An object thrown upward follows a parabolic height, because gravity applies a constant downward acceleration. The vertex gives the greatest height and when it occurs; the positive root gives the landing time; the y-intercept is the starting height. Context limits the domain.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
What the roots mean
h = −5t² + 20t = −5t(t − 4): roots t = 0 and t = 4. The first is the launch, the second the landing. A negative root, in another model, is mathematically valid and physically meaningless.
From a height
h = −5t² + 20t + 25 starts at 25 m. Landing: −5t² + 20t + 25 = 0, t² − 4t − 5 = 0, (t − 5)(t + 1) = 0, t = 5. The peak is still at t = 2, now 45 m.
Area problems
40 m of fence for a rectangle: w and 20 − w. A = w(20 − w) = −w² + 20w. Vertex at w = 10: a 10 by 10 square, area 100 m². A fixed perimeter with a variable side gives a quadratic area.
Against a wall
Three sides fenced, the wall as the fourth: w + 2h = 40, w = 40 − 2h. A = h(40 − 2h) = −2h² + 40h. Vertex at h = 10, w = 20: area 200 m², twice the enclosed square.
Context limits the domain
Negative time and negative lengths do not exist. For the ball, 0 ≤ t ≤ 4. For the fence, 0 < w < 20. Say which part of the parabola actually applies; the rest is algebra, not the situation.
Ask what the question wants
A maximum needs the vertex; a landing time needs a root; a height at a given time needs substitution; a time at a given height needs solving. Different points on the same parabola.
Step 2: Try It Yourself
Tap and try it out.
A rocket follows h = −5t² + 30t + 35. Find the starting height, the peak, when it lands, and when it is at 60 m
- 1h(0) = 35 mthe starting height: c
Step 0 of 4
Step 3: In Real Life
A fenced garden
With 40 meters of fence against a wall, the area is w(40 − 2w). The vertex gives the best width, 10 meters, and the largest garden, 200 square meters.
Step 4: Watch an Example
One step at a time.
Watch Sana Find a Maximum Height
A ball follows h = −5t² + 20t, with h in meters and t in seconds.
- Step 1
The leading coefficient is negative, so the vertex is a maximum, at t = −b ÷ 2a = −20 ÷ (−10) = 2 seconds.
Step 5: 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 meters?
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 meters?
5 of 8
A rectangle has perimeter 40. What side length maximizes 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 maximizes the area?
Step 6: 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; the intercept is the start.
- Context restricts which part of the parabola applies.
- Ask what the question wants before computing: vertex, root, or a substitution.