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Solving equations with graphs

Use graph intersections to solve equations, count real solutions, recognise tangencies and rearrange an equation to use a known graph.

Before you startCubic, quadratic and reciprocal graphs; solving equations

01 / Intersections

Equal outputs give a solution.

To solve f(x) = g(x), draw y = f(x) and y = g(x) on the same axes. The x-coordinate of each intersection is a solution. Substitute back if the whole coordinate pair is required.

In the model, change the horizontal line y = c. Then switch to the difference graph y = x³ − 3x − c. Its zeros have exactly the same x-coordinates as the intersections.

f(x) = g(x) ⇔ f(x) − g(x) = 0

All notes remain here while you choose a view. A touching point is an intersection too.

y = x³ − 3x and y = 0Count solutions

Level c

The graphs y = x³ − 3x and y = 0 meet at three points. Their x-coordinates are −√3, 0 and √3.

Watch intersections become zeros of a difference

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Exact coordinates

Use the sketch to guide the algebra.

A sketch indicates how many solutions to expect and roughly where they lie. Exact answers need algebra unless the question requests a graphical estimate.

When a common factor appears, move everything to one side and factorise. Dividing by x can discard an intersection at x = 0.

A cubic meets a quadraticWorked example

y = x³ − x and y = x² + x

Equate the two outputs.

x³ − x² − 2x = 0
x(x − 2)(x + 1) = 0

Keep the factor x.

x = −1, 0, 2

Find the matching y-values in either equation.

Intersections:
(−1, 0), (0, 0), (2, 6)

Substitution checks each pair in both graphs.

03 / Use a known graph

Rearrange toward the graph you already have.

Suppose y = x³ − 3x is already drawn. To solve x³ − 4x − 1 = 0, rearrange to x³ − 3x = x + 1 and add the line y = x + 1.

You could instead plot y = x³ − 4x − 1 and read its roots, but that would require a different cubic. Choose a rearrangement that uses the given curve.

Given y = x², solve x² − 2x − 3 = 0:
draw y = 2x + 3.
Given y = x³, solve x³ + 2x = 5:
draw y = 5 − 2x.

An equation can have several valid graphical interpretations. State both graph equations and explain why their intersection inputs solve the original equation.

04 / Reciprocal equations

Clear a denominator without forgetting the domain.

For 6/x = x + 1, the input must satisfy x ≠ 0. Multiplying by x is valid on that domain and gives x² + x − 6 = 0.

(x + 3)(x − 2) = 0
Intersections: (−3, −2), (2, 3)

Both candidate inputs are allowed and satisfy the original equation. If multiplication produces an excluded input in another problem, reject it.

Graphs with unknown positive constants

For a, b > 0, y = −a/x and y = (x − b)² have exactly one intersection. There are none for x > 0 because the first is negative and the second non-negative. For x < 0, −a/x increases from 0 towards +∞, while (x − b)² decreases from +∞ towards b². Continuity and these opposite monotonic directions give exactly one meeting.

Thus x(x − b)² = −a has one real solution, and it is negative. x = 0 cannot solve it because a > 0.

The squared denominator changes the countWorked example

4/x² = x − 1, x ≠ 0

The left side is positive, so a solution requires x > 1.

For x > 1:
4/x² decreases; x − 1 increases

The first starts above the second, and ends below it.

Exactly one intersection

Continuity gives a meeting; opposite monotonic directions rule out a second.

Check x = 2: 4/4 = 2 − 1

The exact intersection is (2, 1).

05 / Count carefully

A tangency still counts once.

The cubic y = x³ − 3x has a local maximum at (−1, 2) and a local minimum at (1, −2). These labelled turning points can be used without doing calculus here.

−2 < c < 2: three distinct intersections with y = c.
c = −2 or c = 2: two distinct intersections, including a tangency.
c < −2 or c > 2: one intersection.

x³ − 3x = 2
x³ − 3x − 2 = (x − 2)(x + 1)²

Inputs −1 and 2 give two distinct real solutions. The repeated factor at −1 counts twice algebraically but is a single point.

A rough sketch alone cannot certify a count when two curves are almost tangent or intersections lie outside the window. Support it using signs, monotonic behaviour, factorisation or a discriminant where appropriate.

06 / Estimate and check

A decimal read from a graph is approximate.

For x³ − 3x = 3, the model shows one crossing near x = 2.10. It is an estimate, not the exact value.

9/x² = x − 2

Hint

Any solution needs x > 2.

Worked solution

For x ≤ 2 with x ≠ 0, the right side is non-positive and the left positive. For x > 2, the left side decreases from 9/4 towards 0 and the right side increases from 0 without bound. There is exactly one meeting. It is (3, 1), because 9/3² = 3 − 2.

06 · A parameter and tangency

How many intersections do y = x² + t and y = 4x have?

Hint

Complete the square in their difference.

Worked solution

x² − 4x + t = 0
(x − 2)² = 4 − t

Two for t < 4; one for t = 4; none for t > 4. At tangency the point is (2, 8).

07 · Exact surd pairs

y = x² − 1
y = 2x + 3

Hint

Solve the equality and substitute each root.

Worked solution

x² − 2x − 4 = 0
x = 1 ± √5

The intersections are (1 + √5, 5 + 2√5) and (1 − √5, 5 − 2√5). Each y sign stays paired with the matching x sign.

08 / Recap

An intersection is an equality made visible.

Equate outputs at the same input.
Rearrange to use the supplied graph.
Include tangencies; do not discard a zero factor.
Retain denominator restrictions.
Use algebra for exact coordinates and label estimates honestly.

Next: graph transformations →

Section 1 of 8 · Intersections