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What are the defining features of an exponential function, and how do its base and initial value shape its graph?

Topic 2.3 Exponential Functions: define exponential functions, describe how the base and initial value determine growth or decay, and analyze the domain, range and horizontal asymptote of the graph.

A focused answer to AP Precalculus Topic 2.3, covering the form of an exponential function, growth versus decay, the horizontal asymptote, domain and range, and the natural base e.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. The form of an exponential function
  3. Growth versus decay
  4. Domain, range and the asymptote
  5. The natural base e
  6. Try this

What this topic is asking

The College Board (Topic 2.3) wants you to define an exponential function and analyze its graph. You must identify the initial value and the base, decide whether the function shows growth or decay, and state the domain, range and horizontal asymptote. You should also recognize the natural base ee.

The form of an exponential function

The base b=1b = 1 is excluded because 1x1^x is the constant function, and negative bases are excluded because they fail to give real outputs for many inputs.

Growth versus decay

The base alone decides growth or decay; the initial value aa only scales the graph vertically and fixes the yy-intercept.

Domain, range and the asymptote

An exponential function abxa \cdot b^x accepts every real input, so its domain is all real numbers. With a>0a > 0 the outputs are always positive, so the range is y>0y > 0, and the graph approaches but never reaches the horizontal asymptote y=0y = 0. A vertical shift abx+ka \cdot b^x + k moves the asymptote to y=ky = k and the range to y>ky > k (or y<ky < k if a<0a < 0).

The natural base e

The base e2.71828e \approx 2.71828 is the natural base, and f(x)=exf(x) = e^x is the natural exponential function. It models continuous growth (compound interest taken to its limit, continuous population growth, radioactive decay written as ekte^{kt}). Because e>1e > 1, exe^x grows; because 0<e1<10 < e^{-1} < 1, exe^{-x} decays. The natural exponential is the function whose inverse is the natural logarithm ln\ln, which is why it threads through the rest of the unit.

A point worth making once is that an exponential function never reaches its horizontal asymptote, no matter how far out you go. The outputs get arbitrarily close to y=0y = 0 but stay strictly positive, because a positive base raised to any power is positive. This is the graphical meaning of the range being y>0y > 0 rather than y0y \geq 0, and it is what distinguishes the unbounded-then-flattening shape of an exponential from a polynomial. Recognizing that the asymptote is approached but never touched also explains why exponential decay models, such as drug concentration or cooling, predict a quantity that nears but never equals zero.

Try this

Q1. Does f(x)=21.5xf(x) = 2 \cdot 1.5^{x} grow or decay, and what is its yy-intercept? [1 point]

  • Cue. Base 1.5>11.5 > 1, so it grows; the yy-intercept is f(0)=2f(0) = 2.

Q2. What is the range of g(x)=3x+4g(x) = 3^x + 4? [1 point]

  • Cue. The vertical shift raises the asymptote to y=4y = 4, so the range is y>4y > 4.

Exam-style practice questions

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

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). For f(x)=5(12)xf(x) = 5 \cdot \left(\tfrac{1}{2}\right)^{x}, which statement is true? (A) ff grows, with horizontal asymptote y=5y = 5 (B) ff decays, with horizontal asymptote y=0y = 0 (C) ff decays, with horizontal asymptote y=5y = 5 (D) ff grows, with horizontal asymptote y=0y = 0
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The correct answer is (B), decays with horizontal asymptote y=0y = 0.

The base is 12\frac{1}{2}, which is between 00 and 11, so ff is exponential decay. The standard exponential abxa \cdot b^x has horizontal asymptote y=0y = 0 (no vertical shift here), and the initial value a=5a = 5 is the yy-intercept, not the asymptote.

AP 2024 (style)3 marksSection II (free response, calculator allowed). A function is f(x)=32xf(x) = 3 \cdot 2^{x}. (a) State the initial value, the base, and whether ff grows or decays. (b) State the domain, range, and the equation of the horizontal asymptote.
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A 3-point question on the features of an exponential function.

(a) Features (1 point): initial value a=3a = 3 (the value at x=0x = 0), base b=2b = 2, and since b>1b > 1 the function grows.
(b) Domain, range, asymptote (1 point for domain/range, 1 point for asymptote): the domain is all real numbers; the range is y>0y > 0 (positive outputs only); the horizontal asymptote is y=0y = 0, which the graph approaches as xx \to -\infty.

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