Model real-world situations with quadratic equations and interpret the solutions, including projectile motion and area problems (TN A1.A.REI.B.4, A1.A.CED.A.1).

What this topic is asking

Quadratic applications combine the solving methods with the modeling standards: build or use a quadratic model, solve it, and interpret the result in context. The two staples are projectile motion (height over time) and area problems. The key decisions are whether the question asks for a vertex (a maximum or minimum) or a zero (a start or end), and which solutions are viable.

Projectile motion

A falling-or-thrown object's height in feet follows $h(t) = -16t^2 + v_0 t + h_0$, where $v_0$ is the initial upward velocity and $h_0$ the initial height. The $-16$ comes from gravity. Two questions dominate:

• When does it land (or reach a height)? Set $h(t)$ equal to that height (often $0$) and solve the quadratic.
• What is the maximum height and when? Find the vertex: the time is $t = \frac{-b}{2a}$, and the height is $h$ at that time.

Area problems

For an area model, translate "the length is $5$ more than the width" into $w$ and $w + 5$, set the product equal to the area, and solve. A negative dimension is always rejected.

How TNReady examines this topic

• Numeric response. Find a landing time, a maximum height, or a dimension.
• Multiple choice. Identify the vertex time, the maximum value, or which root is viable.
• Multi-part items. Find a value, then interpret what it represents.

A clarifying idea is that the words choose the tool: "how high," "maximum," "least" point to the vertex; "when does it land," "hits the ground," "breaks even" point to a zero. Reading the question for these cues prevents solving for the wrong thing.

Why a root gets rejected

A quadratic almost always has two solutions, but a real situation rarely allows both. Time before launch is negative, and a negative length or count cannot exist, so one root is discarded as nonviable. This is not optional cleanup; it is part of the interpretive credit the EOC awards. For the landing problem above, the formula produces a negative time as well, but only the positive time is physically meaningful, so it is the answer. The habit to build is: solve fully, then ask of each root, "could this happen in the real scenario?" and report only those that can, with units. A surprising number of lost points in this topic come from reporting a mathematically correct but physically impossible root, or from reporting the launch time $t = 0$ when the question asked for the landing time.

Revenue and other optimization models

Beyond motion and area, the EOC uses quadratics for revenue and other optimization. When revenue is price times quantity and one of them depends linearly on the other, the product is a quadratic whose vertex gives the maximum. For example, if a shop sells $q = 100 - 5p$ items at price $p$, revenue is $R = p(100 - 5p) = 100p - 5p^2$, and the price that maximizes revenue is the vertex at $p = \frac{-100}{2(-5)} = 10$ dollars. The structure is the same as projectile motion: a downward parabola whose peak answers the "maximum" question. Recognizing that "maximum revenue," "maximum area for a fixed perimeter," and "maximum height" are all vertex questions lets you reuse one method across very different contexts.

Try this

Q1. A rocket's height is $h(t) = -16t^2 + 64t$. When does it land? [2 points]

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

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 (reject $-8$).