Why is the coefficient always -16 in these height problems?

It comes from gravity. The distance an object falls under gravity is 1/2gt^2, and g 32\ ft/s^2, so half of that is 16. It is negative because gravity pulls downward while positive height points upward. If a problem uses meters, gravity is about 9.8\ m/s^2 and the coefficient becomes -4.9.

How do I know whether to find the vertex or solve for the roots?

Look at what the question asks. Words like maximum, greatest, highest, peak, or "best dimensions" point to the vertex. Words like lands, hits the ground, or "when is the height zero" point to the roots. If the question names a specific nonzero height or area, set the function equal to that number instead and solve.

Can a projectile problem have two valid times as an answer?

Yes, and it usually does when you ask for a specific height below the peak. The object passes that height once going up and once coming down, so h(t) = 40 typically has two solutions in the feasible domain, and both should be reported. Only when the height equals the maximum is there a single time.

How do I write the feasible domain for an area problem?

Every dimension in the picture must be positive. Write each dimension as an expression in your variable, set each one greater than zero, and combine the results. For a pen with sides x and 60-2x, you need x>0 and 60-2x>0, giving 0<x<30.

ALG1-8.5

Quadratic Models & Word Problems

Learn to build quadratic models for projectile height and area problems, interpret vertex, roots, and y-intercept in context, and reject impossible solutions.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Quadratic Models & Word Problems, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

You already know how to graph a parabola, factor a quadratic, complete the square, and use the quadratic formula. This lesson is where all of that pays off: you take a real situation — a ball thrown off a bleacher, a rectangular dog run built against a garage wall — write a quadratic that describes it, and then translate the algebra back into sentences about the world.

The key shift is that every number in your answer now carries a meaning and a unit. The vertex is not just a point, it is the highest the ball ever gets and the moment it gets there. A root is not just a solution, it is the instant the ball lands. And because time cannot be negative and a fence cannot have negative length, some perfectly good algebraic solutions must be thrown out. Learning when to keep an answer and when to reject it is the heart of this lesson.

Building a Projectile-Height Model

Near Earth's surface, an object that is thrown, dropped, or launched and then left alone follows the height model

h(t) = -16t^2 + v_0 t + h_0

where

h

is height in feet,

t

is time in seconds after release,

v_0

is the initial upward velocity in feet per second, and

h_0

is the initial height in feet. The

-16

comes from gravity (half of

32\ \mathrm{ft/s^2}

) and never changes in feet-and-seconds problems. If the problem uses meters, the leading coefficient is

-4.9

instead.

Reading a word problem means matching each phrase to a slot. "Thrown upward at 48 feet per second" gives

v_0 = 48

. "From the top of a 64-foot cliff" gives

h_0 = 64

. "Dropped" means

v_0 = 0

. "Thrown downward at 20 feet per second" means

v_0 = -20

, because the initial motion opposes the positive (upward) direction.

Two mistakes show up constantly. The first is swapping

v_0

and

h_0

— the number attached to "per second" is always the velocity, the number attached to a place (roof, cliff, hand height) is always the initial height. The second is thinking the graph shows the path of the ball through the air. It does not. If the ball goes straight up and comes straight down, the parabola is a picture of height versus time, not a picture of the ball's flight path. A ball thrown straight up has a parabolic

h

-versus-

t

graph even though it travels along a vertical line.

What the Vertex, Roots, and y-Intercept Mean

Once the model is written, three features carry almost all the meaning. Because the leading coefficient is negative in height problems, the parabola opens down and the vertex is a maximum.

Feature	How to find it	Meaning in a height problem
$y$ -intercept $(0, h_0)$	Evaluate $h(0)$	Height at the moment of release
Vertex $\left(-\frac{b}{2a},\ h\!\left(-\frac{b}{2a}\right)\right)$	Axis of symmetry, then substitute	Time of the peak and the maximum height
Positive root	Solve $h(t)=0$	Time the object hits the ground
Negative root	Solve $h(t)=0$	No physical meaning — reject it

The single most common error is giving the wrong coordinate. If a question asks "what is the maximum height?", the answer is the

y

-coordinate of the vertex. If it asks "when does the object reach its maximum height?", the answer is the

x

-coordinate. Saying "the maximum height is 1.5 seconds" is not a complete answer, and it signals that the units were never checked.

Solving

h(t) = 0

answers "when does it land?" only because ground level is height zero. If the object lands on a 10-foot-high roof instead, you solve

h(t) = 10

. Similarly, "when is the ball at least 50 feet high?" is a question about the interval between the two solutions of

h(t) = 50

, not about the roots.

Always attach units to the final sentence: "The ball reaches a maximum height of 100 feet, 1.5 seconds after release."

Area and Enclosure Models

The other classic quadratic situation is area. Area is a product of two lengths, and when a fixed amount of fencing or border material ties those lengths together, the product becomes quadratic.

Suppose you have 40 feet of fencing to enclose a rectangular garden against a barn, so only three sides need fence. Let

x

be the length of each side perpendicular to the barn. Then the two perpendicular sides use

2x

feet, leaving

40 - 2x

feet for the side parallel to the barn. The area is

A(x) = x(40 - 2x) = -2x^2 + 40x.

Notice the model is already in factored form, so the roots are visible:

x = 0

and

x = 20

. Both give zero area, which makes sense — a garden with no width or no length encloses nothing. The vertex sits halfway between the roots at

x = 10

, giving the maximum area

A(10) = 200

square feet.

Border problems work the same way. If a 6-by-8-foot photo gets a uniform border of width

x

, the outer dimensions are

6 + 2x

and

8 + 2x

, so the total area is

(6+2x)(8+2x)

and the border area alone is

(6+2x)(8+2x) - 48

. The factor of

2x

trips students up: the border is added to both ends of each dimension.

Defining your variable in writing before you build the expression prevents most errors here. "Let

x

= the width in feet of each side perpendicular to the barn" tells you and your teacher exactly what the answer 10 refers to.

The Domain of the Situation and Rejecting Solutions

An algebraic quadratic accepts every real input. A real situation does not. The feasible domain is the set of inputs that actually make sense, and part of a complete answer is stating it and using it.

For

h(t) = -16t^2 + 48t + 64

, time starts at release, so

t \ge 0

, and the model stops describing anything after the object hits the ground, so the domain is

0 \le t \le 4

. When the quadratic formula hands you

t = 4

and

t = -1

, you keep 4 and reject

-1

with a reason: negative time is before the ball was thrown.

For the fencing model

A(x) = -2x^2 + 40x

, a side length must be positive and the parallel side

40 - 2x

must also be positive, so

0 < x < 20

. If solving

A(x) = 150

gave

x = 5

and

x = 15

, both survive the domain check, and both are genuine answers describing two different gardens with the same area.

The habit to build is this: solve completely, then test each solution against the situation, then write a sentence. Rejecting a solution requires a reason, not just crossing it out. Common reasons include negative time, negative length, a dimension larger than the available material, and a width that would make another dimension negative.

Students also lose track of whether a rounded answer still fits. If a landing time comes out as

t \approx 4.27

seconds, do not round to 4 and claim the ball lands exactly at 4 seconds — check

h(4)

and see that the ball is still in the air.

A Checklist for Any Quadratic Word Problem

The same five moves work whether the topic is height, area, or revenue.

First, define the variable in words with units. Second, write the equation, either by plugging into

h(t) = -16t^2 + v_0t + h_0

or by building a product of expressions. Third, decide what the question is asking for — a maximum (vertex), a specific height or area (set the function equal to that number), a landing time (set equal to zero), or a starting value (evaluate at zero). Fourth, solve using whichever tool from this unit is cleanest: factoring when the numbers cooperate, square roots when there is no linear term, the quadratic formula always. Fifth, check the feasible domain and write a sentence with units.

Question wording	Set up	Answer is
"How high does it start?"	Evaluate at $t=0$	A height
"What is the maximum height?"	Vertex	$y$ -coordinate
"When does it peak?"	Vertex	$x$ -coordinate
"When does it hit the ground?"	$h(t)=0$	Positive root
"When is it 40 feet up?"	$h(t)=40$	Usually two times
"What dimensions maximize area?"	Vertex of $A(x)$	$x$ -value, then the other side

A question like "when is the ball 40 feet high?" often has two answers — once on the way up and once on the way down — and a complete response reports both. If the quadratic formula produces a negative discriminant for a target height, that height is never reached, which is itself a meaningful answer about the situation.

Key terms

Quadratic model.: A function of the form $f(x)=ax^2+bx+c$ with $a \ne 0$ used to describe a real situation such as projectile height or enclosed area.
Initial height ( $h_0$ ).: The height of an object at $t=0$ , the moment of release; it is the $y$ -intercept of the height model.
Initial velocity ( $v_0$ ).: The speed and direction of the object at release, in feet per second; positive when thrown upward, negative when thrown downward, zero when dropped.
Vertex.: The turning point of a parabola, located at $x=-\frac{b}{2a}$ . For a downward-opening height model it gives the time of the peak and the maximum height.
Root (zero).: An input that makes the function equal zero. In a height model the positive root is the landing time; in an area model the roots are dimensions giving zero area.
Feasible domain.: The set of input values that make sense in the situation, such as $0 \le t \le 4$ for a flight or $0 < x < 20$ for a fenced side length.
Rejected solution (out of domain).: An algebraically valid solution that falls outside the feasible domain, such as a negative time or a negative length, and must be discarded with a stated reason. This is not the same as an extraneous solution: a rejected solution genuinely satisfies the equation and is thrown out because it does not fit the situation, whereas an extraneous solution does not satisfy the original equation at all.
Axis of symmetry.: The vertical line $x=-\frac{b}{2a}$ through the vertex; it also sits exactly halfway between the two roots.

Worked example

A ball is thrown upward at 48 feet per second from the top of a 64-foot-tall cliff. Its height in feet after

t

seconds is modeled by

h(t) = -16t^2 + 48t + 64

. (a) What is the ball's height at the moment of release? (b) When does the ball reach its maximum height, and how high is it? (c) When does the ball hit the ground at the base of the cliff? (d) State the feasible domain of the model.

(a) Height at release. Evaluate at

t=0

h(0) = -16(0)^2 + 48(0) + 64 = 64

. The ball starts 64 feet above the ground, which matches the cliff height and is the

y

-intercept of the graph.

(b) Maximum height. Here

a=-16

and

b=48

, so the axis of symmetry is

t = -\frac{b}{2a} = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5.

Substitute back:

h(1.5) = -16(2.25) + 48(1.5) + 64 = -36 + 72 + 64 = 100

. The ball reaches a maximum height of 100 feet, 1.5 seconds after it is thrown. Since

a

is negative the parabola opens down, so this really is a maximum.

(c) Landing time. Ground level is height zero, so solve

-16t^2 + 48t + 64 = 0

. Factor out

-16

-16(t^2 - 3t - 4) = 0

, so

t^2 - 3t - 4 = 0

and

(t-4)(t+1) = 0

. The solutions are

t=4

and

t=-1

. Reject

t=-1

: negative time would be before the ball was thrown, so it has no meaning here. The ball hits the ground 4 seconds after release.

(d) Feasible domain. The model describes the ball from release until it lands, so the domain is

0 \le t \le 4

seconds. Beyond

t=4

the formula returns negative heights, which would mean the ball is underground.

Practice questions

A rocket's height in feet is modeled by

h(t) = -16t^2 + 80t + 6

. The vertex of the graph is

(2.5, 106)

. Which statement correctly interprets the vertex?

The rocket travels 2.5 feet horizontally before reaching its greatest height of 106 feet.
The rocket reaches a maximum height of 106 feet, 2.5 seconds after launch.
The rocket lands after 2.5 seconds, having risen 106 feet.
The rocket is launched from a height of 2.5 feet and rises at 106 feet per second.

Answer: The rocket reaches a maximum height of 106 feet, 2.5 seconds after launch.

In a height-versus-time model the input is time and the output is height, so the vertex

(2.5, 106)

means "at

t = 2.5

seconds the height is 106 feet." Since

a = -16 < 0

, the parabola opens down and that height is the maximum. The graph is not a picture of the rocket's path, so nothing horizontal is involved. Landing time comes from

h(t)=0

, not from the vertex, and the launch height is

h(0)=6

feet.

A farmer has 60 feet of fencing to build a rectangular pen with one side against a long barn wall, so only three sides need fence. Write a function for the enclosed area, state the feasible domain, and find the dimensions that maximize the area.

Answer: $A(x) = x(60-2x) = -2x^2 + 60x$ , with feasible domain $0 < x < 30$ ; the maximum area of 450 square feet occurs when the two perpendicular sides are 15 feet each and the side parallel to the barn is 30 feet.

Let

x

be the length in feet of each side perpendicular to the barn. Those two sides use

2x

feet of fence, leaving

60 - 2x

feet for the parallel side, so

A(x) = x(60-2x)

. Both

x

and

60-2x

must be positive, which gives

0 < x < 30

. The roots of

A

are

x=0

and

x=30

, so the vertex lies halfway between them at

x = 15

— you can also use

-\frac{b}{2a} = -\frac{60}{2(-2)} = 15

. Then

A(15) = 15(60-30) = 450

square feet. Reporting only

x=15

is incomplete; the question asks for dimensions, so give both 15 feet and 30 feet.

A diver's height above the water is

h(t) = -16t^2 + 8t + 24

. Solving

h(t)=0

gives

t = 1.5

and

t = -1

. Explain which solution describes the dive and why the other is rejected.

Answer: The diver enters the water at $t = 1.5$ seconds; $t = -1$ is rejected because negative time occurs before the dive begins and is outside the feasible domain $0 \le t \le 1.5$ .

Both values genuinely satisfy the equation, so neither is an algebra mistake — the issue is meaning. Time in this model is measured from the instant the diver leaves the board, so

t

cannot be negative. A complete answer names the rejected solution and gives the physical reason rather than just crossing it out. Checking

h(0) = 24

confirms the board is 24 feet above the water, which is consistent with a short dive lasting 1.5 seconds.

FAQ

Why is the coefficient always $-16$ in these height problems?: It comes from gravity. The distance an object falls under gravity is $\frac{1}{2}gt^2$ , and $g \approx 32\ \mathrm{ft/s^2}$ , so half of that is 16. It is negative because gravity pulls downward while positive height points upward. If a problem uses meters, gravity is about $9.8\ \mathrm{m/s^2}$ and the coefficient becomes $-4.9$ .
How do I know whether to find the vertex or solve for the roots?: Look at what the question asks. Words like maximum, greatest, highest, peak, or "best dimensions" point to the vertex. Words like lands, hits the ground, or "when is the height zero" point to the roots. If the question names a specific nonzero height or area, set the function equal to that number instead and solve.
Can a projectile problem have two valid times as an answer?: Yes, and it usually does when you ask for a specific height below the peak. The object passes that height once going up and once coming down, so $h(t) = 40$ typically has two solutions in the feasible domain, and both should be reported. Only when the height equals the maximum is there a single time.
How do I write the feasible domain for an area problem?: Every dimension in the picture must be positive. Write each dimension as an expression in your variable, set each one greater than zero, and combine the results. For a pen with sides $x$ and $60-2x$ , you need $x>0$ and $60-2x>0$ , giving $0<x<30$ .

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