What does the Functions and Graphs strand of O-Level E-Maths cover?

This part of SEAB 4052 covers the straight-line graph and the form y equals mx plus c (gradient and intercept); the quadratic function and its parabola, with intercepts, turning point and line of symmetry; the standard non-linear shapes (cubic, reciprocal and exponential) and their features; solving equations graphically by intersection; and interpreting distance-time and speed-time travel graphs using the gradient and the area under the graph. It turns algebra into pictures and lets you read solutions off a curve.

What do the gradient and area under a travel graph represent?

On a distance-time graph the gradient gives the speed, and a horizontal section means the object is stationary. On a speed-time graph the gradient gives the acceleration (a negative gradient is deceleration), and the area under the graph gives the distance travelled. Knowing which quantity comes from the gradient and which from the area is the single most important idea for travel-graph questions.

How do you find the equation of a straight line?

Use y equals mx plus c. Find the gradient m as the change in y divided by the change in x between two points, then substitute one point to find the intercept c. If you are given the gradient and one point, substitute straight away. Parallel lines have equal gradients, which lets you find a line parallel to a given one through a stated point.

What are the key features of a quadratic graph?

The graph of a quadratic is a parabola. It is u-shaped when the coefficient of x squared is positive and n-shaped (an inverted u) when it is negative. The x-intercepts are the roots of the equation (where y equals zero), the y-intercept is the constant term, and the curve has a vertical line of symmetry through the turning point, which is the minimum or maximum value of the function.

How do you solve an equation graphically?

Draw the graph or graphs involved and read off where they meet. To solve f of x equals zero, find where the curve crosses the x-axis. To solve f of x equals g of x, draw both graphs and read the x-coordinates of their intersection points. Because you are reading from a drawn graph, answers are estimates, so give them to a sensible accuracy and use a straight edge for tangents when estimating a gradient.

SingaporeMaths

O-Level E-Maths Functions and Graphs: straight-line graphs, quadratic functions, standard non-linear curves, solving equations graphically, and travel graphs

An overview of the O-Level E-Maths Functions and Graphs strand (SEAB 4052). Straight-line graphs and the equation y equals mx plus c, the parabola of a quadratic function, the standard cubic, reciprocal and exponential curves, solving equations graphically, and reading speed, acceleration and distance from travel graphs, with links to every dot point.

Generated by Claude Opus 4.88 min readSEAB-4052Updated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this strand is about
The straight line
Quadratic functions and standard curves
Reading solutions from graphs
Travel graphs
How the strand is examined
Check your knowledge

What this strand is about

Functions and graphs turns the algebra of the previous strands into pictures. Being able to picture a curve, its intercepts, symmetry and gradient, makes equation-solving and modelling far easier, and travel graphs connect mathematics directly to motion in the real world. This overview ties the strand together and links to every dot point, each with worked answers and practice.

See the full set of dot points at /sg-o-level/mathematics/syllabus.

The straight line

The strand starts with linear functions and straight-line graphs. The equation $y = mx + c$ packs two facts into one line: $m$ is the gradient (the steepness, change in $y$ over change in $x$ ) and $c$ is the $y$ -intercept (where the line crosses the $y$ -axis). You find a line's equation from two points or from a point and the gradient, and parallel lines share the same gradient.

Quadratic functions and standard curves

Quadratic functions and their graphs introduces the parabola: u-shaped when the coefficient of $x^2$ is positive, n-shaped when negative. You find the $x$ -intercepts (the roots), the $y$ -intercept, the turning point and the line of symmetry. Graphs of functions and curve sketching extends the gallery to cubic, reciprocal and exponential curves, with their characteristic shapes, asymptotes and symmetry, so you can recognise a function from its graph and vice versa.

Reading solutions from graphs

Graphical solution of equations uses these curves to solve equations: the roots of $f(x) = 0$ are where the curve crosses the $x$ -axis, and the solutions of $f(x) = g(x)$ are the $x$ -coordinates of the intersection of the two graphs. Because the answers are read off a drawn graph, they are estimates given to a sensible accuracy, and a tangent drawn with a straight edge lets you estimate the gradient at a point.

Travel graphs

Gradient and area under graphs is the applied finish. On a distance-time graph the gradient is the speed; on a speed-time graph the gradient is the acceleration and the area under the graph is the distance travelled. Keeping straight which quantity comes from the gradient and which from the area is the heart of these questions.

How the strand is examined

Label gradient and intercept correctly. In $y = mx + c$ , identify which number is the gradient and which is the intercept before writing an equation.
Find every key feature of a curve. For a quadratic, give intercepts, turning point and line of symmetry; for other curves, note asymptotes and symmetry.
State graphical answers as estimates. Reading from a drawn graph gives approximate values; quote them to a reasonable accuracy and draw tangents carefully when finding a gradient.

Check your knowledge

Attempt these, then check the solutions.

Find the gradient and $y$ -intercept of the line $2y = 6x - 8$ . (2 marks)
The quadratic $y = x^2 - 4x + 3$ is given. Find its $x$ -intercepts and the line of symmetry. (3 marks)
A distance-time graph shows a journey of $120\ \text{m}$ covered in $30\ \text{s}$ at constant speed. Find the speed. (2 marks)
On a speed-time graph, a car accelerates uniformly from rest to $20\ \text{m/s}$ in $8\ \text{s}$ . Find the distance travelled in that time. (2 marks)
State the shape of the graph of $y = \dfrac{6}{x}$ for positive $x$ , and name its asymptotes. (2 marks)

Solutions

Step 1: Rearrange into $y = mx + c$ to read off gradient and intercept (Q1)

The equation $2y = 6x - 8$ is not yet in the standard form $y = mx + c$ , so divide every term by $2$ first. Once in this form, the coefficient of $x$ is the gradient and the constant term is the $y$ -intercept.

2y = 6x - 8 \implies y = 3x - 4.

The gradient is $3$ and the $y$ -intercept is $-4$ .

Final answer: gradient $= 3$ ; $y$ -intercept $= -4$

Step 2: Factorise to find $x$ -intercepts, then use symmetry for the turning point (Q2)

The $x$ -intercepts occur where $y = 0$ , so factorise the quadratic and apply the zero-product property. The line of symmetry of a parabola lies exactly halfway between the two $x$ -intercepts, so average their values.

x^2 - 4x + 3 = (x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3.

Line of symmetry: $x = \dfrac{1 + 3}{2} = 2$ .

Final answer: $x$ -intercepts at $x = 1$ and $x = 3$ ; line of symmetry $x = 2$

Step 3: Read the speed as the gradient of a distance-time graph (Q3)

On a distance-time graph the gradient represents the speed, because speed is distance divided by time. A constant speed produces a straight-line graph, and its gradient equals that constant speed.

\text{speed} = \frac{120}{30} = 4\ \text{m/s}.

Final answer: $4\ \text{m/s}$

Step 4: Find the distance as the area under a speed-time graph (Q4)

On a speed-time graph the area enclosed between the graph and the time axis equals the distance travelled. Uniform acceleration from rest produces a straight line, so the enclosed region is a triangle whose area is one half times base times height.

\text{distance} = \frac{1}{2} \times 8 \times 20 = 80\ \text{m}.

Final answer: $80\ \text{m}$

Step 5: Identify the shape and asymptotes of a reciprocal curve (Q5)

The graph of $y = \dfrac{6}{x}$ is a rectangular hyperbola. As $x$ increases without bound, $y$ approaches zero but never reaches it, so the $x$ -axis is a horizontal asymptote. As $x$ approaches zero from the positive side, $y$ increases without bound, so the $y$ -axis is a vertical asymptote.

Final answer: The graph is a reciprocal curve (hyperbola) in the first quadrant, decreasing for positive $x$ . Asymptotes are the $x$ -axis ( $y = 0$ ) and the $y$ -axis ( $x = 0$ ).

Sources & how we know this

Singapore-Cambridge GCE O-Level Mathematics (Syllabus 4052) — Singapore Examinations and Assessment Board (2026)