Cogito
Integrated Math 2 · Chapter 1 · Lesson 3
Modelling with Quadratics
Projectiles, areas, and maximum values.
12 problems · about 21 minutes · F-IF.B.4, A-CED.A.1
What this lesson teaches
The student builds quadratic models and interprets the vertex and intercepts in context.
- Projectile height and fixed-perimeter area are both quadratic.
- The vertex gives a maximum or minimum; the roots give where the value is zero.
- Context restricts which part of the parabola applies.
Warm Up
Straightforward practice. Get the method working first.
5 problemsh = −5t² + 40t. At what time is the height greatest, in seconds?
Answer 4
Why 4 seconds.
What does the vertex give in a projectile model?
Answer The greatest height and the time it occurs.
Why The maximum height.
h = −5t² + 20t at t = 1. What is h, in metres?
Answer 15
Why −5 + 20.
h = −5t² + 20t at t = 2. What is h?
Answer 20
Why −20 + 40.
h = −5t² + 30t. At what time is the height greatest, in seconds?
Answer 3
Why −30 ÷ −10.
Build It Up
The same ideas with more to keep track of.
3 problemsh = −5t² + 10t + 8. What is the starting height, in metres?
Answer 8
Why At t = 0.
A rectangle has perimeter 40. What side length maximises the area?
Answer 10
Why A square.
That maximum area, in square units?
Answer 100
Why 10 × 10.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsMatch each feature with its meaning in a height model.
Answer The vertex → Greatest height and when it happens; The positive root → When the object lands; The y-intercept → The starting height
Why Each feature answers a different question about the flight.
A rectangle has perimeter 24. What side maximises the area?
Answer 6
Why A square.
The Peak: h = −5t² + 20t. At what time, in seconds, is the height greatest?
Answer 2 seconds
Why 2 seconds.
The Landing: Same model. At what time does the ball return to the ground, other than t = 0?
Answer 4 seconds
Why 4 seconds.