Exponential Functions and Geometric Sequences Comprehensive Study Guide

Intercepts, Domain, Range, and Asymptotes of Exponential Functions

• General Definition of Exponential Functions:

  • An exponential function is characterized by a constant base raised to a variable exponent, generally modeled by the parent form $f(x) = a(b)^x$ or the transformed form $f(x) = a(b)^{x-h} + k$, where $a \neq 0$, $b > 0$, and $b \neq 1$.
  • The parameter $a$ represents the initial value (or vertical stretch/shrink factor), $b$ represents the growth/decay factor (or common ratio), $h$ represents the horizontal shift, and $k$ represents the vertical shift.

• Algebraic Determination of the $y$-Intercept:

  • The $y$-intercept of any function occurs precisely where the input variable is zero ($x = 0$), yielding the coordinate pair $(0, f(0))$.
  • Function without Vertical Shift: For $f(x) = 7(5)^x$:
    • Evaluate at $x = 0$: $f(0) = 7(5)^0$.
    • Apply the zero-exponent rule ($b^0 = 1$ for any $b \neq 0$): $5^0 = 1$.
    • Multiply by the coefficient: $7 \times 1 = 7$.
    • The $y$-intercept is $(0, 7)$.
  • Function with Vertical Shift: For $h(x) = 2(3)^x - 5$:
    • Evaluate at $x = 0$: $h(0) = 2(3)^0 - 5$.
    • Simplify exponent and multiplication: $2(1) - 5 = 2 - 5 = -3$.
    • The $y$-intercept is $(0, -3)$.
    • Analytical Consideration: The constant $k = -5$ shifts the parent intercept $(0, 2)$ vertically downward by $5$ units to $(0, -3)$.

• Domain, Range, and Horizontal Asymptotes:

  • The horizontal asymptote is a horizontal line $y = k$ that the curve approaches as $x \rightarrow \theta$ or $x \rightarrow -\theta$. The graph approaches but never crosses this boundary under standard basic transformations.
  • Standard Exponential Function: For $y = 3^x - 2$:
    • Vertical Shift: Shifted down $2$ units ($k = -2$).
    • Domain: $(-\theta, \theta)$ (all real numbers, as any real exponent is valid).
    • Horizontal Asymptote: $y = -2$.
    • Range: $(-2, \theta)$ in interval notation, because the base function $3^x > 0$ for all $x$, meaning $3^x - 2 > -2$.
    • $y$-Intercept: $y = 3^0 - 2 = 1 - 2 = -1$, giving the point $(0, -1)$.
  • Reflected Exponential Function: For $y = -2(4)^x + 6$:
    • Transformations Identified: A vertical stretch by a factor of $2$, a vertical reflection across the $x$-axis (caused by the negative sign in front), and a vertical shift up $6$ units ($k = 6$).
    • Domain: $(-\theta, \theta)$.
    • Horizontal Asymptote: $y = 6$.
    • Effect of Reflection on Range: Because the leading coefficient is negative ($a = -2 < 0$), the graph opens downward beneath the asymptote instead of above it. Hence, the range is $(-\theta, 6)$.
    • $y$-Intercept: $y = -2(4)^0 + 6 = -2(1) + 6 = -2 + 6 = 4$, yielding the coordinate pair $(0, 4)$.

Constructing Exponential Functions from Data and Tables

• Methodology for Constructing $f(x) = a(b)^x$ from Given Coordinates:

  • Identify consecutive integer inputs ($x$-values) to confirm uniform step size ($\Delta x = 1$).
  • Determine the base (growth factor or common ratio $b$) by computing the ratio between consecutive output values: $b = \frac{f(x+1)}{f(x)}$.
  • Identify the initial value $a$ corresponding to the output at $x = 0$, such that $f(0) = a$.
  • If $x = 0$ is not explicitly provided in the data set, apply the inverse operation (divide by $b$) to step backward from positive inputs to determine the value at $x = 0$.

• Worked Table Formulations:

  • Direct Identification with Positive Growth: Given table values for $x = 0, 1, 2, 3$ with outputs $f(x) = 4, 12, 36, 108$:
    • Calculate the successive ratios: $\frac{12}{4} = 3$, $\frac{36}{12} = 3$, $\frac{108}{36} = 3$. The common ratio is $b = 3$.
    • Identify the starting value: At $x = 0$, $f(0) = 4$, meaning $a = 4$.
    • The explicit exponential function is $f(x) = 4(3)^x$.
  • Working Backward to Find Initial Value: Given table values for $x = 1, 2, 3$ with outputs $f(x) = 6, 18, 54$:
    • Calculate the successive ratios: $\frac{18}{6} = 3$, $\frac{54}{18} = 3$. The common ratio is $b = 3$.
    • Step backward to $x = 0$: Since multiplication by $3$ moves forward, division by $3$ moves backward: $a = f(0) = \frac{f(1)}{3} = \frac{6}{3} = 2$.
    • The resulting $y$-intercept is $(0, 2)$.
    • The explicit exponential function is $f(x) = 2(3)^x$.
  • Exponential Decay Table: Given table values for $x = 0, 1, 2, 3$ with outputs $f(x) = 64, 16, 4, 1$:
    • Calculate the successive ratios: $\frac{16}{64} = \frac{1}{4}$, $\frac{4}{16} = \frac{1}{4}$, $\frac{1}{4} = \frac{1}{4}$. The common ratio is $b = \frac{1}{4}$.
    • Initial value: At $x = 0$, $f(0) = 64$, so $a = 64$.
    • The explicit exponential function is $f(x) = 64\left(\frac{1}{4}\right)^x$.
  • Table Spanning Negative Inputs: Given values for $x = -2, -1, 0, 1$ with outputs $f(x) = 2, 6, 18, 54$:
    • Calculate the successive ratios: $\frac{6}{2} = 3$, $\frac{18}{6} = 3$, $\frac{54}{18} = 3$. The common ratio is $b = 3$.
    • Initial value: Locate the value at input $x = 0$, which is $f(0) = 18$, so $a = 18$.
    • The explicit exponential function is $f(x) = 18(3)^x$.

Tabular and Graphical Analysis of Base Exponential Functions

• Analysis of $y = 2^x$ (Exponential Growth):

  • Table of Values over the Domain $x \in \{-2, -1, 0, 1, 2\}$:
    • For $x = -2$: $y = 2^{-2} = \frac{1}{2^2} = \frac{1}{4} = 0.25$
    • For $x = -1$: $y = 2^{-1} = \frac{1}{2^1} = \frac{1}{2} = 0.5$
    • For $x = 0$: $y = 2^0 = 1$
    • For $x = 1$: $y = 2^1 = 2$
    • For $x = 2$: $y = 2^2 = 4$
  • Key Graphical Features:
    • $y$-Intercept: $(0, 1)$.
    • Horizontal Asymptote: The line $y = 0$ (the $x$-axis).
    • Direction: Strictly increasing across the entire domain $(-\theta, \theta)$.

• Analysis of $y = \left(\frac{1}{2}\right)^x$ (Exponential Decay):

  • Table of Values over the Domain $x \in \{-2, -1, 0, 1, 2\}$:
    • For $x = -2$: $y = \left(\frac{1}{2}\right)^{-2} = 2^2 = 4$
    • For $x = -1$: $y = \left(\frac{1}{2}\right)^{-1} = 2^1 = 2$
    • For $x = 0$: $y = \left(\frac{1}{2}\right)^0 = 1$
    • For $x = 1$: $y = \left(\frac{1}{2}\right)^1 = \frac{1}{2} = 0.5$
    • For $x = 2$: $y = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25$
  • Key Graphical Features:
    • $y$-Intercept: $(0, 1)$.
    • Horizontal Asymptote: The line $y = 0$ (the $x$-axis).
    • Direction: Strictly decreasing across the entire domain $(-\theta, \theta)$.

• Comparative Synthesis ($y = \left(\frac{1}{2}\right)^x$ vs. $y = 2^x$):

  • Direction Comparison: While $y = 2^x$ displays exponential growth (increasing monotonically from left to right), $y = \left(\frac{1}{2}\right)^x$ exhibits exponential decay (decreasing monotonically from left to right).
  • $y$-Intercept Comparison: Both functions share an identical $y$-intercept at $(0, 1)$.
  • Asymptote Comparison: Both functions share the exact same horizontal asymptote at $y = 0$.
  • Symmetry and Reflection Property: Because $\left(\frac{1}{2}\right)^x = (2^{-1})^x = 2^{-x}$, the graph of $y = \left(\frac{1}{2}\right)^x$ is the precise geometric reflection of $y = 2^x$ across the vertical $y$-axis.

Geometric Sequences: Explicit and Recursive Representations

• Definitions and General Mathematical Forms:

  • A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio ($r$).
  • Recursive Definition: Expresses the $n$-th term ($g_n$ or $a_n$) as a function of the preceding term ($g_{n-1}$ or $a_{n-1}$):         $g_n = r \cdot g_{n-1} \quad \text{or} \quad g_n = g_{n-1}(r)$         A complete recursive specification strictly requires stating the initial first term ($g_1$ or $a_1$).
  • Explicit Formula: Expresses the $n$-th term directly as a function of the term index $n$:         $g_n = g_1(r)^{n-1} \quad \text{or} \quad a_n = a_1(r)^{n-1}$

• Conversion from Recursive to Explicit Form:

  • Problem 6A: Given recursive form $g_1 = 5$, $g_n = 2 \cdot g_{n-1}$.
    • Identify first term: $g_1 = 5$.
    • Identify common ratio: $r = 2$.
    • Convert to explicit formula: $g_n = 5(2)^{n-1}$.
  • Problem 6B: Given recursive form $a_1 = 81$, $a_n = \left(\frac{1}{3}\right) \cdot a_{n-1}$.
    • Identify parameters: $a_1 = 81$, $r = \frac{1}{3}$.
    • Convert to explicit formula: $a_n = 81\left(\frac{1}{3}\right)^{n-1}$.
    • Calculate the 5th term ($a_5$):             $a_5 = 81\left(\frac{1}{3}\right)^{5-1} = 81\left(\frac{1}{3}\right)^4 = 81 \times \frac{1}{81} = 1$

• Deriving Sequences from Lists of Values:

  • Ascending Sequence ($2, 10, 50, 250, \dots$):
    • Compute common ratio: $r = \frac{10}{2} = \frac{50}{10} = \frac{250}{50} = 5$.
    • First term: $g_1 = 2$.
    • Recursive definition: $g_n = 5(g_{n-1})$ with $g_1 = 2$.
    • Explicit formula: $g_n = 2(5)^{n-1}$.
  • Descending Sequence ($96, 48, 24, 12, \dots$):
    • Compute common ratio: $r = \frac{48}{96} = \frac{24}{48} = \frac{12}{24} = \frac{1}{2}$.
    • First term: $g_1 = 96$.
    • Recursive definition: $g_n = \frac{1}{2}(g_{n-1})$ with $g_1 = 96$.
    • Explicit formula: $g_n = 96\left(\frac{1}{2}\right)^{n-1}$.

• Evaluating Specific Terms:

  • Problem 8A: Given the explicit sequence $g_n = 3(2)^{n-1}$, evaluate the 10th term ($g_{10}$):
    • Substitute $n = 10$: $g_{10} = 3(2)^{10-1} = 3(2)^9$.
    • Evaluate the exponential power: $2^9 = 512$.
    • Compute product: $g_{10} = 3 \times 512 = 1536$.
  • Problem 8B: Given the geometric sequence $1000, 100, 10, \dots$, evaluate the 6th term ($g_6$):
    • Determine first term: $g_1 = 1000$.
    • Compute common ratio: $r = \frac{100}{1000} = \frac{1}{10} = 0.1$.
    • Construct explicit formula: $g_n = 1000\left(\frac{1}{10}\right)^{n-1}$.
    • Substitute $n = 6$:             $g_6 = 1000\left(\frac{1}{10}\right)^{6-1} = 1000\left(\frac{1}{10}\right)^5 = 1000 \times \frac{1}{100000} = \frac{1}{100} = 0.01$

Transformations of Exponential Functions

• General Transformation Framework:

  • For a parent function $f(x) = b^x$, the comprehensive transformed function is modeled by:         $g(x) = a \cdot b^{c(x - h)} + k$
  • Vertical Shift ($k$): Moves the curve up if $k > 0$ and down if $k < 0$. Adjusts the horizontal asymptote to $y = k$.
  • Horizontal Shift ($h$): Replaces $x$ with $x - h$. Moves the curve right by $h$ units if $(x - h)$ and left by $h$ units if $(x + h)$.
  • Vertical Dilation ($a$): Vertically stretches the curve by a factor of $|a|$ if $|a| > 1$, and vertically compresses (shrinks) the curve by a factor of $|a|$ if $0 < |a| < 1$.
  • Vertical Reflection ($-a$): A negative sign outside the base reflects the entire graph over the $x$-axis.
  • Horizontal Reflection ($-x$): A negative sign on the exponent variable reflects the graph over the $y$-axis.
  • Horizontal Dilation ($c$): Replaces $x$ with $c \cdot x$. The curve is horizontally compressed by a factor of $\frac{1}{c}$ if $c > 1$, and horizontally stretched by a factor of $\frac{1}{c}$ if $0 < c < 1$.

• Describing Multi-Step Transformations from Equations:

  • Transformation of $f(x) = 2^x$ to produce $g(x) = 2^{x+5} - 3$:
    • Horizontal translation: Shifted left 5 units (due to the $x + 5$ exponent).
    • Vertical translation: Shifted down 3 units (due to the constant $-3$).
  • Transformation of $f(x) = 5^x$ to produce $g(x) = -(5)^{-x} + 4$:
    • Vertical reflection: Reflected over the $x$-axis (due to the negative sign in front of the base).
    • Horizontal reflection: Reflected over the $y$-axis (due to the negative exponent $-x$).
    • Vertical translation: Shifted up 4 units (due to the constant $+4$).

• Constructing Transformed Equations from Written Descriptions:

  • Shift up 5, left 2, vertical stretch by 4 on $f(x) = 3^x$:
    • Vertical stretch: Multiply by $4$.
    • Shift left 2: Replace $x$ with $x + 2$.
    • Shift up 5: Add $5$.
    • Resulting equation: $g(x) = 4(3)^{x+2} + 5$.
  • Shift down 1, right 4, vertical shrink by factor of $\frac{1}{2}$ on $f(x) = 2^x$:
    • Vertical shrink: Multiply by $\frac{1}{2}$.
    • Shift right 4: Replace $x$ with $x - 4$.
    • Shift down 1: Subtract $1$.
    • Resulting equation: $g(x) = \frac{1}{2}(2)^{x-4} - 1$.
  • Horizontal Compression by a factor of $\frac{1}{3}$ on $f(x) = 2^x$:
    • A horizontal compression by a factor of $\frac{1}{3}$ corresponds to multiplying the input variable $x$ by the reciprocal of the compression factor ($c = 3$).
    • Resulting equation: $g(x) = 2^{3x}$.
  • Horizontal Stretch by a factor of $4$ on $f(x) = 3^x$:
    • A horizontal stretch by a factor of $4$ corresponds to multiplying the input variable $x$ by the reciprocal of the stretch factor ($c = \frac{1}{4}$).
    • Resulting equation: $g(x) = 3^{\frac{1}{4}x}$.
  • Reflection over $x$-axis, right 5, vertical stretch by 3 on $f(x) = 2^x$:
    • Reflect over $x$-axis and vertical stretch: Coefficient is $-3$.
    • Move right 5: Exponent is $x - 5$.
    • Resulting equation: $g(x) = -3(2)^{x-5}$.
  • Reflection over $x$-axis, left 2, vertical stretch by 6, up 1 on $f(x) = 4^x$:
    • Reflect over $x$-axis and vertical stretch: Coefficient is $-6$.
    • Move left 2: Exponent is $x + 2$.
    • Move up 1: Constant term is $+1$.
    • Resulting equation: $g(x) = -6(4)^{x+2} + 1$.

Modeling Exponential Growth and Decay in Real-World Contexts

• Standard Real-World Models:

  • Exponential growth is modeled by $y = a(1 + r)^t$, where $a$ is the initial amount, $r > 0$ is the fractional growth rate, and $(1 + r)$ is the growth factor.
  • Exponential decay is modeled by $y = a(1 - r)^t$, where $a$ is the initial amount, $0 < r < 1$ is the fractional decay rate, and $(1 - r)$ is the decay factor.

• Population and Depreciation Word Problems:

  • Town Population Growth: A town of $12000$ residents grows at a rate of $4\%$ per year.
    • Initial quantity: $a = 12000$.
    • Rate: $r = 0.04$, giving growth factor $1 + r = 1 + 0.04 = 1.04$.
    • Model equation: $y = 12000(1 + 0.04)^t = 12000(1.04)^t$.
    • Evaluation after $t = 10$ years:             $y = 12000(1.04)^{10} \approx 12000 \times 1.480244 = 17762.93$
    • Rounding to the nearest whole number yields $17763$ people.
  • Automobile Value Depreciation: A vehicle worth $\$28000$ loses $12\%$ of its value annually.
    • Initial quantity: $a = 28000$.
    • Rate: $r = 0.12$, giving decay factor $1 - r = 1 - 0.12 = 0.88$.
    • Model equation: $y = 28000(1 - 0.12)^t = 28000(0.88)^t$.
    • Evaluation after $t = 5$ years:             $y = 28000(0.88)^5 \approx 28000 \times 0.527732 = 14776.49$
    • Rounding to the nearest whole dollar yields $\$14776$.

• Direct Growth and Decay Model Construction:

  • Financial Account Growth: Initial balance of $\$250$ increasing by $15\%$ each year:
    • $a = 250$, $r = 0.15$.
    • Function: $y = 250(1 + 0.15)^t = 250(1.15)^t$.
  • School Enrollment Loss: Initial enrollment of $800$ students losing $35\%$ each year:
    • $a = 800$, $r = 0.35$.
    • Function: $y = 800(1 - 0.35)^t = 800(0.65)^t$.

• Theoretical Criteria for Growth vs. Decay:

  • Given an equation $f(x) = a(b)^x$:
    • If the base $b > 1$, it models growth.
    • If the base $0 < b < 1$, it models decay.
  • Specific Problem Evaluations:
    • $f(x) = 0.5(1.8)^x$ models growth because base $b = 1.8 > 1$ (representing $1 + 0.8$, or an $80\%$ increase per unit).
    • $g(x) = 7(0.95)^x$ models decay because base $b = 0.95 < 1$ (representing $1 - 0.05$, or a $5\%$ decrease per unit).
  • Rate of Change Parameter ($r$) Analysis: For $f(x) = 10(1 + r)^x$ where $r$ is expressed as a decimal:
    • Growth: Occurs when $r > 0$, making the base $(1 + r) > 1$.
    • Decay: Occurs when $-1 < r < 0$ (or $r < 0$), making the base $0 < (1 + r) < 1$.

• Integer Multiple and Half-Life Rate Models:

  • Bacterial Tripling Growth: A culture begins with $5$ cells and triples every hour.
    • Model: $y = 5(3)^t$, where $t$ is measured in hours.
    • Evaluation after $t = 6$ hours:             $y = 5(3)^6 = 5 \times 729 = 3645 \text{ cells}$
  • Half-Life Mass Decay: A sample of $3200\text{ g}$ loses half its mass every day.
    • Model: $y = 3200\left(\frac{1}{2}\right)^t$, where $t$ is time in days.
    • Evaluation after $t = 5$ days:             $y = 3200\left(\frac{1}{2}\right)^5 = 3200 \times \frac{1}{32} = 100\text{ grams}$

Compound Interest Calculations

• Compound Interest Mathematical Formula:

  • The standard formula for interest compounded at regular intervals is:         $A = P\left(1 + \frac{r}{n}\right)^{nt}$
  • $A$: Total accrued balance after $t$ years.
  • $P$: Initial principal deposited.
  • $r$: Annual nominal interest rate (expressed as a decimal).
  • $n$: Number of times interest is compounded per calendar year.
  • $t$: Time invested in years.
  • Standard Compounding Period Frequencies ($n$):
    • Annually: $n = 1$
    • Semiannually: $n = 2$
    • Quarterly: $n = 4$
    • Monthly: $n = 12$
    • Daily: $n = 365$

• Analysis of Compound Interest Models:

  • Deconstruction of Given Formula: Given $y = 500\left(1 + \frac{0.06}{12}\right)^{12t}$.
    • Compounding Frequency: Because the denominator and exponent multiplier are $n = 12$, interest is compounded monthly.
    • Underlying Terms: Principal $P = \$500$, annual interest rate $r = 0.06 = 6\%$.
    • Balance Evaluation after $t = 3$ Years:$y = 500\left(1 + \frac{0.06}{12}\right)^{12 \times 3} = 500(1 + 0.005)^{36} = 500(1.005)^{36}$$y \approx 500 \times 1.1966805 = 598.34026$
    • Rounding to the nearest cent yields $\$598.34$.
  • Quarterly Compounding Problem Formulation: An investment of $\$1200$ earning $5\%$ interest compounded quarterly.
    • Parameters: $P = 1200$, $r = 0.05$, $n = 4$.
    • Equation:$y = 1200\left(1 + \frac{0.05}{4}\right)^{4t} = 1200(1 + 0.0125)^{4t} = 1200(1.0125)^{4t}$
    • Balance Evaluation after $t = 8$ Years:$y = 1200\left(1 + \frac{0.05}{4}\right)^{4 \times 8} = 1200(1.0125)^{32}$$y \approx 1200 \times 1.488131 = 1785.757$
    • Rounding to the nearest cent yields $\$1785.76$.

Classification of Linear, Exponential, and Non-Linear Data

• Mathematical Identification Criteria:

  • Linear Relationships: Exhibit a constant first difference ($\Delta y = y_{k+1} - y_k$) between successive outputs for uniform increments of $x$. Modeled by $y = mx + b$, indicating a constant additive rate of change.
  • Exponential Relationships: Exhibit a constant ratio ($\frac{y_{k+1}}{y_k} = b$) between successive outputs for uniform increments of $x$. Modeled by $y = a(b)^x$, indicating a constant multiplicative rate of change.
  • Neither Linear nor Exponential: Displays neither a constant first difference nor a constant common ratio (e.g., quadratic sequences where second differences are constant, or arbitrary irregular sequences).

• Tabular Data Analysis:

  • Data Set 1: Inputs $x = 0, 1, 2, 3$ with outputs $y = 5, 10, 20, 40$.
    • Calculate differences: $10 - 5 = 5$, $20 - 10 = 10$, $40 - 20 = 20$ (not constant).
    • Calculate ratios: $\frac{10}{5} = 2$, $\frac{20}{10} = 2$, $\frac{40}{20} = 2$ (constant ratio of $2$).
    • Classification: Exponential.
    • Function: $y = 5(2)^x$.
  • Data Set 2: Inputs $x = 0, 1, 2, 3$ with outputs $y = 100, 85, 70, 55$.
    • Calculate differences: $85 - 100 = -15$, $70 - 85 = -15$, $55 - 70 = -15$ (constant difference of $-15$).
    • Classification: Linear.
    • Function: Slope $m = -15$, initial value at $x = 0$ is $b = 100$, yielding $y = -15x + 100$.

• Comparative Table Evaluation (Table P vs. Table Q for $x = 0, 1, 2, 3$):

  • Table P ($y = 1, 4, 9, 16$):
    • Differences: $4 - 1 = 3$, $9 - 4 = 5$, $16 - 9 = 7$ (not constant).
    • Ratios: $\frac{4}{1} = 4$, $\frac{9}{4} = 2.25$, $\frac{16}{9} \approx 1.78$ (not constant).
    • Classification: Neither linear nor exponential because there is no common difference and no common ratio. (This data represents the quadratic relationship $y = (x + 1)^2$).
  • Table Q ($y = 3, 6, 12, 24$):
    • Ratios: $\frac{6}{3} = 2$, $\frac{12}{6} = 2$, $\frac{24}{12} = 2$ (constant ratio of $2$).
    • Classification: Exponential.

• Detailed Table Construction for $f(x) = 2(3)^x$:

  • Evaluate inputs over the domain $x \in \{-2, -1, 0, 1, 2\}$
    • For $x = -2$: $f(-2) = 2(3)^{-2} = \frac{2}{3^2} = \frac{2}{9}$
    • For $x = -1$: $f(-1) = 2(3)^{-1} = \frac{2}{3}$
    • For $x = 0$: $f(0) = 2(3)^0 = 2(1) = 2$
    • For $x = 1$: $f(1) = 2(3)^1 = 6$
    • For $x = 2$: $f(2) = 2(3)^2 = 2(9) = 18$
  • Exponential Justification: Testing consecutive output ratios confirms that:         $\frac{2/3}{2/9} = 3, \quad \frac{2}{2/3} = 3, \quad \frac{6}{2} = 3, \quad \frac{18}{6} = 3$         Because the common ratio is consistently $3$, each successive output is generated by multiplying the preceding value by $3$, mathematically proving the function is exponential.

• Real-World Scenario Classification:

  • A plant grows $3\text{ cm}$ every week: Classified as linear because a constant quantity ($3\text{ cm}$) is added each week (constant additive rate of change: $+3$).
  • Shares of a post triple every hour: Classified as exponential because the total count is multiplied by a constant factor of $3$ every single hour (multiplicative growth: $\times 3$).
  • A savings account earns $4\%$ interest each year: Classified as exponential because the balance is multiplied by a constant factor of $(1 + 0.04) = 1.04$ each year (percentage-based compounding).
  • Gym Membership: Charges a $\$30$ sign-up fee plus $\$20$ per month.
    • Classification: Linear (adds a constant $\$20$ each month).
    • Function: $y = 20x + 30$, where $x$ represents months.
  • Social Application User Base: Begins with $1000$ users and expands at a rate of $10\%$ per month.
    • Classification: Exponential (multiplies by $1 + 0.10 = 1.10$ each month).
    • Function: $y = 1000(1 + 0.10)^x = 1000(1.10)^x$, where $x$ represents months.