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.
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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 , where is the initial value and is the growth factor. The percentage growth and decay models, and , 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 starts at and doubles each step; starts at 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 means the quantity increases (growth); a base between and means it decreases (decay). Converting a percent: add for growth (), subtract for decay (). A base of is growth; a base of is decay (since ).
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 grows by more dollars each year as the balance rises, because 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 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 ?" almost always favors the exponential once 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 \120010%$ per year. [2 points]
- Cue. Base , so .
Q2. For , what is ? [1 point]
- Cue. , so .
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 \5004\%A(t)tA(3)$ rounded to the nearest dollar.Show worked answer →
, and A(3) \approx \562$.
Percentage growth uses with initial value and rate , so the base is and . Evaluate at (Part 2, calculator): , about \562r1 + r$.
Ohio Algebra I EOC (style)2 marksMultiple choice. A car worth \20{,}00015\%20000(0.85)^t20000(1.15)^t20000(0.15)^t20000 - 0.15t$Show worked answer →
The correct answer is (A).
Percentage decay uses . Losing means , so the base is and the value is . Option (B) would be growth, (C) keeps only 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 of the value each year.
Related dot points
- Distinguish linear, quadratic, and exponential functions from tables, graphs, and contexts using constant differences and ratios, and compare their long-run growth (Ohio F-LE.1, F-LE.3, F-IF.4).
An Ohio Algebra I answer on comparing function families (F-LE.1, F-LE.3): constant first differences for linear, constant second differences for quadratic, constant ratios for exponential, and why exponential growth eventually overtakes the others.
- Recognize, extend, and write rules for arithmetic and geometric sequences, using the explicit formulas, and treat sequences as functions of the term number (Ohio F-IF.3, F-BF.2, F-LE.2).
An Ohio Algebra I answer on sequences (F-BF.2, F-IF.3): telling arithmetic from geometric, the explicit formulas on the reference sheet, finding the nth term, and seeing sequences as functions whose domain is the whole numbers.
- Build a function that models a relationship, write a linear function from a context, table, or two points, and interpret its parameters in context (Ohio F-BF.1, F-LE.2, F-IF.7).
An Ohio Algebra I answer on building functions (F-BF.1, F-LE.2): writing a linear function from a verbal description, a table, or two points, interpreting the slope and intercept as rate and starting value, and using the function to predict.
- Apply the properties of exponents to simplify expressions, including rational exponents interpreted as radicals (Ohio N-RN.1, N-RN.2).
An Ohio Algebra I answer on the exponent rules and radicals (N-RN.1, N-RN.2): the product, quotient, power, zero, and negative rules, and rewriting rational exponents as radicals such as x to the one-half equals the square root of x.
- Use function notation to evaluate and interpret functions, decide whether a relation is a function, and identify domain and range from equations, tables, and graphs (Ohio F-IF.1, F-IF.2, F-IF.5).
An Ohio Algebra I answer on functions (F-IF.1, F-IF.2): the definition of a function, the vertical line test, evaluating f(x), solving f(x) = k, and reading domain and range from graphs and tables.
Sources & how we know this
- Ohio's Learning Standards for Mathematics: Algebra 1 — Ohio Department of Education and Workforce (2024)
- Algebra I course resources (blueprint, reference sheet, released items) — Ohio Department of Education and Workforce (2024)