Solve real-world problems modeled by exponential functions, including population growth, depreciation, and compound interest, evaluate the model, and use technology to find an exponential best fit (TEKS A.9B, A.9E).

What this topic is asking

TEKS A.9B and A.9E ask you to apply exponential functions to real situations, population growth, depreciation, and compound interest, to evaluate the model at a given time, and to use technology to find an exponential best fit for data. These are the higher-value items in the Exponential Functions category and they tie directly to the compound-interest dot point in the Number and Algebraic Methods category.

Evaluating a growth or decay model

The core skill is substituting a time into $f(x) = ab^x$ and computing. The exponent does the work: the factor $b$ is applied $t$ times, which is $b^t$, not $b \times t$.

For a population $P(t) = 500(1.04)^t$ after 5 years: $P(5) = 500(1.04)^5 = 500(1.2167) \approx 608$. The growth compounds, so the result is more than five separate 4% increases would suggest if added linearly.

Depreciation and decay

Depreciation is exponential decay: an asset loses a fixed percent of its current value each period, so $b = 1 - r$.

Compound interest as an exponential

Compound interest $A = P(1 + r)^t$ is an exponential model in disguise: the principal $P$ is the initial value $a$, and $1 + r$ is the base $b$. So a \$1,000 deposit at 5$A = 1000(1.05)^t$, the same form as any growth model. This is why memorizing the exponential form makes interest problems routine.

Exponential best fit with technology

When given a data set that grows or decays by a roughly constant factor, a calculator's exponential regression returns $f(x) = ab^x$. A.9E allows technology here. The interpretive step is reading $a$ (initial value) and $b$ (growth or decay factor, and so the percent rate $b - 1$) from the fitted model and using it to predict.

How STAAR examines this topic

• Multiple choice. Evaluate a model after several periods; the "linear instead of exponential" and "one period only" answers are standard distractors.
• Number entry. Compute a future value or population, rounding at the end.
• Inline choice. Identify the rate or factor, or whether the situation is growth or decay.

A clarifying idea is that the exponent is the number of periods: each period multiplies by $b$, so after $t$ periods the factor is $b^t$. Treating the change as $b \times t$ (linear) instead of $b^t$ (exponential) is the single biggest error in applications.

Matching the period to the rate

A detail that trips students is keeping the time unit consistent with the rate. If a rate is "per year", then $t$ counts years, and an answer "after 18 months" means $t = 1.5$. If a quantity halves "every 6 hours" and you want the amount after 24 hours, that is $\frac{24}{6} = 4$ halving periods, so $b^4 = (0.5)^4$. Reading how many full periods have elapsed, rather than plugging in the raw clock time, is essential whenever the period is not a single unit. STAAR applications usually keep the period equal to one time unit, but the half-life and doubling-time variants make the period explicit, so always confirm what one step of the exponent represents before evaluating.

Try this

Q1. A town of 5,000 grows 2% per year. Find the population after 3 years, to the nearest whole number. [2 points]

• Cue. $5000(1.02)^3 = 5000(1.061208) \approx 5306$.

Q2. A $4,000 machine depreciates 25% per year. Value after 2 years? [2 points]

• Cue. $4000(0.75)^2 = 4000(0.5625) = 2250$.