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What is an exponential function, how do the growth and decay models work, and how do you read the starting value and the rate?

Build and interpret exponential functions of the form f(x) = ab^x, including growth y = a(1+r)^t and decay y = a(1-r)^t, identifying the initial value and the rate (Ohio F-LE.1, F-LE.2, F-IF.8).

An Ohio Algebra I answer on exponential functions (F-LE.2, F-IF.8): the form f(x) = ab^x, the growth and decay percentage models, reading the initial value a and base b, and when growth beats linear change.

Generated by Claude Opus 4.812 min answer

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Jump to a section
  1. What this topic is asking
  2. The base form
  3. The percentage growth and decay models
  4. Reading growth versus decay
  5. How Ohio examines this topic
  6. Why exponential change is multiplicative
  7. Why growth eventually beats linear
  8. Try this

What this topic is asking

Ohio standards F-LE.2 and F-IF.8 ask you to build and interpret exponential functions. The base form is f(x)=abxf(x) = ab^x, where aa is the initial value and bb is the growth factor. The percentage growth and decay models, y=a(1+r)ty = a(1 + r)^t and y=a(1r)ty = a(1 - r)^t, are not on the reference sheet, so you must memorize them. Exponentials sit in the Functions category and appear especially on the calculator part.

The base form

Every exponential function has the same two-part structure.

So f(x)=32xf(x) = 3 \cdot 2^x starts at 33 and doubles each step; g(x)=80(0.5)xg(x) = 80 \cdot (0.5)^x starts at 8080 and halves each step.

The percentage growth and decay models

When a quantity changes by a percent each period, convert the percent to the base.

Reading growth versus decay

The base alone tells the story. A base greater than 11 means the quantity increases (growth); a base between 00 and 11 means it decreases (decay). Converting a percent: add for growth (1+r1 + r), subtract for decay (1r1 - r). A base of 1.071.07 is 7%7\% growth; a base of 0.920.92 is 8%8\% decay (since 10.92=0.081 - 0.92 = 0.08).

How Ohio examines this topic

  • Equation response. Write the growth or decay function from a context, then evaluate it (often on the calculator part).
  • Multiple choice and multiple-select. Pick the model whose base matches the percent, distinguishing growth from decay and from a linear model.
  • Tables. Identify a constant ratio between outputs as the signature of exponential change.

Why exponential change is multiplicative

The defining feature of an exponential function is that it multiplies by a constant factor each step, while a linear function adds a constant amount each step. This is why a constant percent change is exponential: keeping or adding the same fraction of the current amount means multiplying by a fixed base, not adding a fixed number. A savings account earning 4%4\% grows by more dollars each year as the balance rises, because 4%4\% of a larger amount is larger, that increasing dollar growth is the hallmark of multiplication. Spotting "per year by a percent" (exponential) versus "per year by a fixed amount" (linear) is the key modeling decision, and the test builds distractors around exactly this difference.

Why growth eventually beats linear

Over a short interval a steep line can stay ahead of a slowly growing exponential, but in the long run exponential growth overtakes any linear function. Multiplying by a base greater than 11 compounds: the increases themselves keep getting bigger, so the curve bends upward and accelerates, while a line rises at a fixed rate forever. This is why population, compound interest, and viral spread are modeled exponentially, and why "which is larger for large tt?" almost always favors the exponential once tt is big enough. Understanding this contrast is the heart of comparing function families, and it explains why a constant-ratio table eventually dwarfs a constant-difference one.

Try this

Q1. Write a decay model for \1200losing losing 10%$ per year. [2 points]

  • Cue. Base 10.10=0.901 - 0.10 = 0.90, so V(t)=1200(0.90)tV(t) = 1200(0.90)^t.

Q2. For f(x)=53xf(x) = 5 \cdot 3^x, what is f(0)f(0)? [1 point]

  • Cue. 30=13^0 = 1, so f(0)=5f(0) = 5.

Exam-style practice questions

Practice questions written in the style of ODEW exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Ohio Algebra I EOC (style)3 marksEquation response. A \500investmentgrows investment grows 4\%peryear.Writeafunction per year. Write a function A(t)foritsvalueafter for its value after tyears,thenfind years, then find A(3)$ rounded to the nearest dollar.
Show worked answer →

A(t)=500(1.04)tA(t) = 500(1.04)^t, and A(3) \approx \562$.

Percentage growth uses y=a(1+r)ty = a(1 + r)^t with initial value a=500a = 500 and rate r=0.04r = 0.04, so the base is 1+0.04=1.041 + 0.04 = 1.04 and A(t)=500(1.04)tA(t) = 500(1.04)^t. Evaluate at t=3t = 3 (Part 2, calculator): 500(1.04)3=500(1.124864)562.43500(1.04)^3 = 500(1.124864) \approx 562.43, about \562.Thegrowthmodelisnotonthereferencesheet,soyoumustknowthat"grows. The growth model is **not** on the reference sheet, so you must know that "grows rpercent"makesthebase percent" makes the base 1 + r$.

Ohio Algebra I EOC (style)2 marksMultiple choice. A car worth \20{,}000loses loses 15\%ofitsvalueeachyear.Whichfunctionmodelsitsvalue?(A) of its value each year. Which function models its value? (A) 20000(0.85)^t(B) (B) 20000(1.15)^t(C) (C) 20000(0.15)^t(D) (D) 20000 - 0.15t$
Show worked answer →

The correct answer is (A).

Percentage decay uses y=a(1r)ty = a(1 - r)^t. Losing 15%15\% means r=0.15r = 0.15, so the base is 10.15=0.851 - 0.15 = 0.85 and the value is 20000(0.85)t20000(0.85)^t. Option (B) would be 15%15\% growth, (C) keeps only 15%15\% each year, and (D) is a linear model, which is wrong because a constant percent loss (not a constant dollar loss) is exponential. Decay keeps 1r1 - r of the value each year.

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