Are any Quadratic Functions and Equations in One Variable formulae given in the exam?
This chapter has no formula on the exam formula sheet, the working is expected from memory and method.
When should I factorise and when should I use the formula?
Try factorising first when the roots look like whole numbers or simple fractions. If the factors do not appear within a few seconds, switch to the quadratic formula, it always works and is on the formula sheet.
A clue: if the question asks for the answer to two decimal places, the roots are not whole numbers, so go straight to the formula.
How do I know if the parabola opens up or down?
Look only at a, the coefficient of x². If a is positive, the curve opens upward like a valley and its turning point is a minimum.
If a is negative, it opens downward like a hill and the turning point is a maximum. The b and c terms shift the curve around but never change which way it opens.
My discriminant came out negative, did I make a mistake?
Not necessarily. A negative b² − 4ac means the equation has no real roots, so the parabola never touches the x-axis.
That is a perfectly valid answer in SPM, write 'no real roots'. But a sign slip is the most common cause, so recheck your values of a, b and c before you commit to it.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)· SPM: Format Pentaksiran mulai 2021, Matematik (1449)
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