8.4 Building Functions & Interpreting Parameters in Context
Key Takeaways
- Building a function means writing a rule that describes how one quantity depends on another, starting from a verbal description, a table, or two data points.
- In a linear model f(x) = mx + b, b is the initial value or fixed amount and m is the constant rate of change, and TABE asks what each means in the situation.
- In an exponential model f(x) = a(b)^x, a is the starting amount and b is the growth or decay factor, where b = 1 + r for growth and b = 1 - r for decay.
- A parameter interpretation must include the unit and the direction: "the tank loses 3 gallons per minute," not merely "the slope is -3."
Building Functions & Interpreting Parameters in Context
Sections 8.2 and 8.3 taught you to work with linear and exponential functions. Level A asks for two further skills: write a function that describes a relationship between two quantities, and interpret the parameters in a linear or exponential function in terms of a context. Those are modeling skills, and they are what TABE actually puts in front of adult learners.
Building From a Verbal Description
Three questions convert almost any description into a function.
- What is the input? Name it and give it a unit.
- What is the starting value — the amount when the input is 0? That is $b$ (linear) or $a$ (exponential).
- What happens per unit of input? A fixed amount added means linear; a fixed percent means exponential.
| Description | Type | Function |
|---|---|---|
| Membership: $40 to join, $22 a month | linear | $C(m) = 22m + 40$ |
| Tank holds 90 gal, drains 3 gal/min | linear, decreasing | $V(t) = 90 - 3t$ |
| Population 8,000, grows 4% a year | exponential growth | $P(t) = 8000(1.04)^t$ |
| Machine worth $60,000, loses 15% a year | exponential decay | $V(t) = 60000(0.85)^t$ |
| Technician earns $21/hr, no base pay | linear, proportional | $E(h) = 21h$ |
The single fastest diagnostic: does the situation add the same number each step, or multiply by the same percent each step? Same number → linear. Same percent → exponential.
Building From a Table
| Weeks $w$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Balance $B$ | 900 | 825 | 750 | 675 |
First differences: $-75$, $-75$, $-75$. Constant, so the model is linear. The value at $w = 0$ is 900.
Now a second table:
| Years $t$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Value $V$ | 5,000 | 5,500 | 6,050 | 6,655 |
First differences are $500$, $550$, $605$ — not constant. Check ratios instead: $5500 \div 5000 = 1.1$, $6050 \div 5500 = 1.1$, $6655 \div 6050 = 1.1$. Constant ratio, so the model is exponential with $a = 5000$ and $b = 1.1$.
Building From Two Points
Two data points determine a line. Slope first, then back out the intercept.
A plumber's bill is $205 for 2 hours and $355 for 5 hours.
Interpreting the Parameters — The Part TABE Grades
A parameter interpretation is a sentence with a unit and a direction, not a number.
Linear: $f(x) = mx + b$
| Parameter | Generic meaning | In $C(h) = 50h + 105$ |
|---|---|---|
| $b$ | value when input is 0; fixed, one-time, or starting amount | the $105 service-call fee charged before any work is done |
| $m$ | change in output per one-unit change in input | the labor rate of $50 for each additional hour |
| sign of $m$ | increasing or decreasing | positive, so the bill rises with time worked |
In $V(t) = 90 - 3t$: the 90 is the initial 90 gallons in the tank, and the $-3$ is a loss of 3 gallons every minute. Saying "the slope is $-3$" earns nothing; saying "the tank loses 3 gallons per minute" is the answer.
Exponential: $f(x) = a(b)^x$
| Parameter | Meaning | Reading the rate |
|---|---|---|
| $a$ | initial amount, at input 0 | $60,000 purchase price |
| $b$ | growth or decay factor per period | $0.85$ |
| $b > 1$ | growth | $b = 1.04 \Rightarrow$ 4% growth per year |
| $0 < b < 1$ | decay | $b = 0.85 \Rightarrow$ 15% loss per year |
Converting factor to rate is the tested step. From $b$, the percent rate is $|b - 1| \times 100%$:
- $b = 1.07$ → 7% increase
- $b = 0.92$ → 8% decrease
- $b = 1.5$ → 50% increase
- $b = 0.5$ → 50% decrease (half each period)
The classic trap. A machine "depreciates 12% per year" is modeled by $(0.88)^t$, not $(0.12)^t$. The factor $0.12$ would mean the machine retains only 12% of its value annually — a collapse, not depreciation.
Writing Functions With Constraints Attached
A model is only valid over a sensible domain, and TABE asks about it.
$V(t) = 90 - 3t$ models a draining tank.
The function keeps producing values forever, but the tank stops at empty: $90 - 3t = 0$ at $t = 30$. So the realistic domain is $0 \le t \le 30$ minutes. Feeding in $t = 40$ gives $-30$ gallons, which is arithmetic, not reality.
Putting It Together
Problem. A community college's enrollment was 4,200 in 2020 and has grown about 6% per year since. Write a function for enrollment $E$ after $t$ years, state what each parameter means, and estimate enrollment in 2025.
- Type: constant percent growth → exponential.
- Function: $E(t) = 4200(1.06)^t$.
- Parameters: 4,200 is the enrollment in 2020, the base year where $t = 0$; 1.06 is the annual growth factor, meaning enrollment rises 6% each year.
- Evaluate: 2025 is $t = 5$, and $1.06^5 \approx 1.3382$, so $E(5) \approx 4200 \times 1.3382 \approx \mathbf{5{,}620}$ students.
A car purchased for $32,000 loses 18% of its value each year. Which function models its value V after t years?
A cell plan is modeled by C(g) = 12g + 45, where g is gigabytes used and C is the monthly cost in dollars. What does the 45 represent in this context?
A technician records that a chemical bath contains 4,800 mL at the start and 4,320 mL after one hour, 3,888 mL after two hours, and 3,499.2 mL after three hours. Which function models the volume?