Quadratic Function Applications
Model projectile motion, area optimization, and revenue/profit scenarios with quadratic functions. Set up the quadratic model, find the vertex for the max/min, and solve for zeros to find break-even points or landing times.
A lesson of Algebra 1, in Quadratic Functions, about 35 minutes.
Quadratic Applications
By the end of this lesson, you will be able to build a quadratic model from a described situation, decide whether the question wants the vertex or the zeros, compute the right one, and reject the answers that make no physical sense.
Two Questions, Two Tools
Almost every applied quadratic asks one of two things, and knowing which decides everything you do next.
"What is the greatest or least?" wants the vertex.
$$x = -\frac{b}{2a} \quad \text{then substitute}$$
Maximum height, maximum profit, largest area, minimum cost. The vertex is where the curve turns, so it is the only candidate.
"When is it zero?" wants the zeros.
$$ax^2 + bx + c = 0$$
When does the ball land, when does profit break even, when does the population die out. Solve by factoring or the formula.
Read the question before doing any algebra. A problem asking "what is the maximum height" and one asking "when does it hit the ground" use the same model and different halves of it, and the commonest lost mark in the topic is answering the other one.
Sort It: Vertex or Zeros
Projectile Motion
Anything thrown, dropped or launched follows the same model, because gravity pulls at a constant rate.
$$h(t) = -16t^2 + v_0 t + h_0 \qquad \text{(feet, seconds)}$$
$$h(t) = -4.9t^2 + v_0 t + h_0 \qquad \text{(meters, seconds)}$$
- $v_0$ is the starting upward speed
- $h_0$ is the starting height, so $h(0) = h_0$ as it should
- the leading coefficient is gravity, and it is negative, which is why the parabola opens down and there is a maximum rather than a minimum
The $-16$ is half of $32$ ft/s², the acceleration of gravity. It is not a number chosen to make the algebra tidy; it is physics arriving in the coefficient, and it is why $a$ is the same in every one of these problems.
$$t = -\frac{v_0}{2(-16)} = \frac{v_0}{32}$$
That is the time at the top. And note carefully: that is a time, not a height. Substitute it back to get how high.
Example 1: A Thrown Ball
Here is every step on one problem, worked all the way through.
A ball is thrown upward at 64 ft/s from ground level, so $h(t) = -16t^2 + 64t$.
- What is a? -16
- What is b? 64
- Find -b over 2a. At what time, in seconds, is the ball highest? 2
- Substitute t = 2. What is the maximum height, in feet? 64