Building a Function Model — AP® Precalculus

Part 1 · Hook

Why this matters

A garden center tracks how many of a new spring plant it sells each month. Three numbers from last year are enough to build a model, predict next spring, and say when the model stops making sense.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Which system gives $y = ax^2 + bx + c$ through $(1, 3)$ , $(2, 4)$ and $(3, 7)$ ?

$a + b + c = 3$ , $4a + 2b + c = 4$ , $9a + 3b + c = 7$
$a + b + c = 3$ , $2a + 2b + c = 4$ , $3a + 3b + c = 7$
$3a + 3b + c = 1$ , $4a + 4b + c = 2$ , $7a + 7b + c = 3$
$a + b = 3$ , $4a + 2b = 4$ , $9a + 3b = 7$

Show the answer

Substitute each point; the input is squared for $a$ .

Correct: $a + b + c = 3$ , $4a + 2b + c = 4$ , $9a + 3b + c = 7$ :
$a + b + c = 3$ , $2a + 2b + c = 4$ , $3a + 3b + c = 7$ :
$3a + 3b + c = 1$ , $4a + 4b + c = 2$ , $7a + 7b + c = 3$ :
$a + b = 3$ , $4a + 2b = 4$ , $9a + 3b = 7$ :

2. What are the domain restrictions a model of a tossed ball's height needs?

From the toss until the ball lands
All real times
Only the times in the data
Until the ball reaches its highest point

Show the answer

Time starts at the toss, and the model stops describing the ball once it lands.

Correct: From the toss until the ball lands:
All real times:
Only the times in the data:
Until the ball reaches its highest point:

Part 4 · See it

See it first

Plant sales S, in hundreds, against month t: the data points (1, 14), (3, 30) and (5, 30) on the parabola S(t) = -2t^2 + 16t, drawn from t = 0 to t = 8 where the sales return to 0. — Three data points determine the quadratic $S(t) = -2t^2 + 16t$ . The context allows $0 \le t \le 8$ , the months when the model gives sales that are not negative. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Choose the model type from the data and the contexta form with unknown constants, such as $at^2 + bt + c$
Substitute data points, or run a regression on the calculatorthe constants, stored at full precision
Evaluate the model and its average rates of changepredictions with units, inside the domain the context allows

Part 6 · Key ideas

Key ideas

A model is built from points (a system), from a known feature and transformations (vertex form), or from all the data at once (a regression).
Inverse proportion: $y = \frac{k}{x}$ ; inverse-square laws: $y = \frac{k}{x^2}$ . One data point gives $k$ .
Report predictions and rates with units: an average rate of change is output units per input unit, such as hundred plants per month.

Part 7 · Misconception

A common mistake

The wrong idea: Once a model fits the data, it can predict any input.

What actually happens: A model is supported only near its data and only where the context allows; past that, its predictions can be impossible, such as negative sales.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Monthly sales of a seasonal plant

Plant sales S, in hundreds of plants, during month t, where t = 1 is March
$t$ (months)	1	3	5
$S$ (hundreds of plants)	14	30	30

1. A quadratic model $S(t) = at^2 + bt + c$ passes through the three data points. Write and solve the system. What is $a$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The points give $a + b + c = 14$ , $9a + 3b + c = 30$ and $25a + 5b + c = 30$ . Subtracting in pairs: $8a + 2b = 16$ and $16a + 2b = 0$ . Subtracting again: $8a = -16$ , so $a = -2$ ; then $b = 16$ and $c = 0$ .

Answer: -2

2. Using the data, what is the average rate of change of $S$ on $[1, 3]$ , in hundreds of plants per month?

No calculator. Type the exact value (a fraction, π, e or a root is fine), in hundred plants per month.

Show the answer

$\frac{S(3) - S(1)}{3 - 1}$ $= \frac{30 - 14}{2}$ $= 8$ hundred plants per month, the slope of the secant line through $(1, 14)$ and $(3, 30)$ .

Answer: 8 hundred plants per month

3. For which months does the context allow the model $S(t) = -2t^2 + 16t$ to be used?

$0 \le t \le 8$ , since sales after $t = 8$ would be negative
$1 \le t \le 5$ , because those are the months in the table
$0 \le t \le 4$ , because sales start to decrease after $t = 4$
$t \ge 0$ , because a month number is not negative

Show the answer

$S(t) = -2t(t - 8)$ is $0$ at $t = 0$ and $t = 8$ and negative after $t = 8$ . Sales cannot be negative, so the model can be used for $0 \le t \le 8$ .

Correct: $0 \le t \le 8$ , since sales after $t = 8$ would be negative: The model gives negative sales past $t = 8$ .
$1 \le t \le 5$ , because those are the months in the table: A model can be used between and somewhat beyond its data if the context allows.
$0 \le t \le 4$ , because sales start to decrease after $t = 4$ : Monthly sales can go down; a decrease does not break the context.
$t \ge 0$ , because a month number is not negative: This misses the upper bound from negative sales.

4. For the light in the previous situation, write $I$ as a function of $d$ .

Type an expression for $I(d) =$ .

Show the answer

$I(d) = \frac{k}{d^2}$ , and $200 = \frac{k}{3^2}$ gives $k = 1800$ . So $I(d) = \frac{1800}{d^2}$ .

Answer: $I(d) =$ $\frac{1800}{d^{2}}$

5. A quadratic model has its maximum at $(4, 32)$ and passes through $(0, 0)$ . Write the model $S(t)$ using transformations of $t^2$ .

Type an expression for $S(t) =$ .

Show the answer

Vertex form: $S(t) = a(t - 4)^2 + 32$ . Through $(0, 0)$ : $0 = 16a + 32$ , so $a = -2$ and $S(t) = -2(t - 4)^2 + 32$ , which equals $-2t^2 + 16t$ .

Answer: $S(t) =$ $-2 (t - 4)^{2} + 32$

6. A candle is $20$ cm tall. After burning for $3$ hours it is $14$ cm tall. Assuming it burns at a constant rate, what is the rate of change of its height?

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$\frac{14 - 20}{3 - 0} = -2$ cm per hour. The linear model is $L(t) = 20 - 2t$ .

Answer: -2 cm per hour

Part 9 · Summary

Summary

Build a model from data points (a system), from features and transformations, or by regression; store its constants; use it to predict and to find average rates of change with units, only for inputs the context allows.

Part 10 · Up next

What comes next

Part 11 · Connections