IGCSE · Practice questions · Solving equations graphically

Draw the line, read the answer.

Six original Cambridge-style questions on solving equations graphically: roots, horizontal lines, choosing a line, intersections and interpreting solutions.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Match the curve, draw the line, read the \(x\)-values.

Roots: \(f(x) = 0\) is solved where \(y = f(x)\) meets the \(x\)-axis.
Equal to a number: \(f(x) = k\) is solved where the curve meets the horizontal line \(y = k\).
New equation: add or subtract the same terms on both sides until one side is exactly the drawn curve.
The other side is the line to draw; plot it with two or three points.
Answer with the \(x\)-coordinates of every crossing, to the accuracy of the graph (usually 1 d.p.).

Recall

[3 marks]

(a) Complete a table of values for \(y = x^2 - x - 6\) for \(x = -3, -2, \dots, 4\).

(b) Use your graph to solve \(x^2 - x - 6 = 0\).

(a) \(y\): \(6, 0, -4, -6, -6, -4, 0, 6\). ✓
(b) The curve crosses the \(x\)-axis at \(x = -2\) and \(x = 3\). ✓

Recall

[2 marks]

The graph of \(y = x^2 - x - 6\) is drawn. Write down the equation of the line you would draw to solve (a) \(x^2 - x - 6 = 4\) (b) \(x^2 - 3x - 2 = 0\).

(a) The left side is the curve, so draw \(y = 4\). ✓
(b) Add \(2x - 4\) to both sides: \(x^2 - 3x - 2 + 2x - 4 = 2x - 4\), so \(x^2 - x - 6 = 2x - 4\). Draw \(y = 2x - 4\). ✓

Calculation

[3 marks]

The line \(y = 2x - 2\) is drawn on the graph of \(y = x^2 - x - 6\). Find, by algebra, the coordinates of the points where they meet, so that you can check your graph.

\(x^2 - x - 6 = 2x - 2\), so \(x^2 - 3x - 4 = 0\). ✓
\((x - 4)(x + 1) = 0\), so \(x = 4\) or \(x = -1\). ✓
\(y = 2(4) - 2 = 6\) and \(y = 2(-1) - 2 = -4\): the points are \((4, 6)\) and \((-1, -4)\). ✓

Analysis

[2 marks]

The lowest point of the curve \(y = x^2 - x - 6\) is \((0.5, -6.25)\).

Find the value of \(k\) for which \(x^2 - x - 6 = k\) has exactly one solution, and state the number of solutions when \(k = -7\).

Exactly one solution when the line \(y = k\) just touches the lowest point: \(k = -6.25\). ✓
\(k = -7\) is below the lowest point, so there are no solutions. ✓

Analysis

[3 marks]

(Extended) The graphs of \(y = \dfrac{4}{x}\) and \(y = 5 - x\) are drawn for \(x > 0\).

Show that at the points where they meet, \(x^2 - 5x + 4 = 0\). Hence find the coordinates of these points.

\(\dfrac{4}{x} = 5 - x\). Multiply by \(x\): \(4 = 5x - x^2\). ✓
Rearrange: \(x^2 - 5x + 4 = 0\). ✓
\((x - 1)(x - 4) = 0\), so \(x = 1\) or \(x = 4\). ✓
\(y = 5 - 1 = 4\) and \(y = 5 - 4 = 1\): the points are \((1, 4)\) and \((4, 1)\). ✓

Analysis

[4 marks]

A ball is thrown upwards. Its height, \(h\) metres, after \(t\) seconds is \(h = 20t - 5t^2\).

(a) Complete a table of values for \(t = 0, 1, 2, 3, 4\) and draw the graph.

(b) Draw the line \(h = 15\) and use it to find for how long the ball is more than 15 m high.

(a) \(h\): \(0, 15, 20, 15, 0\). The graph is a ∩-shaped parabola with its highest point at \((2, 20)\). ✓
(b) The line \(h = 15\) meets the curve at \(t = 1\) and \(t = 3\) (check: \(5t^2 - 20t + 15 = 0\), \(t^2 - 4t + 3 = 0\), \((t - 1)(t - 3) = 0\)). ✓
The ball is above 15 m between these times: \(3 - 1 = 2\) seconds. ✓
(c) \(h = 0\) again at \(t = 4\), so the ball lands after 4 seconds. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.