4.4 Points of Intersection

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Learning Objectives

  • Understand what it means for two graphs to intersect
  • Find the points of intersection algebraically by solving equations
  • Use substitution or simultaneous equations to find coordinates
  • Interpret intersection points in context (e.g. solutions to equations)
  • Sketch graphs and mark intersections clearly

Key Concepts

What Are Points of Intersection?

Points of intersection are where two graphs meet — that is, they have the same x and y values at the same time.

Algebraically, we find them by **solving the equations simultaneously**.

Graphically, they are the points where the curves or lines cross.

Finding Points of Intersection Algebraically

Set the two equations equal to each other, since you're looking for when they have the same output.

Then solve the resulting equation to find the x-values, and substitute back to get the y-values.

Example 1

Find the points of intersection between y = x² and y = 2x + 3.

Solution:

Step 1: Set the equations equal: x² = 2x + 3Step 2: Rearrange: x² − 2x − 3 = 0Step 3: Factor: (x − 3)(x + 1) = 0 ⇒ x = 3 or x = −1Step 4: Substitute: y = 2(3) + 3 = 9 and y = 2(−1) + 3 = 1Points of intersection: (3, 9) and (−1, 1)

Using Substitution or Elimination

You can use substitution (plug one equation into the other) or elimination (combine equations) to solve algebraically.

Substitution is particularly useful when one equation is already in terms of y or x.

Example 1

Find the point of intersection between y = x + 4 and y = −x + 2

Solution:

Set x + 4 = −x + 2 ⇒ 2x = −2 ⇒ x = −1Substitute: y = −1 + 4 = 3Intersection point: (−1, 3)

Applications of Points of Intersection

Points of intersection often represent **solutions to equations** or **real-world crossover points** (e.g. when two quantities are equal).

You can also estimate them from graphs when exact algebraic solving is too difficult.

Example 1

A ball is thrown and its height is modelled by y = −5t² + 20t. A platform is at height y = 15. When does the ball hit the platform?

Solution:

Set the equations equal: −5t² + 20t = 15Rearrange: −5t² + 20t − 15 = 0Solve: t = 0.84 or t = 3.56 (approx.)So the ball hits the platform at t ≈ 0.84s and again at t ≈ 3.56s

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