Home / Graphical Methods / Solving equations graphically
3.11c · Graphical Methods · Sub-skill
Use the intersections of graphs to solve equations — including choosing which line to draw.
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Build it up, step by step
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To solve $f(x) = g(x)$, draw $y = f(x)$ and $y = g(x)$. The $x$-coordinates of the points where they cross are the solutions. For $f(x) = 0$, the solutions are where the curve crosses the $x$-axis (the roots).
Given the graph of $y = x^2 - 2x - 3$, to solve $x^2 - 2x - 3 = 2$ you draw the horizontal line $y = 2$. To solve a different equation, rearrange it so one side matches the curve, e.g. $x^2 - 2x - 8 = 0$ becomes $x^2 - 2x - 3 = 5$, so draw $y = 5$.
Count the intersections: a horizontal line can cut a $\cup$-shaped parabola twice (above the minimum), once (at the minimum) or not at all (below it). A line can cut a cubic up to three times.
Answers read from a graph are approximate — give them to a sensible accuracy (usually 1 or 2 decimal places) and say 'from the graph'. A larger, more careful graph gives better accuracy.
Use the graph of $y = x^2 - 2x - 3$ to solve $x^2 - 2x - 3 = 2$.
Draw the line $y = 2$. It crosses the curve at $x \approx -1.45$ and $x \approx 3.45$.
So the solutions are $x \approx -1.45$ and $x \approx 3.45$. (Exactly, $x = 1 \pm \sqrt{6}$.)
Test yourself
The graph of $y = x^2 - 4x + 1$ is drawn.
(a) Use the graph to estimate the solutions of $x^2 - 4x + 1 = 0$.
(b) By drawing a suitable straight line, solve $x^2 - 4x - 2 = 0$. [5 marks]
(a) Where the curve crosses the $x$-axis: $x \approx$ 0.27 and 3.73. (2 marks)
(b) Rearrange: $x^2 - 4x + 1 = 3$. Draw the line $y = 3$; it meets the curve at $x \approx$ −0.45 and 4.45. (3 marks: rearranging, line, solutions)
Practice
Five short questions. Work through them, then reveal the mark scheme to check.