Solving equations graphically: intersections of graphs

What is it about?

An equation such as $x^2 - 1 = 0.5x + 2$ compares two expressions. Read both sides as functions, $f(x) = x^2 - 1$ and $g(x) = 0.5x + 2$ : you are looking for all $x$ at which both have the same value – exactly where the graphs of $f$ and $g$ intersect (or touch).

Draw the graphs of both sides.
Find the points of intersection.
Read off their $x$ -coordinates – these are the solutions: here $S = \{-1.5,\ 2\}$ .
Check by substituting into both sides: $f(2) = 2^2 - 1 = 3 = 0.5 \cdot 2 + 2 = g(2)$ ✓.

Two intersections mean two solutions, a touching point gives one, no common point gives none and coinciding graphs give infinitely many. Equivalently, the solutions are the zeros of the difference function $h(x) = f(x) - g(x)$ : subtracting $g(x)$ moves every intersection straight down onto the $x$ -axis.

When a solution is not on the grid, as for $x^2 = 2$ , bisection traps it: $h(1) < 0 < h(2)$ , so a solution lies in $[1, 2]$ ; testing the midpoint halves the interval at every step. At a touching point the sign does not change, so bisection fails there.

Try it

Drag the vertex of the parabola or the points on the line. When are there two, one or no solutions? Try a touching point.
Switch on As zeros of h(x) = f(x) − g(x) and watch the intersections drop to the axis. Load example
Load x² = 2 and press Approximate x₂ / Halve the interval several times. The magnifier shows where you are.
Explore the fractional equation 2/(x − 1) = x. Why can $x = 1$ never be a solution?

Tasks

Task 1: Read off and check

Load task Solve $x^2 - 4 = x + 2$ graphically and check.

Show solution

The graphs intersect at $(-2, 0)$ and $(3, 5)$ , so $S = \{-2,\ 3\}$ . Check: $(-2)^2 - 4 = 0 = -2 + 2$ ✓ and $3^2 - 4 = 5 = 3 + 2$ ✓.

Task 2: Fractional equation

Load task State the domain and solve $\dfrac{4}{x} = x + 3$ .

Show solution

$D = \mathbb{R} \setminus \{0\}$ ; the solutions are $-4$ and $1$ (from $x^2 + 3x - 4 = (x + 4)(x - 1) = 0$ ).

Task 3: Trapping a root

Load task Use bisection on $x^2 = 3$ until the first decimal place is certain.

Show solution

$[1, 2] \to [1.5, 2] \to [1.5, 1.75] \to [1.625, 1.75] \to [1.6875, 1.75]$ : not certain yet, since $1.6875$ rounds to $1.7$ but $1.75$ to $1.8$ . Two more steps: $\to [1.71875, 1.75] \to [1.71875, 1.734375]$ . Now both bounds round to $1.7$ , so $x \approx 1.7$ (more precisely $\sqrt{3} \approx 1.732$ ).

Ideas for the classroom

Use: solving equations graphically, fractional equations as an intersection problem, quadratic equations (parabola and line), approximating irrational roots by bisection.
Model: each side is a line, parabola $a(x - d)^2 + e$ , hyperbola $\frac{a}{x - d} + e$ or V-shaped graph $a|x - d| + e$ . Solutions are computed exactly (with square roots for quadratics), cubic cases numerically.
Task mode: show=0 hides intersections and solutions in the picture, _hide=1 hides the results.
Common stumbling blocks: giving the point $(2, 3)$ instead of $x = 2$ ; missing a touching point; forgetting the domain of fractional equations.