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Math · Integrated Math 2

Chapter 1: Quadratic Functions

Modelling with Quadratics

Projectiles, areas, and maximum values.

Lesson
3
Time
About 21 minutes
0 of 12 done

Step 1: Let's Learn

Read it, or press Listen and follow the words.

An object thrown upward follows a parabolic height, because gravity applies a constant downward acceleration.

What the vertex means

For a projectile the vertex gives the greatest height and the time it occurs. For an area problem it gives the best dimensions.

What the roots mean

For a height model the positive root is when the object lands. A negative root is mathematically valid and physically meaningless.

What the intercept means

The y-intercept is the starting height, the value before any time has passed.

Area problems

A fixed perimeter with a variable side gives a quadratic area, and its vertex is the maximum.

Context limits the domain

Negative time and negative lengths do not exist. Say which part of the parabola actually applies.

Projectiles are quadratic

Height under constant gravity is a quadratic function of time. The vertex gives the maximum height and the positive root gives the landing time, so both questions come from the same equation.

Area problems produce quadratics

A fixed perimeter with variable dimensions gives area as a quadratic in one side. The vertex gives the maximum area, which turns out to be at the squarest shape available.

The context restricts the domain

A negative time or a negative length is mathematically permitted and physically meaningless. Stating the sensible domain is part of a complete model, and it discards half the roots in most problems.

Interpret each feature

The vertex is the maximum or minimum, the y-intercept is the starting value, and the roots are where the quantity reaches zero. Naming what each means in the situation is the point of modelling.

Step 2: Try It Yourself

Tap and try it out.

Set a negative and b positive. That downward parabola is the height of a thrown object against time.
-8-8-6-6-4-4-2-222446688
y = -1x² + 6x + 0

Step 3: Watch an Example

One step at a time.

Watch Sana Find a Maximum Height

A ball follows h = −5t² + 20t, with h in metres and t in seconds.

  1. Step 1

    The leading coefficient is negative, so the vertex is a maximum.

Step 4: Your Turn

Practice makes it stick.

The Peak

Problem 1 of 2

h = −5t² + 20t. At what time, in seconds, is the height greatest?

seconds

The Landing

Problem 2 of 2

Same model. At what time does the ball return to the ground, other than t = 0?

seconds

Model It

1 of 8

h = −5t² + 20t at t = 1. What is h, in metres?

2 of 8

h = −5t² + 20t at t = 2. What is h?

3 of 8

h = −5t² + 30t. At what time is the height greatest, in seconds?

4 of 8

h = −5t² + 10t + 8. What is the starting height, in metres?

5 of 8

A rectangle has perimeter 40. What side length maximises the area?

6 of 8

That maximum area, in square units?

7 of 8

Match each feature with its meaning in a height model.

Tap a card on the left to start.

8 of 8

A rectangle has perimeter 24. What side maximises the area?

Step 5: Quick Check

Show what you know.

Question 1 of 2

h = −5t² + 40t. At what time is the height greatest, in seconds?

Question 2 of 2

What does the vertex give in a projectile model?

What You Learned

  • 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.