Intersection of Two Graphs

Video lesson · from 38:181 worked example

One lesson video covers all of Polynomials & Quadratics, so it opens at 38:18 for this topic — not from the beginning.

Theory

Given two equations of graphs, we can find their points of intersection. We already have experience of this with straight lines (National 5) and quadratics/lines.

To find the intersection, equate the two expressions $f(x) = g(x)$ and solve for $x$ .

Two graphs could have several points of intersection, one point of intersection, or none:

⚠️ Common Examiner Traps

Set the equations equal: intersections happen where the $y$ values match. Bring everything to one side to get a polynomial equal to zero.
Solve fully, then find the $y$ values: the $x$ values are only half the answer. Substitute each back to give coordinates.
Substitute into the simpler equation: usually the straight line — it is faster and less error-prone.
Watch the signs when subtracting: subtracting a whole polynomial changes every one of its terms.

Worked examples

Example 1

Find the coordinates of the points where the curve with equation $y=x^3-x^2$ intersects the line with equation $y=5x+3$ .

Equate the expressions for $y$ :

x^3 - x^2 = 5x + 3

Rearrange into standard polynomial form $f(x)=0$ :

x^3 - x^2 - 5x - 3 = 0

We need to solve this cubic. Let $f(x) = x^3 - x^2 - 5x - 3$ .

Test factors of -3: $\pm 1, \pm 3$ .

Try $x = -1$ :

f(-1) = (-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0

So $(x+1)$ is a factor. Use synthetic division to find the quotient:

-1	1	-1	-5	-3
	-1	2	3
	1	-2	-3	0

x^3 - x^2 - 5x - 3 = (x+1)(x^2 - 2x - 3) = 0

Factorise the quadratic $x^2 - 2x - 3$ :

(x+1)(x-3)(x+1) = 0

(x+1)^2(x-3) = 0

So the x-coordinates are $x = -1$ and $x = 3$ .

Substitute these $x$ values back into the linear equation $y=5x+3$ to find the $y$ values:

For $x = -1$ :

y = 5(-1) + 3 = -2

For $x = 3$ :

y = 5(3) + 3 = 18

The points of intersection are $(-1, -2)$ and $(3, 18)$ .