Eighth Grade Mathematics: Comparing Functions across Graphs, Tables, and Equations
In real-world problem solving, information is rarely delivered in identical formats. One company might advertise its pricing through an algebraic formula, a competitor might provide a printed table of rates, and a third company might present a marketing graph. To make informed quantitative decisions, you must be able to compare functions that are displayed across completely different representations. In this chapter, you will master extracting rates of change and initial values from equations, tables, graphs, and verbal scenarios to compare functions directly.
The Common Currency of Comparison: Slope and Intercept
Regardless of how a linear function is presented—as a graph, a table of numbers, an algebraic formula, or a written paragraph—it is uniquely determined by two fundamental parameters:
Parameter 1: Rate of Change (Slope, m) ---> Measures speed, steepness, or unit cost.
Parameter 2: Initial Value (y-intercept, b) ---> Measures starting point, base fee, or offset.
To compare two functions presented in different formats, convert both representations into their numerical values of m and b:
Format How to Extract Slope (m) How to Extract Initial Value (b)
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Equation: y = mx + b Read the coefficient m of x Read the constant term b
Graph Calculate Rise / Run on grid Find the y-value where line crosses x = 0
Table Compute (y2 - y1) / (x2 - x1) Find the y-value corresponding to x = 0
Verbal Description Identify the rate per unit Identify the starting or base amount
Once m and b are determined for each function, direct comparisons of speed, cost, and starting conditions become trivial.
Comparing Rates of Change (Steepness)
The function with the greater rate of change increases at a faster pace, resulting in a steeper line on a graph.
Scenario: Comparing Two Water Pumps
Pump A: Modeled by the equation y = 45x (where y is gallons and x is minutes)
Slope mA = 45 gallons per minute.
Pump B: Given by a table of data:
Minutes (x): 2 4 6
Gallons (y): 100 200 300
Slope mB = (200 - 100) / (4 - 2) = 100 / 2 = 50 gallons per minute.
Comparison:
Pump B empties water faster than Pump A because 50 gpm > 45 gpm.
The graph of Pump B is steeper than the graph of Pump A.
Comparing Initial Values and Break-Even Points
When one function has a higher starting value but a lower rate of change than another, the two functions will eventually cross at a break-even point.
Comparing Cellular Phone Plans:
Plan 1 (Equation): C = 0.10m + 30 (m1 = 0.10, b1 = 30)
Plan 2 (Graph): Passes through (0, 15) and (100, 35)
Slope m2 = (35 - 15) / (100 - 0) = 20 / 100 = 0.20
y-intercept b2 = 15
Comparison:
- Initial Value: Plan 1 has a higher base fee ($30 vs $15).
- Rate of Change: Plan 2 has a higher per-minute rate ($0.20/min vs $0.10/min).
Find the Break-Even Point (Set costs equal):
0.10m + 30 = 0.20m + 15
15 = 0.10m ---> m = 150 minutes.
For under 150 minutes, Plan 2 is cheaper. For over 150 minutes, Plan 1 is cheaper.
Chapter Exercises
Exercise 1: Function A is given by the equation y = 3x - 5. Function B is represented by a table containing the points (1, 2), (3, 8), and (5, 14). Determine which function has a greater rate of change and which function has a greater y-intercept.
Exercise 2: Vehicle 1 travels according to the graph of a line passing through (0, 0) and (4, 220). Vehicle 2 travels according to the equation d = 58t, where d is distance in miles and t is time in hours. Determine which vehicle is traveling faster, showing all calculations.
Exercise 3: Two savings accounts are opened on the same day. Account A starts with 150 dollars and receives deposits of 25 dollars per week. Account B is represented by a table where the balance after 2 weeks is 220 dollars and after 6 weeks is 340 dollars. Determine which account started with more money and which account is growing at a faster rate.
Solutions and Explanations
Solution 1: For Function A, y = 3x - 5, the equation is in slope-intercept form, giving a rate of change of mA = 3 and a y-intercept of bA = -5. For Function B, calculate the rate of change using the table points (1, 2) and (3, 8): mB = (8 - 2) / (3 - 1) = 6 / 2 = 3. Check with the next pair: (14 - 8) / (5 - 3) = 6 / 2 = 3. Thus, Function B has a rate of change of mB = 3. To find the y-intercept of Function B, use y = mx + b with point (1, 2): 2 = 3*(1) + b, which gives 2 = 3 + b, so bB = -1. Comparing the parameters: both functions have the exact same rate of change (mA = mB = 3). Function B has a greater y-intercept because -1 is greater than -5 (-1 > -5).
Solution 2: For Vehicle 1, the graph passes through the origin (0, 0) and (4, 220). Its speed is the slope of the line: m1 = (220 - 0) / (4 - 0) = 220 / 4 = 55 miles per hour. For Vehicle 2, the equation is d = 58t, which shows directly that its speed (rate of change) is m2 = 58 miles per hour. Comparing the two speeds: 58 mph is greater than 55 mph (58 > 55). Therefore, Vehicle 2 is traveling faster than Vehicle 1 by 3 miles per hour.
Solution 3: For Account A, the verbal description reveals an initial value of bA = 150 dollars and a weekly rate of change of mA = 25 dollars per week. For Account B, calculate the rate of change using the coordinate points (2, 220) and (6, 340): mB = (340 - 220) / (6 - 2) = 120 / 4 = 30 dollars per week. To find the initial starting balance bB of Account B, substitute mB = 30 and point (2, 220) into y = mx + b: 220 = 30*(2) + bB, which gives 220 = 60 + bB, yielding bB = 160 dollars. Comparing the two accounts: Account B started with more money (160 dollars versus 150 dollars for Account A), and Account B is also growing at a faster rate (30 dollars per week versus 25 dollars per week for Account A).