SingaporeMathsSyllabus dot point

How do we use graphs to solve equations, and how do we read solutions off a curve?

Solve equations graphically by finding intersection points, and estimate solutions and gradients from a drawn curve

A focused answer to the O-Level E-Maths outcome on solving equations from graphs. Reading roots from where a curve crosses the axis, solving by intersection of two graphs, and estimating the gradient of a curve from a tangent.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

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

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

What this dot point is asking

SEAB wants you to solve equations using graphs, by reading where a curve crosses an axis or where two graphs intersect, and to estimate the gradient of a curve at a point by drawing a tangent. Graphical methods give approximate solutions to equations that may be hard to solve exactly.

The answer

Solving f(x) = 0 from a graph

The solutions (roots) of $f(x) = 0$ are the values of $x$ where the graph of $y = f(x)$ crosses the $x$ -axis, because there $y = 0$ . Reading these crossing points off a carefully drawn curve gives the solutions to the accuracy of the graph.

Solving by intersection of two graphs

To solve an equation of the form $f(x) = g(x)$ , draw $y = f(x)$ and $y = g(x)$ on the same axes. The solutions are the $x$ -coordinates of the points where the two graphs intersect, since there the two expressions are equal.

Choosing the line to add

Often a curve $y = f(x)$ is already drawn and you are asked to solve a related equation. Rearrange the equation so that one side equals $f(x)$ and the other side is a simple straight line $y = mx + c$ ; then drawing that line and reading the intersections gives the solutions without redrawing the curve.

Estimating the gradient at a point

The gradient of a curve at a point changes from place to place. To estimate it, draw a tangent (a straight line just touching the curve at that point), pick two clear points on the tangent, and compute its gradient as rise over run. The tangent's gradient is the curve's gradient there.

Worked example

The curve $y = x^2$ is drawn. Show how to solve $x^2 = x + 2$ graphically, and state the expected solutions.

Step 1: Identify the straight line to add

Keep the curve $y = x^2$ and write the right side as a line: $y = x + 2$ . The solutions occur where $x^2 = x + 2$ , that is where the curve meets this line.

Step 2: Draw the line y = x + 2

This line has gradient $1$ and intercept $2$ , crossing the $y$ -axis at $(0, 2)$ .

Step 3: Read the intersection x-coordinates

The curve and line meet at two points. Reading across to the $x$ -axis gives the solutions, which should be near $x = 2$ and $x = -1$ .

Step 4: Confirm algebraically

Solving $x^2 - x - 2 = 0$ gives $(x - 2)(x + 1) = 0$ , so $x = 2$ or $x = -1$ , matching the graphical reading.

Examples in context

Example 1. Break-even point. Plotting a cost line and a revenue line on the same axes, the intersection gives the break-even quantity where cost equals revenue. Reading the meeting point answers a business question graphically.

Example 2. Estimating a rate of change. On a curve showing the volume of water in a tank over time, a tangent drawn at a moment gives the instantaneous rate of flow at that moment. The tangent's gradient estimates how fast the volume is changing right then.

Try this

Q1. The graph of $y = x^2 - 4$ is drawn. State how to read off the solutions of $x^2 - 4 = 0$ . [1 mark]

Cue. Read the $x$ -coordinates where the curve crosses the $x$ -axis, namely $x = 2$ and $x = -2$ .

Q2. To solve $x^2 = 3x$ using the curve $y = x^2$ , what line should you draw? [1 mark]

Cue. The line $y = 3x$ ; the intersections give the solutions $x = 0$ and $x = 3$ .

Q3. A tangent to a curve passes through $(1, 2)$ and $(3, 8)$ . Estimate the gradient of the curve where the tangent touches. [2 marks]

Cue. $\dfrac{8 - 2}{3 - 1} = \dfrac{6}{2} = 3$ .

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 graph of

y = x^2 - 2x - 2

is drawn. (a) Explain how the graph is used to solve

x^2 - 2x - 2 = 0

. (b) By adding a suitable straight line, explain how the same graph could be used to solve

x^2 - 2x - 2 = x

Show worked answer →

(a) The solutions of $x^2 - 2x - 2 = 0$ are the $x$ -coordinates where the curve $y = x^2 - 2x - 2$ crosses the $x$ -axis (the line $y = 0$ ). Read these values off the graph.

(b) Rearranging is not needed if we draw the line $y = x$ on the same axes. The solutions of $x^2 - 2x - 2 = x$ are the $x$ -coordinates of the points where the curve and the line $y = x$ intersect.

Markers reward identifying the roots as the $x$ -axis crossings, and recognising that adding the line $y = x$ and reading the intersections solves the second equation.

Original3 marksA curve is drawn and a tangent is constructed at the point

(2, 5)

. The tangent passes through

(0, 1)

. Estimate the gradient of the curve at the point

(2, 5)

Show worked answer →

The gradient of the curve at a point equals the gradient of the tangent drawn there.

Using the two points on the tangent $(2, 5)$ and $(0, 1)$ : gradient $= \dfrac{5 - 1}{2 - 0} = \dfrac{4}{2} = 2$ .

So the gradient of the curve at $(2, 5)$ is approximately $2$ .

Markers reward using the tangent's two points and computing rise over run to estimate the gradient as $2$ .