Interpret quadratic models and extrema

Your feedback matters:

Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 28 minutes
Domains: Advanced Math
Techniques: Quadratic-modelsContextual-interpretationExtremaRealistic-domain

What you’ll learn

Read a vertex as a real high or low point, like a ball’s greatest height.
Keep its two numbers straight: when or how many, and how high or how much.
Explain what intercepts mean, like a starting height or a landing time.
Explain what the number in front of the square means, and why it isn’t a steady rate.
Throw out inputs that can’t happen, like a negative time.
Use a given model to make a prediction.

Prerequisites

Why this matters on the SAT

Turn a point into a complete sentence

The SAT often hands you a quadratic model and asks what one of its points means. Finding the point is only half the job. You also have to say what each number counts, in which unit, and what’s happening at that moment.

Solution to the example

The model is in vertex form, $a(x-h)^2+k$ , so you can read the vertex straight off it: $(40,3200)$ . The number in front of the square, $-2$ , is negative, so the parabola opens downward and the vertex is a maximum.

Now give each number its job. The input $x$ counts meal kits, and the output $P(x)$ is profit in dollars. So $40$ is a number of kits and $3200$ is dollars: the model predicts a maximum daily profit of $3,200 when $40$ meal kits are sold. The answer is B.

Each wrong choice is a trap you’ll learn to spot: A swaps the numbers’ roles, C calls the peak a minimum, and D treats the $2$ in front of the square as a steady rate. Remember this: a point isn’t an answer until it’s a sentence, with each number given its meaning and its unit.

SAT example

The daily profit $P(x)$ , in dollars, from selling $x$ meal kits is modeled by

P(x)=-2(x-40)^2+3200.

Which choice best interprets the vertex of the graph of $P$ in the $x$ - $P(x)$ plane?

A
The maximum daily profit is $40 when $3{,}200$ meal kits are sold.
B
The maximum daily profit is $3,200 when $40$ meal kits are sold.
C
The minimum daily profit is $3,200 when $40$ meal kits are sold.
D
The daily profit increases by $2 for every additional meal kit sold.

What each point means in context

Picture an object launched into the air. Its height, in meters, $t$ seconds after launch is

h(t)=-3(t-2)^2+48.

The equation defines a full parabola, but the flight uses only inputs from launch through landing.

The solid part of the curve is the flight. The dotted part is math the flight never uses. Vertex form shows you the peak, but not when the object lands, so graph the model and let Desmos find the points. Enter

h(x)=-3(x-2)^2+48

where $x$ stands for the time $t$ . Select the curve, then select its vertex and both $x$ -intercepts. Desmos shows the vertex $(2,48)$ and the intercepts $(-2,0)$ and $(6,0)$ . The inputs where the height is $0$ , $t=-2$ and $t=6$ , are the model’s roots.

Desmos finds every point. You decide which ones belong to the flight. Here’s each point in math words, then in flight words:

Translate each graph feature into the situation

Feature	Mathematical meaning	Contextual meaning
Vertex $(2,48)$	Greatest output is $48$ at input $2$	The object reaches a maximum height of $48$ meters $2$ seconds after launch
Axis $t=2$	Input at the vertex	The time when the maximum height occurs
$h$ -intercept $(0,36)$	$h(0)=36$	The object starts $36$ meters above the ground
Positive $t$ -intercept $(6,0)$	$h(6)=0$	The object reaches the ground $6$ seconds after launch
Negative root $t=-2$	The equation also has output $0$ there	It is not a time after launch, so it does not describe this flight

For more practice finding these points in Desmos, see Read points of interest from a graph.

Calculator loads as you approach

Select the curve, then the vertex and both intercepts. Reset the example when you’re done.

To explain any point on a model, ask four questions:

Which point is it? The vertex, a point where the graph crosses an axis, or some other point on the curve?
What do the axes mean? What do the input and the output stand for?
What are the units? Seconds, meters, dollars, items?
Could that input really happen? You can’t sell $-5$ meal kits, for example.

Check your understanding:

For the flight model above, what is the maximum height, and when does the object reach it?

Common mistake:

Swapping the vertex’s numbers, as in “48 seconds” or “2 meters.” Both look like plain numbers, so it’s an easy slip. Say what each axis means first: the first number takes the input’s unit, and the second takes the output’s. Then sanity-check it. The object lands at $6$ seconds, so a peak at $48$ seconds is impossible.

Is the vertex a maximum or minimum?

Go back to the meal-kit model, $P(x)=-2(x-40)^2+3200$ . At $x=40$ the square is $0$ , so the profit is $3200$ . At $39$ or $41$ kits the square is $1$ , so the profit drops to $3200-2=3198$ . Any other number of kits makes the square positive, and the $-2$ pulls the profit below $3200$ . So $3200$ is the maximum.

The same idea works for any model in vertex form,

f(x)=a(x-h)^2+k.

A square is never negative:

(x-h)^2\ge0.

At $x=h$ the square is $0$ , so the output is $k$ . Everywhere else, the sign of $a$ decides:

If $a<0$ , then $a(x-h)^2$ is negative, so every other output is below $k$ . The parabola is a hill, and $k$ is its top: a maximum.
If $a>0$ , then $a(x-h)^2$ is positive, so every other output is above $k$ . The parabola is a valley, and $k$ is its bottom: a minimum.

So the vertex answers two questions at once:

\boxed{\text{The maximum or minimum is }k\text{, and it happens at }x=h.}

Some questions ask for only one of them, so read the last sentence of the question closely. “What is the maximum profit?” wants the output, $k$ . “How many units maximize profit?” wants the input, $h$ .

Check your understanding:

A delivery company models its daily cost, in dollars, by $C(r)=5(r-12)^2+460$ , where $r$ is the number of routes. What does the vertex mean?

What the coefficients mean, and what they don’t

This idea trips up a lot of students, so let’s test it with numbers. In the flight model

h(t)=-3(t-2)^2+48,

it’s tempting to read $-3$ as “the height drops $3$ meters every second.” Check the heights after the peak at $t=2$ :

$1$ second past the peak: $h(3)=45$ , so $3$ meters below it.
$2$ seconds past: $h(4)=36$ , so $12$ meters below.
$3$ seconds past: $h(5)=21$ , so $27$ meters below.

Second by second, the object falls $3$ meters, then $9$ , then $15$ . The drop keeps growing, so $-3$ can’t be a steady rate.

Now look at the distances below the peak: $3=3\cdot1^2$ , $12=3\cdot2^2$ and $27=3\cdot3^2$ . Each one is $3$ times the square of the seconds from the peak. That’s the real job of $-3$ : twice as far from the peak, four times as far below it.

To write that for any time, call the seconds from the peak $d$ . Then $(t-2)^2=d^2$ , and the model becomes

h(t)=48-3d^2.

The height is $3d^2$ meters below the maximum. That squared time is also why $-3$ has units of meters per second squared.

It works before the peak too. At launch, $t=0$ is also $2$ seconds away, so

h(t)=48-3(2^2)=36.

That matches both $h(0)=36$ and $h(4)=36$ .

The same flight can also be written in standard form. Multiply out $-3(t-2)^2+48$ and you get

h(t)=-3t^2+12t+36.

Now the vertex is hidden, but the constant term still means something: it’s $h(0)=36$ , the starting height again. In any standard-form model, $ax^2+bx+c$ , the constant $c$ is the output when the input is $0$ . In a profit model, it can be the predicted profit when no units are sold.

Don’t read the $12$ as a steady rate either. The height rises $9$ meters in the first second, from $36$ to $45$ , and only $3$ meters in the next, from $45$ to $48$ .

Common mistake:

Calling $a$ or $b$ a constant rate of change. It’s tempting because in a linear model, the number in front of $x$ is the rate. But a quadratic doesn’t change by the same amount over equal steps of input. In vertex form, multiply $a$ by the squared distance from the vertex, as in $3\cdot2^2=12$ meters below the peak. In standard form, read $c$ as $f(0)$ when an input of $0$ makes sense.

Check your understanding:

For $Q(s)=-2(s-15)^2+900$ , where $Q$ is a quality score and $s$ is a machine setting, what does the coefficient $-2$ tell you about a setting $d$ units from $15$ ?

Keep predictions inside a realistic domain

An equation takes any input you give it. The situation doesn’t.

The flight model has two roots, $t=-2$ and $t=6$ . Both make the height $0$ , but $t=-2$ means $2$ seconds before launch, when nothing was flying yet. At the other end, a time after $6$ gives a negative height, as if the object kept falling underground. But it has already landed. So only

0\le t\le6

describes the flight, from launch to landing. The equation doesn’t know the object has landed. You do.

Look for these clues in the wording:

Time after an event: usually $t\ge0$ , often ending at a landing or another event the problem names.
Numbers of items or people: whole numbers, $0$ or more, sometimes up to a capacity.
Lengths and areas: positive only. If a side is $x-3$ meters long, $x$ has to be more than $3$ .
A stated interval: use exactly that interval, even if the equation keeps going past it.

Once the input makes sense, plug it into the model and give the output with its unit. Perfect algebra on an impossible input still gives a wrong answer.

Try it yourself:

For the flight model $h(t)=-3(t-2)^2+48$ , compare $h(4)=36$ with $h(8)=-60$ . Why is the first a real prediction about the flight and the second isn’t, even though both are fine to calculate?

Choose: read it by hand or graph it

Before you open Desmos, check what the equation already shows you.

Read or evaluate by hand when…

the model is in vertex form, like $P(x)=-2(x-40)^2+3200$ . The vertex, $(40,3200)$ , is right there.
you need only $f(0)$ or one quick prediction, like $h(0)=-3(0-2)^2+48=36$ .
you can already see the point you need and what it means.

Graph the model when…

it’s in standard form, like $h(t)=-5t^2+40t+45$ . The vertex is hidden.
you need a start, a landing or a break-even point that the equation doesn’t show.
you need to compare those points with a limit in the story, like a time or a capacity, to see which inputs make sense.

If you can see it, read it. If it’s hidden, graph it and select it. Either way, the question tells you which coordinate, unit and inputs to use.

Example: find a hidden vertex and the flight times

Worked example

The height $h(t)$ , in meters, of a launched object $t$ seconds after launch is modeled by

h(t)=-5t^2+40t+45.

Which choice best interprets the vertex and gives a realistic domain for the flight?

A
The object reaches a maximum height of $125$ meters $4$ seconds after launch, and a realistic domain is $0\le t\le9$ .
B
The object reaches a maximum height of $4$ meters $125$ seconds after launch, and a realistic domain is $t\ge0$ .
C
The object reaches a minimum height of $125$ meters $4$ seconds after launch, and a realistic domain is $0\le t\le9$ .
D
The object reaches a maximum height of $125$ meters $4$ seconds after launch, and every real value of $t$ is realistic.

Step 1

Say what each axis means

The horizontal axis is time $t$ , in seconds. The vertical axis is height $h(t)$ , in meters. The number in front of $t^2$ is negative, so the graph opens downward and its vertex will be a maximum.

Step 2

Graph to find the hidden points

Standard form hides both the vertex and the $t$ -intercepts, and one graph shows them all. Enter

h(x)=-5x^2+40x+45

where $x$ stands for the time $t$ . Select the highest point and the two $x$ -intercepts. Desmos shows

(4,125),\qquad(-1,0),\qquad(9,0).

The vertex gives the greatest height and when it happens. The intercepts are the times when the height is $0$ , and only one of them is part of the flight.

Calculator loads as you approach

Select the vertex and both intercepts. Then ask which ones make sense for the flight.

Step 3

Put the vertex into words

The point $(4,125)$ means the object is $125$ meters high at $t=4$ seconds. The graph opens downward, so that’s the maximum height.

Step 4

Keep only the flight times, then choose

The flight starts at launch, $t=0$ . The negative root, $t=-1$ , is before launch, so drop it. The flight ends when the object hits the ground at the positive root, $t=9$ . So a realistic domain is

0\le t\le9.

The answer is A.

Try it yourself:

The same graph shows $h(0)=45$ . What does that point mean, with units? And $t=-1$ makes the height $0$ , so why doesn’t it belong to the flight?

Practice problems

Try these on your own. Each calculator starts empty.

Interpret a minimum-cost vertex

Practice problem

A company’s daily operating cost $C(r)$ , in dollars, is modeled by

C(r)=2(r-14)^2+320,

where $r$ is the number of delivery routes scheduled. Which choice best interprets the vertex?

Calculator loads as you approach

Vertex form already shows the point you need, so read the vertex by hand first. The calculator is here if you want it.

Choose the landing time and domain

Practice problem

The height $h(t)$ , in meters, of a ball $t$ seconds after it is thrown is modeled by

h(t)=-4(t+1)(t-5).

Which statement is best supported by the model?

Calculator loads as you approach

Graph the model to see both roots and the starting point at

x=0

. Then let the situation pick the realistic interval.

Maximize within a scaled realistic domain

Practice problem

A workshop’s weekly profit $P(x)$ , in thousands of dollars, is modeled by

P(x)=-2x^2+24x-40.

In this model, $x$ is the number of hundreds of tables produced. The workshop can produce at most $500$ tables in one week. According to the model and this capacity, what is the greatest predicted weekly profit, in dollars?

Your answer

Calculator loads as you approach

Graph the model, then look at its outputs only on the interval the workshop can reach,

0\le x\le5

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

A point isn’t an answer until it’s a sentence: each number needs its meaning and its unit.
A vertex $(h,k)$ holds two answers: the maximum or minimum is $k$ , and it happens at input $h$ .
Negative $a$ makes a hill, so the vertex is a maximum. Positive $a$ makes a valley, so it’s a minimum.
An $x$ -intercept is where the output is $0$ , like a landing. The situation may rule some of these roots out.
$f(0)$ , the output at input $0$ , is often a starting value.
In vertex form, $a$ isn’t a steady rate: twice as far from the vertex, four times as far from $k$ .
Use only inputs the situation allows. Time can't be negative or run past the end, like a landing. Lengths must be positive, items come whole, and nothing goes past a capacity or a stated interval.
If you can see it, read it. If it’s hidden, graph it.

Next lesson

Build, identify, and interpret exponential models

Recognize constant-ratio change and interpret the parameters of an exponential model.

Start next lesson

Practice

Practice this lesson

224 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

Start practice