Build and use quadratic models for situations such as projectile motion and area, using the vertex for maximum or minimum values and the zeros for boundary values, and interpreting solutions in context (A.FGR and A.MM, Functional and Graphical Reasoning and Modeling).

What this topic is asking

This standard from Functional and Graphical Reasoning (A.FGR) and Mathematical Modeling (A.MM) asks you to use quadratic functions to model real situations, the two classics being projectile motion (height over time) and area (length times width). The skill is twofold: set up the model, then decide which feature answers the question, the vertex for a maximum or minimum, or the zeros for a boundary like ground level. The Georgia Milestones EOC tests this with two-point constructed-response items that reward shown work, the right feature, rejecting unrealistic solutions, and stating units.

Projectile motion

A projectile's height over time is a downward parabola. A common model (in feet and seconds) is

where $v_0$ is the initial upward velocity and $h_0$ is the initial height. Two features answer the usual questions:

• The maximum height is at the vertex, time $t = \frac{-b}{2a}$, then substitute.
• The object lands (or reaches the ground) when $h(t) = 0$, a zero.

Area models

An area model multiplies two dimensions, often expressed with one variable. A rectangle "4 feet longer than wide" with width $w$ has length $w + 4$ and area $w(w + 4)$. Setting the area equal to a given value gives a quadratic to solve, usually by factoring.

For area 45: $w(w + 4) = 45$, so $w^2 + 4w - 45 = 0$, factoring to $(w + 9)(w - 5) = 0$, giving $w = 5$ (rejecting $w = -9$).

Vertex versus zeros

Choosing the right feature is the heart of the topic.

• Vertex answers an extreme value: maximum height, maximum revenue or profit, minimum cost. Signal words: "how high," "maximum," "least," "greatest."
• Zeros answer a boundary: when something lands, hits the ground, breaks even, or returns to start. Signal words: "when does it land," "hits the ground," "break even."

How the Milestones examines this topic

• Constructed response. Set up a model and find a maximum, a landing time, or a dimension, with full work.
• Numeric entry. Solve an area or projectile equation and report the realistic solution.
• Multiple choice / inline choice. Identify whether the vertex or a zero answers a given question.

Why you reject the unrealistic root

A quadratic almost always has two solutions, but a real context frequently makes one impossible, and discarding it is part of the credit, not an optional nicety. A negative time would be before the ball was thrown, a negative length or width cannot exist, and a negative count of objects is meaningless. So when factoring an area problem gives $w = -9$ or $w = 5$, only $w = 5$ is a valid width. The mathematics produces both roots because the equation is symmetric about its vertex, but the model only applies where the variable makes physical sense. On the EOC, stating which root you reject and why ("a width cannot be negative") demonstrates the interpretive reasoning the constructed-response rubric rewards, and forgetting it costs a point even when the arithmetic is correct.

A full landing-time example

For a complete projectile problem, finding the landing time uses the zeros. With $h(t) = -16t^2 + 48t$, set $h = 0$: factor $-16t(t - 3) = 0$, giving $t = 0$ (the launch) or $t = 3$ (the landing), so the ball lands at 3 seconds. The $t = 0$ root is real but corresponds to the start, so the landing answer is the positive nonzero root. When the quadratic does not factor, use the quadratic formula and keep only the positive time. Recognizing that one zero is often the trivial starting moment, and the other is the event you want, is the modeling judgment that turns a correct solve into a correct answer.

Try this

Q1. For $h(t) = -16t^2 + 64t$, when does the object land? [2 points]

• Cue. $-16t(t - 4) = 0$, so $t = 0$ or $t = 4$; it lands at $t = 4$ seconds.

Q2. A rectangle is 3 m longer than wide with area 40. Find the width. [2 points]

• Cue. $w(w + 3) = 40 \Rightarrow w^2 + 3w - 40 = 0 \Rightarrow (w + 8)(w - 5) = 0$; width 5 m.