SingaporeMathsSyllabus dot point

What shape does a quadratic function make, and how do we read its turning point and intercepts?

Draw graphs of quadratic functions y = ax^2 + bx + c, identify the shape, the turning point and the x-intercepts

A focused answer to the N(A)-Level Mathematics outcome on quadratic graphs. The parabola shape, the effect of the sign of a, the turning point, the line of symmetry, and the x-intercepts.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to draw the graph of a quadratic function $y = ax^2 + bx + c$ , recognise its shape (a parabola), and identify its turning point, line of symmetry and the points where it crosses the $x$ -axis. Quadratic graphs link directly to quadratic equations, because the $x$ -intercepts are the solutions of the equation.

The answer

The shape of a quadratic graph

A quadratic function always graphs to a smooth U-shaped curve called a parabola. The sign of $a$ (the coefficient of $x^2$ ) decides which way it opens:

If $a$ is positive, the parabola opens upward and has a lowest point (a minimum).
If $a$ is negative, the parabola opens downward and has a highest point (a maximum).

Plotting the curve

To draw a parabola, build a table of values across a sensible range of $x$ , then join the points with a smooth curve (not straight segments):

Choose several $x$ values either side of where you expect the turning point.
Work out $y$ for each, taking care with squares of negatives.
Plot and join with a single smooth curve.

For $y = x^2$ : the points $(-2, 4)$ , $(-1, 1)$ , $(0, 0)$ , $(1, 1)$ , $(2, 4)$ trace the basic U-shape.

The turning point and line of symmetry

The turning point is the lowest point of an upward parabola or the highest point of a downward one. The parabola is symmetrical about a vertical line through the turning point, called the line of symmetry. If the curve crosses the $x$ -axis at two points, the line of symmetry sits exactly halfway between them.

The x-intercepts

The curve crosses the $x$ -axis where $y = 0$ . Setting $ax^2 + bx + c = 0$ and solving (by factorising or the formula) gives these crossing points. A parabola may cross the $x$ -axis at two points, touch it at one, or not reach it at all.

The y-intercept

The curve crosses the $y$ -axis where $x = 0$ , which gives $y = c$ . So the constant term is the $y$ -intercept, read off directly.

Worked example

For the curve $y = x^2 - 2x - 3$ , find the $x$ -intercepts, the line of symmetry and the turning point.

Step 1: Find the x-intercepts

Set $y = 0$ : $x^2 - 2x - 3 = 0$ . Factorise to $(x - 3)(x + 1) = 0$ , so $x = 3$ or $x = -1$ . The intercepts are $(3, 0)$ and $(-1, 0)$ .

Step 2: Find the line of symmetry

It is halfway between the intercepts:

x = \frac{3 + (-1)}{2} = \frac{2}{2} = 1.

Step 3: Find the turning point

Substitute $x = 1$ into the equation:

y = 1^2 - 2(1) - 3 = 1 - 2 - 3 = -4.

Step 4: State the result

The turning point is $(1, -4)$ . Since the $x^2$ term is positive, the parabola opens upward and this point is a minimum.

Examples in context

Example 1. A thrown ball. The height of a ball is modelled by $y = -x^2 + 4x$ , where $y$ is height and $x$ is time. Because the $x^2$ term is negative, the graph is a downward parabola with a maximum - the highest point of the throw. Setting $y = 0$ gives $x = 0$ and $x = 4$ , the launch and landing times. Quadratics naturally model paths that rise then fall.

Example 2. Linking graph and equation. To solve $x^2 - x - 6 = 0$ graphically, draw $y = x^2 - x - 6$ and read where it crosses the $x$ -axis. The crossings at $x = -2$ and $x = 3$ are exactly the solutions of the equation, showing that the graph and the equation describe the same thing from two angles.

Try this

Cue. State whether $y = -2x^2 + 3$ has a maximum or minimum. The $x^2$ coefficient is negative, so it has a maximum.
Cue. Find the $x$ -intercepts of $y = x^2 - 9$ . Set $y = 0$ : $(x - 3)(x + 3) = 0$ , so $x = 3$ or $x = -3$ .
Cue. A parabola crosses the $x$ -axis at $x = 2$ and $x = 6$ . Its line of symmetry is at $x = \dfrac{2 + 6}{2} = 4$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksThe curve

y = x^2 - 4x + 3

is a parabola. (a) Find where it crosses the

x

-axis. (b) State whether it has a minimum or maximum point.

Show worked answer →

(a) The curve crosses the $x$ -axis where $y = 0$ , so solve $x^2 - 4x + 3 = 0$ .

Factorise: $(x - 1)(x - 3) = 0$ , so $x = 1$ or $x = 3$ .

The curve crosses the $x$ -axis at $(1, 0)$ and $(3, 0)$ .

(b) The coefficient of $x^2$ is positive, so the parabola opens upward and has a minimum point.

What markers reward: setting $y = 0$ and solving for the intercepts, both crossing points written as coordinates, and using the sign of the $x^2$ coefficient to decide minimum versus maximum.

Original3 marksA parabola

y = x^2 - 6x + 8

has its line of symmetry halfway between its

x

-intercepts at

x = 2

and

x = 4

. Find the coordinates of its turning point.

Show worked answer →

The line of symmetry is halfway between the intercepts: $x = \dfrac{2 + 4}{2} = 3$ .

Find $y$ at $x = 3$ : $y = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1$ .

So the turning point is $(3, -1)$ . Since the $x^2$ term is positive, this is a minimum point.

What markers reward: the line of symmetry as the midpoint of the intercepts, substituting to find the $y$ -coordinate, and the turning point as a pair of coordinates.