8.4 Building Functions & Interpreting Parameters in Context

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.

1. What is the input? Name it and give it a unit.
2. What is the starting value — the amount when the input is 0? That is $b$ (linear) or $a$ (exponential).
3. What happens per unit of input? A fixed amount added means linear; a fixed percent means exponential.

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

First differences: $-75$, $-75$, $-75$. Constant, so the model is linear. The value at $w = 0$ is 900.

$$B(w) = 900 - 75w$$

Now a second table:

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 $b = 1.1$.

$$V(t) = 5000(1.1)^t$$

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.

$$m = \frac{355 - 205}{5 - 2} = \frac{150}{3} = 50 \text{ dollars per hour}$$ $$205 = 50(2) + b \;\Rightarrow\; b = 105$$ $$\mathbf{C(h) = 50h + 105}$$

Interpreting the Parameters — The Part TABE Grades

A parameter interpretation is a sentence with a unit and a direction, not a number.

Linear: $f(x) = mx + b$

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: $f(x) = a(b)^x$

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.

1. Type: constant percent growth → exponential.
2. Function: $E(t) = 4200(1.06)^t$.
3. 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.
4. 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.