Year 12 Functions and relations: Intersection of two graphs

An intersection point is where two graphs share the same x-value and the same y-value. That means solving an intersection problem usually comes down to setting the two function rules equal and solving the resulting equation. The graph gives the geometric meaning, but the algebra gives the exact answer. Year 12 questions often use this idea to connect graphing, solving equations, and interpretation in one step.

If the graphs of

y = f (x)

and

y = g (x)

intersect, then

f (x) = g (x)

at those intersection x-values.

Main method

Set the two expressions equal.
Solve the resulting equation for x.
Substitute back to find the matching y-value.
Interpret how many intersections the graphs have.

Worked example 1

Problem: Find the points of intersection of

y = x^{2}

and

y = x + 2

Set the functions equal: $x^{2} = x + 2$ .
Rearrange: $x^{2} - x - 2 = 0$ .
Factor: $(x - 2) (x + 1) = 0$ .
So $x = 2$ or $x = - 1$ .
Substitute into $y = x + 2$ .
For $x = 2$ , $y = 4$ .
For $x = - 1$ , $y = 1$ .

Answer: the intersection points are

(2, 4)

and

(- 1, 1)

Worked example 2

Problem: Determine whether

y = x^{2} + 1

and

y = - x + 1

intersect.

Set the functions equal: $x^{2} + 1 = - x + 1$ .
Simplify: $x^{2} + x = 0$ .
Factor: $x (x + 1) = 0$ .
So $x = 0$ or $x = - 1$ .
Substitute into $y = - x + 1$ .
This gives points $(0, 1)$ and $(- 1, 2)$ .

Answer: yes, the graphs intersect at

(0, 1)

and

(- 1, 2)

Tangent intersections and repeated solutions

Not every intersection question produces two different points. Sometimes one graph just touches another before turning away. Algebraically, this often appears as a repeated factor or repeated root. Graphically, it means the two functions share a point without properly crossing there. Recognising this possibility is important because it explains why an equation can have one repeated solution even when both graphs are still meeting.

Worked example 3

Problem: Find the intersection of

y = x^{2}

and

y = 2 x - 1

Set the rules equal: $x^{2} = 2 x - 1$ .
Rearrange: $x^{2} - 2 x + 1 = 0$ .
Factor: ${(x - 1)}^{2} = 0$ .
So the only x-value is $x = 1$ .
Substitute back: $y = 2 (1) - 1 = 1$ .
The repeated factor shows the line is tangent to the parabola at that point.

Answer: the graphs touch at exactly one point,

(1, 1)

Common traps

Solving for x but forgetting to find the corresponding y-values.
Substituting into the wrong formula and creating an arithmetic inconsistency.
Reading a graph as if the intersection count must match what "looks" right without algebra.

Revision focus

Intersection questions are one of the clearest examples of functions moving between graph and equation. The graph tells you how many solutions may exist, while the equation gives the exact coordinates. A good solution respects both views. If the algebra gives two intersections but your sketch only allows one, one of them needs to be checked again.

It is also useful to remember that intersection problems are often disguised equation problems. If two curves meet, they are sharing a common output. Setting those outputs equal is the whole idea.

Practice links

Intersection of two graphs