What does the Functions and Graphs strand of N(A)-Level Mathematics cover?

This strand of SEAB 4045 (Mathematics Syllabus A) covers the linear function $y = mx + c$, where $m$ is the gradient and $c$ the $y$-intercept, plotted from a table of values and read off the equation; the quadratic function $y = ax^2 + bx + c$, whose graph is a parabola with intercepts, a line of symmetry and a turning point; and distance-time and travel graphs, where the gradient of the line represents speed.

What do $m$ and $c$ mean in $y = mx + c$?

In the straight-line equation $y = mx + c$, the letter $m$ is the gradient, which measures steepness as the change in $y$ divided by the change in $x$: a positive $m$ rises from left to right and a negative $m$ falls. The letter $c$ is the $y$-intercept, the value of $y$ where the line crosses the $y$-axis (where $x = 0$). You can read both directly from the equation.

How do I tell whether a parabola opens upward or downward?

Look at the coefficient $a$ in $y = ax^2 + bx + c$. If $a$ is positive, the parabola opens upward and has a minimum turning point (a valley shape). If $a$ is negative, it opens downward and has a maximum turning point (a hill shape). The larger the size of $a$, the narrower the curve.

What does the gradient of a distance-time graph tell you?

On a distance-time graph, time is on the horizontal axis and distance on the vertical axis, so the gradient (change in distance over change in time) is the speed. An upward slope means moving away from the start, a flat horizontal line means resting, a downward slope means returning, and a steeper line means a faster speed.

How do I find the average speed for a whole journey?

Average speed for an entire journey is the total distance travelled divided by the total time taken, including any time spent resting. It is not the average of the separate speeds. Remember to convert minutes to hours before quoting a speed in kilometres per hour, since $30$ minutes is $0.5$ hours, not $0.30$ hours.

SingaporeMaths

N(A)-Level Mathematics Functions and Graphs: linear graphs and gradient, quadratic graphs, and distance-time and travel graphs

An overview of the N(A)-Level Mathematics Functions and Graphs strand (SEAB 4045). Plotting and reading linear graphs $y = mx + c$ and their gradient and intercept, sketching quadratic parabolas $y = ax^2 + bx + c$ , and interpreting distance-time and travel graphs where gradient is speed, with links to every dot point.

Generated by Claude Opus 4.89 min readSEAB-4045Updated 2026-06-13

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

Jump to a section

Why graphs matter
Linear graphs and gradient
Quadratic graphs
Distance-time and travel graphs
Check your knowledge

Why graphs matter

Graphs turn algebra into a picture you can read. In N(A)-Level Mathematics (SEAB 4045, Mathematics Syllabus A), the Functions and Graphs strand teaches you to move both ways: to plot a function from its equation, and to read meaning back out of a curve, whether that is the speed of a car or the solutions of an equation. This overview links to every dot point in the module, each with its own worked answers and practice.

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

Linear graphs and gradient

Linear graphs and gradient cover the straight-line function $y = mx + c$ . The gradient $m$ is the change in $y$ over the change in $x$ (positive for a rising line, negative for a falling one), and $c$ is the $y$ -intercept where the line crosses the $y$ -axis. To draw the line, make a small table of values, plot the points, and join them with a ruler. To read its equation, take the gradient and the intercept straight off the line.

Quadratic graphs

Quadratic graphs cover the function $y = ax^2 + bx + c$ , whose graph is a U-shaped parabola. It opens upward (a minimum point) when $a$ is positive and downward (a maximum point) when $a$ is negative. The $x$ -intercepts are the solutions of $ax^2 + bx + c = 0$ , the $y$ -intercept is $c$ , and the turning point lies on the line of symmetry exactly halfway between the two $x$ -intercepts. This is the visual link back to the quadratic equations you learned to solve.

Distance-time and travel graphs

Distance-time and travel graphs put time on the horizontal axis and distance on the vertical axis, so the gradient of each segment is a speed. An upward slope is travelling away, a flat line is resting, a downward slope is returning, and a steeper line is faster. The average speed for the whole journey is the total distance divided by the total time, rests included.

Reading speed from a distance-time graph

A cyclist rides from home, covering $12$ km in the first $30$ minutes, then rests for $15$ minutes, before returning the $12$ km home in $45$ minutes. Find the average speed for the whole journey in km/h.

Step 1: Find the total distance

The cyclist travels $12$ km out and $12$ km back, so the total distance is $12 + 12 = 24$ km.

Step 2: Find the total time in hours

The journey takes $30 + 15 + 45 = 90$ minutes. Converting, $90$ minutes $= \dfrac{90}{60} = 1.5$ hours.

Step 3: Divide total distance by total time

\text{average speed} = \frac{24}{1.5} = 16 \text{ km/h}.

The average speed for the whole journey, including the rest, is

16

km/h.

Check your knowledge

A mix of linear, quadratic and travel-graph questions covering the strand. Attempt them, then check the solutions.

State the gradient and $y$ -intercept of $y = 3x - 4$ . (2 marks)
A line passes through $(0, 2)$ and $(4, 10)$ . Find its equation. (2 marks)
For $y = x^2 - 4$ , state the coordinates of the $y$ -intercept and both $x$ -intercepts. (3 marks)
Does the graph of $y = -2x^2 + 5$ have a maximum or a minimum point? Explain. (1 mark)
A car travels $90$ km in $1.5$ hours, rests for $0.5$ hours, then travels a further $60$ km in $1$ hour. Find the average speed for the whole journey. (2 marks)

Solutions

These five questions cover linear equations, quadratic intercepts, and average speed. Work through each part, then check your reasoning against the steps below.

Step 1: Read the gradient and intercept from Q1

The equation $y = 3x - 4$ is already in the form $y = mx + c$ . The number multiplying $x$ is the gradient and the constant term is the $y$ -intercept, so you can read both values directly without any rearrangement.

m = 3, \quad c = -4.

Final answer (Q1): gradient $3$ , $y$ -intercept $(0, -4)$ .

Step 2: Find the equation of the line in Q2

The $y$ -intercept is the value of $y$ when $x = 0$ , which is the point $(0, 2)$ , so $c = 2$ . The gradient is the rise divided by the run between the two given points:

m = \frac{10 - 2}{4 - 0} = \frac{8}{4} = 2.

Substituting into $y = mx + c$ gives the equation.

Final answer (Q2): $y = 2x + 2$ .

Step 3: Find the intercepts of the quadratic in Q3

For the $y$ -intercept, substitute $x = 0$ into $y = x^2 - 4$ :

y = 0^2 - 4 = -4, \quad \text{so the } y\text{-intercept is } (0, -4).

For the $x$ -intercepts, set $y = 0$ and solve $x^2 - 4 = 0$ , which gives $x^2 = 4$ , so $x = \pm 2$ .

Final answer (Q3): $y$ -intercept $(0, -4)$ ; $x$ -intercepts $(-2, 0)$ and $(2, 0)$ .

Step 4: Determine whether Q4 has a maximum or minimum

A parabola $y = ax^2 + bx + c$ opens upward (minimum) when $a > 0$ and downward (maximum) when $a < 0$ . Here $a = -2$ , which is negative, so the parabola opens downward and the turning point is at the top.

Final answer (Q4): maximum point, because $a = -2 < 0$ .

Step 5: Compute the average speed in Q5

Average speed is total distance divided by total time for the whole journey, including any rests. Add up all the distances and all the time intervals:

\text{total distance} = 90 + 60 = 150 \text{ km}, \qquad \text{total time} = 1.5 + 0.5 + 1 = 3 \text{ hours}.

\text{average speed} = \frac{150}{3} = 50 \text{ km/h}.

Final answer (Q5): $50$ km/h.

Sources & how we know this

Singapore-Cambridge GCE N(A)-Level Mathematics (Syllabus A, 4045) — Singapore Examinations and Assessment Board (2026)