Comprehensive Guide to Exponential Growth, Decay, and Mathematical Modeling

General Definitions and Forms of Exponential Functions# An exponential function $f$ is defined by the form $f(x) = b^x$, where $x$ is any real number, b > 0, and $b \neq 1$.# A more general form is $y = a(b)^x$, where $a$ represents the initial value (also the y-intercept) and $b$ is the common multiplying factor.# Identification of Growth vs. Decay based on the factor $b$:# If b > 1, the function represents Exponential Growth.# If 0 < b < 1, the function represents Exponential Decay.# Comparison with Other Function Types:# Linear Functions: The form is $y = mx + b$. These show a constant/same rate of change (slope) where the values grow or shrink by adding/subtracting the same amount each time.# Quadratic Functions: The form is $Ax^2 + Bx + C$.# Exponential Functions: These grow or decay at a rate (multiplicative factor), often doubling ($b=2$) or growing by a specific percentage.# Exponential Growth Functions# A function of the form $y = a(1 + r)^t$ is an exponential growth function, where a > 0 and r > 0.# Component Definitions:# $y$: Final amount.# $a$: Initial amount or start value.# $r$: Rate of growth (always expressed in decimal form).# $t$: Time.# $(1 + r)$: The growth factor, which comes from simplifying the term inside the parentheses.# Exponential Decay Functions# A function of the form $y = a(1 - r)^t$ is an exponential decay function, where a > 0 and 0 < r < 1.# Component Definitions:# $y$: Final amount.# $a$: Initial amount.# $r$: Rate of decay (expressed in decimal form).# $t$: Time.# $(1 - r)$: The decay factor.# Percent to Decimal Conversion Rules# Percent means "per hundred." To convert, divide by 100 or move the decimal two places to the left.# Examples:# $75\% = 0.75$# $5\% = 0.05$# $15.8\% = 0.158$# $0.65\% = 0.0065$# $1.9\% = 0.019$# Detailed Practical Applications and Worked Examples# Example 1: Vehicle Depreciation (Labeled as Decrease/Decay)# Task: Use an exponential function to find the value of a car initially worth $18,000 depreciating at a rate of 12% per year after 10 years.# Variables:# $a = 18,000$ (Initial)# $r = -12\% = -0.12$ (Rate)# $t = 10$ (Time)# Equation Setup: $y = a(1 + r)^t$# Execution: $y = 18,000(1 + 0.12)^{10}$# Simplified: $y = 18,000(1.12)^{10}$# Result: $y = 55,905.26$# Conclusion in Notes: After 10 years, the value of the car goes down by $55,905.26 (Note: The calculation provided in the transcript uses the growth formula despite the decay label).# Example 2: Teacher Salary Growth# Task: Ms. Acosta starts a job with a salary of $34,000 and receives a 1.5% increase annually. How much will she earn in 7 years?# Variables:# $a = 34,000$# $r = 1.5\% = 0.015$# $t = 7$# Equation: $y = a(1 + r)^t$# Execution: $y = 34,000(1 + 0.015)^7$# Simplified: $y = 34,000(1.015)^7$# Result: $y = 37,734.72$# Conclusion: Ms. Acosta will earn $37,734.72 in 7 years.# Example 3: High School Enrollment Decline# Context: In 2000, 2200 students attended Polaris High School. Enrollment declines 2% annually.# Equation for t years after 2000: $y = 2200(1 - 0.02)^t$# Prediction for 2015 ($t = 15$):# Variables: $a = 2200$, $r = 0.02$, $t = 15$# Execution: $y = 2200(1 - 0.02)^{15}$# Simplified: $y = 2200(0.98)^{15}$# Result: 1624# Conclusion: In 2015, there will be 1624 students still enrolled in the school.# Example 4: Investment Growth# Task: Find the value of a $1400 investment after 25 years if it increases by 9% each year.# Variables: $a = 1400$, $r = 0.09$, $t = 25$# Execution: $y = 1400(1 + 0.09)^{25}$# Simplified: $y = 1400(1.09)^{25}$# Result: $12,072.31$# Conclusion: After 25 years, the investment value is $12,072.31.# Identification Exercises for Growth and Decay# Parameters to determine:# 1. Initial value# 2. Growth or Decay type# 3. Factor# Examples provided by student (Tomisi Bakare):# $f(x) = 4(0.5)^x$: Initial value: 4; Decay (D); Factor: 0.5.# $y = 8(\frac{1}{2})^x$: Initial value: 8; Decay (D); Factor: $\frac{1}{2}$.# $f(x) = 6(2)^x$: Initial value: 6; Growth (G); Factor: 2 (doubles).# $y = 5^x$: Initial value: 1 (implied); Growth (G); Factor: 5.# $f(x) = 2(3)^x$: Initial value: 2; Growth (G); Factor: 3.# $y = 4(3)^x$: Growth.# $y = 5(1)^x$: N/A (b cannot be 1).# $y = 0.01(2)^{30}$: Growing exponentially.# $f(x) = 1200(1 + 0.7)^x$: Initial: 1200; Growth; Factor: 1.7 (Handwritten note says factor 6.7, likely error).# $f(x) = 87(1 - 0.8)^x$: Initial: 87; Decay; Factor: 0.2.# Mathematical Modeling and Regression Analysis# Weight and Fuel Economy Study (Tomish Bahare):# Data Table for Sport-Utility Vehicles:# Weight (tons) vs. Fuel Economy (MPG):# 1.875 tons: 36.8 MPG# 2.0 tons: 28.4 MPG# 2.0 tons: 28.4 MPG# 2.125 tons: 26.7 MPG# 2.25 tons: 24.8 MPG# 2.5 tons: 23.3 MPG# 2.75 tons: 19.7 MPG# 3.0 tons: 20.4 MPG# 3.25 tons: 19.6 MPG# Linear Model Function: $f(w) = -10.139w + 49.993$# Comparative Evaluation:# For 1.875 tons: $f(1.875) = -10.139(1.875) + 49.993 = 30.982$ MPG (vs actual 36.8).# For 3.25 tons: $f(3.25) = -10.139(3.25) + 49.993 = 17.041$ MPG (vs actual 19.6).# Desmos Better Fit/Regression Function: $f(w) = -8.69w + 36.68$# Linear Graphing and Transformation Exercise (Page 5)# Problem: Determine the equation for a line passing through (5,0) and (0,-7).# Slope Calculation ($m$): $\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-7)}{5 - 0} = \frac{7}{5}$# Equation: $f(x) = \frac{7}{5}x - 7$# Reflection Note: Calculating $6(3) + 11 = 18 + 11 = 29$.# Quadratic Functions and Vertex Form (Jomisin Bakare)# Given Standard Form: $f(x) = -3x^2 + 18x - 21$# To convert to Vertex Form $f(x) = a(x - h)^2 + k$:# Find $h$: $h = -\frac{b}{2a} = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$# Find $k$: $f(3) = -3(3)^2 + 18(3) - 21 = -3(9) + 54 - 21 = -27 + 54 - 21 = 6$# Vertex Form: $f(x) = -3(x - 3)^2 + 6$# True Statements:# A. In vertex form, $f(x) = -3(x - 3)^2 + 6$ is true.# E. The vertex of $f(x)$ is located at (3, 6).# Quadrant Analysis for Transformations:# Function $f(x) = (x + 6)^2 - 3$ has a vertex at (-6, -3).# Function $g(x) = f(x) + 5 = (x + 6)^2 - 3 + 5 = (x + 6)^2 + 2$.# New vertex is at (-6, 2).# Since it opens up and the vertex is in Quadrant II, the graph stays in Quadrants I and II only.# Athlete Training and Linear Modeling# Data Table for Distance to Run Each Week:# Week 1: 13 miles# Week 2: 15.5 miles# Week 3: 15.5 miles# Week 4: 22.5 miles# Week 5: 23 miles# Week 6: 30 miles# Equation Model: $y = 3.5x + 8.5$# Analysis: For which week is the equation value greater than the actual distance?# Checking Week 5: $y = 3.5(5) + 8.5 = 17.5 + 8.5 = 26$. Since 26 > 23 (actual), Week 5 is the correct answer.# Bank Account Balance Modeling (Exponential)# Data Table for Bank Account:# 0 years: $10,000.00# 1 year: $10,130.00# 2 years: $10,261.69# Representative Function Identification:# The starting amount is 10,000. The growth factor is calculated by $\frac{10,130}{10,000} = 1.013$.# Correct Function: $f(x) = 10,000(1.013)^x$# Systems of Inequalities and Desmos Usage Strategy# Strategy: Go to Desmos and type the inequalities manually to see the intersection.# Interpretation of the Plane Graph:# Line boundaries that are unequal to (strictly greater than or less than) will be represented by broken lines.# Shading "below" a line corresponds to the "less than" symbol (y <).# Equation forms provided for checking:# y > -ax - b# y < -ax + b# y > ax + b# y < ax - b# Guidelines for Solving:# 1. Go to Desmos.# 2. Go to the table.# 3. Input points.# 4. Click exponential regression options.

General Definitions

An exponential function $f$ is defined by the form $f(x) = b^x$, where $x$ is any real number, b > 0, and $b \neq 1$.

A more general form is $y = a(b)^x$, where $a$ represents the initial value (also the y-intercept) and $b$ is the common multiplying factor.

Identification of Growth vs. Decay

Based on the factor $b$:

• If b > 1, the function represents Exponential Growth.

• If 0 < b < 1, the function represents Exponential Decay.

Comparison with Other Function Types

• Linear Functions: The form is $y = mx + b$. These show a constant/same rate of change (slope) where the values grow or shrink by adding/subtracting the same amount each time.

• Quadratic Functions: The form is $Ax^2 + Bx + C$.

• Exponential Functions: These grow or decay at a rate (multiplicative factor), often doubling ($b=2$) or growing by a specific percentage.

Exponential Growth Functions

A function of the form $y = a(1 + r)^t$ is an exponential growth function, where a > 0 and r > 0.

Component Definitions:

• $y$: Final amount.

• $a$: Initial amount or start value.

• $r$: Rate of growth (always expressed in decimal form).

• $t$: Time.

• $(1 + r)$: The growth factor, which comes from simplifying the term inside the parentheses.

Exponential Decay Functions

A function of the form $y = a(1 - r)^t$ is an exponential decay function, where a > 0 and 0 < r < 1.

Component Definitions:

• $y$: Final amount.

• $a$: Initial amount.

• $r$: Rate of decay (expressed in decimal form).

• $t$: Time.

• $(1 - r)$: The decay factor.

Percent to Decimal Conversion Rules

Percent means "per hundred." To convert, divide by 100 or move the decimal two places to the left.

Examples:

• $75\% = 0.75$

• $5\% = 0.05$

• $15.8\% = 0.158$

• $0.65\% = 0.0065$

• $1.9\% = 0.019$

Detailed Practical Applications and Worked Examples

Example 1: Vehicle Depreciation (Labeled as Decrease/Decay)

• Task: Use an exponential function to find the value of a car initially worth $18,000 depreciating at a rate of 12% per year after 10 years.

• Variables:

  • $a = 18,000$ (Initial)

  • $r = -12\% = -0.12$ (Rate)

  • $t = 10$ (Time)

• Equation Setup: $y = a(1 + r)^t$

• Execution: $y = 18,000(1 + 0.12)^{10}$

• Simplified: $y = 18,000(1.12)^{10}$

• Result: $y = 55,905.26$

• Conclusion in Notes: After 10 years, the value of the car goes down by $55,905.26 (Note: The calculation provided in the transcript uses the growth formula despite the decay label).

Example 2: Teacher Salary Growth

• Task: Ms. Acosta starts a job with a salary of $34,000 and receives a 1.5% increase annually. How much will she earn in 7 years?

• Variables:

  • $a = 34,000$

  • $r = 1.5\% = 0.015$

  • $t = 7$

• Equation: $y = a(1 + r)^t$

• Execution: $y = 34,000(1 + 0.015)^7$

• Simplified: $y = 34,000(1.015)^7$

• Result: $y = 37,734.72$

• Conclusion: Ms. Acosta will earn $37,734.72 in 7 years.

Example 3: High School Enrollment Decline

• Context: In 2000, 2200 students attended Polaris High School. Enrollment declines 2% annually.

• Equation for t years after 2000: $y = 2200(1 - 0.02)^t$

• Prediction for 2015 ($t = 15$):

• Variables: $a = 2200$, $r = 0.02$, $t = 15$

• Execution: $y = 2200(1 - 0.02)^{15}$

• Simplified: $y = 2200(0.98)^{15}$

• Result: 1624

• Conclusion: In 2015, there will be 1624 students still enrolled in the school.

Example 4: Investment Growth

• Task: Find the value of a $1400 investment after 25 years if it increases by 9% each year.

• Variables: $a = 1400$, $r = 0.09$, $t = 25$

• Execution: $y = 1400(1 + 0.09)^{25}$

• Simplified: $y = 1400(1.09)^{25}$

• Result: $12,072.31$

• Conclusion: After 25 years, the investment value is $12,072.31.

Identification Exercises for Growth and Decay

Parameters to determine:

1. Initial value

2. Growth or Decay type

3. Factor

Examples provided by student (Tomisi Bakare):

• $f(x) = 4(0.5)^x$: Initial value: 4; Decay (D); Factor: 0.5.

• $y = 8(\frac{1}{2})^x$: Initial value: 8; Decay (D); Factor: $\frac{1}{2}$.

• $f(x) = 6(2)^x$: Initial value: 6; Growth (G); Factor: 2 (doubles).

• $y = 5^x$: Initial value: 1 (implied); Growth (G); Factor: 5.

• $f(x) = 2(3)^x$: Initial value: 2; Growth (G); Factor: 3.

• $y = 4(3)^x$: Growth.

• $y = 5(1)^x$: N/A (b cannot be 1).

• $y = 0.01(2)^{30}$: Growing exponentially.

• $f(x) = 1200(1 + 0.7)^x$: Initial: 1200; Growth; Factor: 1.7 (Handwritten note says factor 6.7, likely error).

• $f(x) = 87(1 - 0.8)^x$: Initial: 87; Decay; Factor: 0.2.

Mathematical Modeling and Regression Analysis

Weight and Fuel Economy Study (Tomish Bahare):

• Data Table for Sport-Utility Vehicles:

  • Weight (tons) vs. Fuel Economy (MPG):

  • 1.875 tons: 36.8 MPG

  • 2.0 tons: 28.4 MPG

  • 2.0 tons: 28.4 MPG

  • 2.125 tons: 26.7 MPG

  • 2.25 tons: 24.8 MPG

  • 2.5 tons: 23.3 MPG

  • 2.75 tons: 19.7 MPG

  • 3.0 tons: 20.4 MPG

  • 3.25 tons: 19.6 MPG

• Linear Model Function: $f(w) = -10.139w + 49.993$

Comparative Evaluation:

• For 1.875 tons: $f(1.875) = -10.139(1.875) + 49.993 = 30.982$ MPG (vs actual 36.8).

• For 3.25 tons: $f(3.25) = -10.139(3.25) + 49.993 = 17.041$ MPG (vs actual 19.6).

Desmos Better Fit/Regression Function:

$f(w) = -8.69w + 36.68$

Linear Graphing and Transformation Exercise (Page 5)

• Problem: Determine the equation for a line passing through (5,0) and (0,-7).

Slope Calculation ($m$):

$\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-7)}{5 - 0} = \frac{7}{5}$

Equation:

$f(x) = \frac{7}{5}x - 7$

Reflection Note:

Calculating $6(3) + 11 = 18 + 11 = 29$.

Quadratic Functions and Vertex Form (Jomisin Bakare)

• Given Standard Form: $f(x) = -3x^2 + 18x - 21$

• To convert to Vertex Form $f(x) = a(x - h)^2 + k$:

Find $h$:

$h = -\frac{b}{2a} = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$

Find $k$:

$f(3) = -3(3)^2 + 18(3) - 21 = -3(9) + 54 - 21 = -27 + 54 - 21 = 6$

Vertex Form:

$f(x) = -3(x - 3)^2 + 6$

True Statements:

• A. In vertex form, $f(x) = -3(x - 3)^2 + 6$ is true.

• E. The vertex of $f(x)$ is located at (3, 6).

Quadrant Analysis for Transformations

• Function $f(x) = (x + 6)^2 - 3$ has a vertex at (-6, -3).

• Function $g(x) = f(x) + 5 = (x + 6)^2 - 3 + 5 = (x + 6)^2 + 2$.

• New vertex is at (-6, 2).

• Since it opens up and the vertex is in Quadrant II, the graph stays in Quadrants I and II only.

Athlete Training and Linear Modeling

Data Table for Distance to Run Each Week:

• Week 1: 13 miles

• Week 2: 15.5 miles

• Week 3: 15.5 miles

• Week 4: 22.5 miles

• Week 5: 23 miles

• Week 6: 30 miles

• Equation Model: $y = 3.5x + 8.5$

Analysis:

• For which week is the equation value greater than the actual distance?

• Checking Week 5: $y = 3.5(5) + 8.5 = 17.5 + 8.5 = 26$. Since 26 > 23 (actual), Week 5 is the correct answer.

Bank Account Balance Modeling (Exponential)

Data Table for Bank Account:

• 0 years: $10,000.00

• 1 year: $10,130.00

• 2 years: $10,261.69

• Representative Function Identification:

• The starting amount is 10,000. The growth factor is calculated by $\frac{10,130}{10,000} = 1.013$.

• Correct Function: $f(x) = 10,000(1.013)^x$

Systems of Inequalities and Desmos Usage Strategy

Strategy:

• Go to Desmos and type the inequalities manually to see the intersection.

Interpretation of the Plane Graph:

• Line boundaries that are unequal to (strictly greater than or less than) will be represented by broken lines.

• Shading "below" a line corresponds to the "less than" symbol (y <).

Equation forms provided for checking:

• y> -ax - b

• y < -ax + b

• y > ax + b

• y < ax - b

Guidelines for Solving:

1. Go to Desmos.

2. Go to the table.

3. Input points.

4. Click exponential regression options.