Lesson 2: Modeling with Linear Functions
Estimated time: 35-45 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Interpret slope as rate of change in real-world contexts
- Interpret y-intercept as an initial value or starting point
- Write linear equations from word problems and real-world scenarios
- Solve application problems involving cost, distance, temperature, and more
- Understand and apply direct variation models
- Make predictions using linear models
- Determine when a linear model is appropriate for a given situation
Interpreting Slope and Intercept in Context
In Lesson 1, we learned about slope and y-intercept as mathematical concepts. Now we'll see how these translate into real-world meanings. The key is understanding what the variables x and y represent in each situation.
Slope in Context: The slope represents the rate of change—how much y changes for each 1-unit increase in x.
Y-Intercept in Context: The y-intercept represents the initial value—the value of y when x = 0.
Real-World Interpretation Guide
Cost Problems
Slope: Cost per item or per unit of time
Y-intercept: Fixed cost, initial fee, or setup charge
Distance Problems
Slope: Speed or velocity
Y-intercept: Starting distance or initial position
Temperature Conversion
Slope: Conversion rate between units
Y-intercept: Offset between scales
Population/Growth
Slope: Rate of increase or decrease per time period
Y-intercept: Initial population or amount
Example 1: Phone Plan Cost
A cell phone plan costs $30 per month plus $0.10 per text message. Write a linear equation for the total monthly cost.
Solution:
Step 1: Identify the variables
Let x = number of text messages
Let C = total monthly cost in dollars
Step 2: Identify slope and y-intercept
Slope (m) = $0.10 per message (rate of change)
Y-intercept (b) = $30 (fixed monthly charge)
Step 3: Write the equation
C = 0.10x + 30
Answer: C = 0.10x + 30
Interpretation:
The slope of 0.10 means each additional text message increases the cost by $0.10. The y-intercept of 30 means even with zero text messages, you pay $30 for the basic service.
Using the model: How much would 200 text messages cost?
C = 0.10(200) + 30 = 20 + 30 = $50
Example 2: Car Rental
A car rental company charges a flat fee of $45 plus $0.25 per mile driven.
(a) Write a linear function for the total cost.
(b) What does the slope represent?
(c) How much would it cost to drive 150 miles?
Solution:
(a) Let x = miles driven, C = total cost
C = 0.25x + 45
(b) The slope of 0.25 represents the cost per mile. For every additional mile driven, the cost increases by $0.25.
(c) Cost for 150 miles:
C = 0.25(150) + 45
C = 37.50 + 45
C = $82.50
Distance and Motion Problems
Linear functions are perfect for modeling constant-speed motion. The fundamental relationship is:
Distance = Rate × Time
In linear form: d = rt + d₀
where d = distance, r = rate (speed), t = time, d₀ = starting distance
Example 3: Road Trip
You're on a road trip, already 50 miles from home, traveling at a constant speed of 60 miles per hour.
(a) Write an equation for your distance from home.
(b) How far from home will you be after 3 hours?
(c) When will you be 200 miles from home?
Solution:
(a) Let t = hours traveled, d = distance from home in miles
Slope (rate) = 60 mph
Y-intercept (initial distance) = 50 miles
d = 60t + 50
(b) After 3 hours:
d = 60(3) + 50
d = 180 + 50
d = 230 miles
(c) When d = 200:
200 = 60t + 50
150 = 60t
t = 2.5 hours
You'll be 200 miles from home after 2.5 hours (2 hours and 30 minutes).
Example 4: Two Cars Meeting
Two cities are 300 miles apart. A car leaves City A traveling toward City B at 50 mph. At the same time, another car leaves City B traveling toward City A at 70 mph. When will they meet?
Solution:
Step 1: Set up equations
Car from A: d₁ = 50t (starts at position 0)
Car from B: d₂ = 300 - 70t (starts at 300, moving toward 0)
Step 2: They meet when d₁ = d₂
50t = 300 - 70t
120t = 300
t = 2.5 hours
Answer: The cars will meet after 2.5 hours.
Check: Position where they meet:
Car A: 50(2.5) = 125 miles from City A
Car B: 70(2.5) = 175 miles traveled = 300 - 175 = 125 miles from City A
Temperature Conversion
Temperature conversions between Fahrenheit and Celsius provide a perfect example of linear functions that don't start at the origin.
Celsius to Fahrenheit Conversion:
F = (9/5)C + 32
Slope = 9/5 = 1.8 (Fahrenheit degrees per Celsius degree)
Y-intercept = 32 (freezing point offset)
Example 5: Temperature Conversion
(a) Convert 25°C to Fahrenheit.
(b) At what temperature are Celsius and Fahrenheit equal?
Solution:
(a) Using F = (9/5)C + 32:
F = (9/5)(25) + 32
F = 45 + 32
F = 77°F
(b) When F = C, substitute C for F:
C = (9/5)C + 32
C - (9/5)C = 32
-4/5·C = 32
C = 32 × (-5/4)
C = -40
Answer: Celsius and Fahrenheit are equal at -40 degrees!
Direct Variation
Direct variation is a special type of linear relationship where two quantities are proportional—one is a constant multiple of the other.
Direct Variation: y varies directly with x if there exists a nonzero constant k such that:
y = kx
k is called the constant of variation or constant of proportionality.
Key Features of Direct Variation
- The equation is in the form y = kx (no y-intercept term)
- The graph passes through the origin (0, 0)
- The ratio y/x is always constant: y/x = k
- If x doubles, y doubles; if x triples, y triples, etc.
Example 6: Wages
Your wage varies directly with the number of hours worked. If you earn $150 for working 12 hours:
(a) Find the constant of variation.
(b) Write the direct variation equation.
(c) How much will you earn for 20 hours?
Solution:
(a) Finding k when W = 150 and h = 12:
W = kh
150 = k(12)
k = 150/12 = 12.50
The constant of variation is $12.50 per hour (your hourly wage).
(b) The equation is:
W = 12.50h
(c) For 20 hours:
W = 12.50(20)
W = $250
Example 7: Distance at Constant Speed
Distance varies directly with time when traveling at constant speed. A car travels 180 miles in 3 hours.
(a) Find the constant of variation.
(b) How far will the car travel in 7 hours?
Solution:
(a) Finding k:
d = kt
180 = k(3)
k = 60 mph
(b) Distance in 7 hours:
d = 60(7) = 420 miles
Writing Equations from Word Problems
Strategy for Word Problems
- Identify the variables: What are x and y?
- Find the rate of change: What's the slope?
- Find the initial value: What's the y-intercept?
- Write the equation: Use y = mx + b or appropriate form
- Answer the question: Substitute and solve
Example 8: Water Tank
A water tank contains 500 gallons of water. Water is being drained at a rate of 15 gallons per minute.
(a) Write an equation for the amount of water in the tank.
(b) How much water is in the tank after 12 minutes?
(c) When will the tank be empty?
Solution:
(a) Setting up:
Let t = time in minutes
Let W = water remaining in gallons
Slope = -15 (negative because water is decreasing)
Y-intercept = 500 (initial amount)
W = -15t + 500
(b) After 12 minutes:
W = -15(12) + 500
W = -180 + 500
W = 320 gallons
(c) When W = 0:
0 = -15t + 500
15t = 500
t = 33.33 minutes
The tank will be empty after approximately 33 minutes and 20 seconds.
Example 9: Gym Membership
A gym charges a $50 enrollment fee plus $35 per month.
(a) Write an equation for total cost based on months of membership.
(b) What is the cost for 1 year of membership?
(c) If you've paid $540 total, how many months have you been a member?
Solution:
(a) Let m = number of months, C = total cost
Slope = $35 per month
Y-intercept = $50 (one-time enrollment)
C = 35m + 50
(b) For 1 year (m = 12):
C = 35(12) + 50
C = 420 + 50
C = $470
(c) When C = 540:
540 = 35m + 50
490 = 35m
m = 14 months
Example 10: Depreciation
A car purchased for $24,000 depreciates (loses value) at $2,000 per year.
(a) Write an equation for the car's value.
(b) What will the car be worth after 5 years?
(c) When will the car be worth $10,000?
Solution:
(a) Let t = years since purchase, V = car value
Slope = -2000 (negative for depreciation)
Y-intercept = 24000 (original value)
V = -2000t + 24000
(b) After 5 years:
V = -2000(5) + 24000
V = -10000 + 24000
V = $14,000
(c) When V = 10000:
10000 = -2000t + 24000
-14000 = -2000t
t = 7 years
Check Your Understanding
Try these questions to see if you've grasped the key concepts:
1. A streaming service costs $8 per month with no signup fee. Write an equation for the total cost after m months. What do the slope and y-intercept represent?
Slope (8): Cost per month ($8/month)
Y-intercept (0): No initial fee or startup cost
Note: This is actually a direct variation since there's no initial fee!
2. You're driving at 65 mph. You're already 30 miles from home. Write an equation for your distance from home, and find how far you'll be after 2 hours.
(where t = hours, d = distance from home)
After 2 hours:
d = 65(2) + 30
d = 130 + 30
d = 160 miles
3. The number of calories burned varies directly with time spent exercising. If you burn 150 calories in 20 minutes, how many calories will you burn in 45 minutes?
C = kt
150 = k(20)
k = 7.5 calories per minute
For 45 minutes:
C = 7.5(45)
C = 337.5 calories
4. A bathtub contains 40 gallons of water. It drains at 2.5 gallons per minute. Write an equation and determine when the tub will be empty.
(negative slope because water is decreasing)
When empty (W = 0):
0 = -2.5t + 40
2.5t = 40
t = 16 minutes
5. A fitness app charges $10 per month plus a one-time setup fee. After 6 months, you've paid $85 total. What was the setup fee?
Using the given information:
85 = 10(6) + b
85 = 60 + b
b = 25
The setup fee was $25
6. Convert 100°F to Celsius. Use the formula C = (5/9)(F - 32).
C = (5/9)(F - 32)
C = (5/9)(100 - 32)
C = (5/9)(68)
C = 340/9
C ≈ 37.78°C
Key Takeaways
- Slope in context represents rate of change (cost per item, speed, dollars per hour, etc.)
- Y-intercept in context represents initial value (fixed fee, starting position, setup cost)
- Direct variation: y = kx (proportional relationship, passes through origin)
- Distance = rate × time: d = rt + d₀ (with starting distance d₀)
- For word problems: identify variables, find slope and intercept, write equation, solve
- Linear models work when there's a constant rate of change
- Always include units in your interpretation (dollars per hour, miles per gallon, etc.)
Ready for More?
Next Lesson
In Lesson 3, we'll move beyond lines to study quadratic functions. You'll learn about parabolas, vertex form, transformations, and how to graph these powerful functions!
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