Comparing linear functions given different ways
One linear function is a table and another is a graph or a sentence. How can you tell which grows faster, which starts higher, and whether one ever catches the other?
Where you are 5 in Linear functions
When you reach for it
Choices often come down to two linear functions set side by side, and they rarely arrive in the same form: one shop sends a table, another a graph; one plan is an equation, another a sentence. A table and a graph cannot be compared just by looking at them.
Every linear function, whatever its form, is settled by two numbers: its rate of change m, the change in output for each one unit of input, and its initial value b, the output at input 0. With them the rule is f(x) = mx + b. So the move is to turn each function into those same two numbers first, in the same units, and only then compare: which grows faster, which starts higher, and whether one ever overtakes the other.
For two relationships that both start at 0, one number settles the comparison, as in Direct variation. Here both functions may have a nonzero start, and the two numbers can pull in opposite directions.
The lesson takes two days. Part 1 (Examples 1 to 3) gets the two numbers out of each form and compares them. Part 2 (Examples 4 to 6) uses them to decide who is ahead, and when, if ever, that changes.
The procedure
Part 1: the same two numbers from every form
From an equation. When the equation already reads y = mx + b, the coefficient of the input is the rate and the constant is the initial value, as in Slope-intercept form. When it is written another way, such as 4x - 2y = 6, solve for the output first: y = 2x - 3, so m = 2 and b = -3. The numbers 4 and -2 in the original are neither, which is why standard form is rewritten before it is read.
From words. Something charged or gained “per”, “each” or “for every” unit of input is the rate. Something paid or held “to join”, “at the start” or “before anything happens” is the output at input 0, the initial value. If the words give two outcomes instead (“after 4 months she had paid 60 dollars”), treat them as two points and work as for a table.
From a table. First check that the table is linear: divide each change in output by the change in input between neighboring columns, as in Constant rate of change and linear tables. Equal quotients mean a constant rate, and that quotient is m. Dividing matters when the input steps are uneven. If input 0 is not in the table, work back from any column (x_1, y_1): since y_1 = mx_1 + b,
From a graph. Read the initial value where the line meets the vertical axis (when the horizontal axis starts at 0), and the rate from two grid corners on the line, turning squares into the axes’ units before dividing, as in Reading a linear function from its graph.
Then compare, in the same units. The function with the larger m grows faster: each unit of input adds more to its output. The function with the larger b starts higher. When one rate is negative, two questions come apart. “Which grows faster” compares the rates themselves, so any rising function grows faster than any falling one. “Which changes faster” compares their sizes, so a fall of 400 a step is faster than a rise of 240. Before comparing, put both rates in the same units (dollars per month with dollars per month, not per year), and make sure input 0 means the same moment for both.
Part 2: who is ahead, and when that changes
The gap is linear too. Take f(x) = m_1x + b_1 and g(x) = m_2x + b_2. Their difference is
which is itself a linear function. Its initial value b_1 - b_2 is how far apart the two start. Its rate m_1 - m_2 is how much f gains on g for each unit of input.
When one overtakes the other. Suppose g starts higher, b_2 > b_1, but f grows faster, m_1 > m_2. Then f is behind by b_2 - b_1 at the start and closes the gap by m_1 - m_2 each unit. The gap is gone after
units of input: the starting gap divided by how fast it closes. There the two outputs are equal, and past it f is ahead and stays ahead, because it keeps gaining. If one function rises and the other falls, the gap closes from both sides at once, at the sum of the two rates’ sizes.
When it never happens. If the rates are equal, the gap never changes, so the two never take the same value unless they are the same function; their graphs are parallel lines, as in Parallel and perpendicular lines. If the function that starts higher also grows faster, it only pulls further ahead, and the formula above gives a negative input, a moment before the start.
The crossing must be in range. A crossing counts only if its input lies inside both functions’ domains, and is a whole number when the input counts whole things such as posters.
Looking ahead. Setting f(x) = g(x) and solving is the same reasoning written as an equation, and the crossing point is the subject of solving systems by graphing.
Worked examples
Two streaming plans: an equation against words
Ignatius is choosing a streaming plan. Plan A posts its cost as A(m) = 8m + 25, the dollars paid in all for the first m months, where the 25 is a one-off fee for the streaming box. Plan B is described in words: “10 dollars to sign up, then 11 dollars for each month.” Give each plan’s rate of change and initial value with their units, and say which plan’s cost grows faster and which starts higher.
SolutionPlan A is already in the form mx + b, so its rate is 8 dollars per month and its initial value is A(0) = 25 dollars, the box fee paid before any month is watched.
Plan B’s words name its numbers. “For each month” marks the rate: 11 dollars per month. “To sign up” is paid at m = 0: the initial value is 10 dollars. So
| Plan | Rate (dollars per month) | Initial value (dollars) |
|---|---|---|
| A | 8 | 25 |
| B | 11 | 10 |
Plan B’s cost grows faster, by 11 - 8 = 3 dollars a month. Plan A starts higher, by 25 - 10 = 15 dollars. The two numbers point opposite ways, so neither alone says which plan is cheaper; that depends on how many months, which is Example 4.
Two poster quotes: a table against a graph
Gethin and Hortensia are candidates in a local election, and each asks a print shop for a quote for campaign posters. Gethin’s shop sends a table of the cost G(n), in dollars, of n posters:
| Posters n | 50 | 120 | 200 | 300 |
|---|---|---|---|---|
| Cost G(n) (dollars) | 40 | 68 | 100 | 140 |
Hortensia’s shop sends the graph of its cost H(n) in Figure 1. Find each quote’s rate of change and initial value, compare them, and say what each number means for a candidate.
Gethin’s table first. The input steps are uneven (70, then 80, then 100 posters), so divide each change in cost by its change in posters:
The quotient is the same each time, so the table is linear with rate 0.4 dollars per poster. Input 0 is not in the table, so work back from (50, 40): b = 40 - 0.4(50) = 40 - 20 = 20. Hence G(n) = 0.4n + 20. Check with the last column: 0.4(300) + 20 = 140.
Hortensia’s graph. The line meets the cost axis at 50, so H(0) = 50 dollars. Each square is 50 posters across and 25 dollars up. From the corner (0, 50) to the corner (200, 100) the cost rises 50 dollars over 200 posters:
So H(n) = 0.25n + 50. Check with the corner (300, 125): 0.25(300) + 50 = 125.
| Quote | Rate (dollars per poster) | Initial value (dollars) |
|---|---|---|
| Gethin’s, G | 0.4 | 20 |
| Hortensia’s, H | 0.25 | 50 |
Gethin’s quote grows faster, by 0.15 dollars on every poster; Hortensia’s starts higher, by 30 dollars. The initial value is the cost of 0 posters: a fixed charge for setting up the job, paid however many are printed. The rate is the cost of each extra poster.
Two airships: one climbing, one descending
Juniper and Kwabena each pilot an airship. At noon Juniper’s is climbing steadily, and its instruments log its height J(t) above sea level, in meters, t hours after noon:
| Time t (hours) | 0.5 | 1.5 | 2.5 | 4 |
|---|---|---|---|---|
| Height J(t) (m) | 1140 | 1380 | 1620 | 1980 |
Kwabena’s is descending steadily toward an airfield where it lands; Figure 2 graphs its height K(t) over the same hours. Compare the two rates and the two initial values. Whose function grows faster, and whose height changes faster?
Juniper’s table. The changes are 240 m over 1 hour, 240 m over 1 hour, and 360 m over 1.5 hours, which is \displaystyle \frac{360}{1.5} = 240. The rate is 240 meters per hour. Working back from (0.5, 1140): b = 1140 - 240(0.5) = 1140 - 120 = 1020. So
Check with (4, 1980): 240(4) + 1020 = 960 + 1020 = 1980.
Kwabena’s graph. The height axis starts at 700, but that does not matter: the initial value is about input 0, and the time axis does start at 0, so the segment’s left end gives K(0) = 2300. Each square is half an hour across and 200 meters up. From (0, 2300) to the corner (1, 1900) the height changes by -400 meters in 1 hour, a rate of -400 meters per hour. So K(t) = -400t + 2300 for 0 \le t \le 3.5. Check with the end of the segment: -400(3.5) + 2300 = 900, the airfield’s height.
Kwabena starts higher, by 2300 - 1020 = 1280 meters. Juniper’s function grows faster: 240 is greater than -400. But Kwabena’s height changes faster: 400 meters every hour against her 240. The two questions have different answers because one compares the signed rates and the other their sizes.
When does the cheaper plan change?
Ignatius’s two plans from Example 1 cost A(m) = 8m + 25 and B(m) = 11m + 10 dollars for m months. Which plan is cheaper for a short subscription, which for a long one, and after how many months do they cost the same?
SolutionPlan A starts 25 - 10 = 15 dollars dearer. Each month A adds 8 dollars and B adds 11, so B gains 3 dollars on A every month, and the 15 dollar gap shrinks by 3 a month. It is gone after
Check: A(5) = 8(5) + 25 = 65 and B(5) = 11(5) + 10 = 65. Both plans cost 65 dollars for 5 months.
Before 5 months plan B is cheaper, since it started lower and has not yet caught up. After 5 months it is dearer, and it keeps pulling away by 3 dollars a month. For a year, A(12) = 96 + 25 = 121 and B(12) = 132 + 10 = 142 dollars. So B suits a short subscription and A a long one.
The gap is itself a linear function, A(m) - B(m) = -3m + 15, with initial value 15 and rate -3, and it reaches 0 at m = 5. The point (5, 65) lies on both graphs, which is where solving a system by graphing begins.
Which poster quote is cheaper?
The two print shops in Example 2 quoted G(n) = 0.4n + 20 (Gethin’s) and H(n) = 0.25n + 50 (Hortensia’s), in dollars for n posters, and either candidate may use either shop. For 150 posters, which quote is cheaper? For 250? For how many posters do the quotes agree?
SolutionHortensia’s quote starts 30 dollars higher, and Gethin’s gains 0.4 - 0.25 = 0.15 dollars on it with every poster. The gap closes after
and there both quotes come to 100 dollars. This was in the data all along: Gethin’s table lists 200 posters at 100 dollars, and Hortensia’s line passes through the corner (200, 100).
For 150 posters, fewer than 200, Gethin’s has not caught up: G(150) = 60 + 20 = 80 dollars against H(150) = 37.5 + 50 = 87.5 dollars, so Gethin’s quote is cheaper. For 250 posters it has passed: G(250) = 100 + 20 = 120 against H(250) = 62.5 + 50 = 112.5, so Hortensia’s is cheaper. Both differences are 7.5 dollars, which is 0.15 \times 50: each order is 50 posters away from 200, and the gap changes by 0.15 dollars a poster on either side.
Will they ever be at the same height?
Juniper’s and Kwabena’s airships from Example 3 have heights J(t) = 240t + 1020 and K(t) = -400t + 2300 for 0 \le t \le 3.5, in meters, t hours after noon.
(a) When are they at the same height above sea level, and what is it?
(b) Lavinia’s airship lifts off at noon from an airfield at 700 meters and climbs steadily at 240 meters per hour. Does it ever reach Juniper’s height? Kwabena’s?
Solution(a) Kwabena starts 2300 - 1020 = 1280 meters higher. Each hour he loses 400 meters and Juniper gains 240, so the gap closes from both ends at once, by 400 + 240 = 640 meters an hour. It closes after
Check: J(2) = 480 + 1020 = 1500 and K(2) = -800 + 2300 = 1500. Both are at 1500 meters at 2 in the afternoon, and t = 2 lies inside Kwabena’s domain, so it really happens.
(b) Lavinia’s height is L(t) = 240t + 700. Against Juniper: the same rate, 240 meters per hour, and a start 1020 - 700 = 320 meters lower. Each hour both gain 240 meters, so the gap stays 320 meters for as long as they climb. She never reaches Juniper’s height; their graphs are parallel lines.
Against Kwabena: a starting gap of 2300 - 700 = 1600 meters, closing at 640 meters an hour, as in (a). It closes after \displaystyle \frac{1600}{640} = 2.5 hours. Check: L(2.5) = 600 + 700 = 1300 and K(2.5) = -1000 + 2300 = 1300, and 2.5 \le 3.5, so this crossing is in range too.
Boundaries
Both functions must be linear. The two-number comparison rests on each function having one rate everywhere. A table whose quotients differ has no single rate; Linear and nonlinear functions deals with such functions.
Same units, same starting moment. Rates can only be compared in the same units, and initial values only when input 0 means the same moment for both. If one airship’s clock started an hour later, its rule must be shifted before the two initial values say anything about each other.
“Grows faster” and “changes faster” differ when a rate is negative. The first compares m_1 with m_2, the second their sizes.
A crossing counts only inside the domains. The reasoning gives an input where the two rules agree; it describes the situation only if that input lies in both domains, and is a whole number when the input counts whole things. A crossing at a negative input lies before the start.
Equal rates never cross. Two linear functions with the same rate and different initial values are never equal, however far the input goes.
Where this goes next
The crossing point found here, written as a pair of numbers, satisfies two equations at once, and finding such points from graphs and by algebra is solving systems. Saying in words what each function’s two numbers mean in a situation is Linear functions in context, and comparing a linear function with one that keeps multiplying is in Linear and nonlinear functions.
Exercises
Part 1: the two numbers from each form
Figure 4 (a) graphs the cost E(m), in dollars, of streaming plan E for m months, and Figure 4 (b) graphs a linear function p.
- Streaming plan C costs C(m) = 13m + 5 dollars for m months. Plan D is described as “25 dollars to join, then 9 dollars for each month.” Give each plan’s rate and initial value with units. Which grows faster, and which starts higher?
- One print shop’s quote T(n), in dollars for n posters, gives T(40) = 37, T(100) = 55, T(160) = 73 and T(250) = 100. Another shop’s is Q(n) = 0.35n + 12. Find the rate and initial value of T, and compare the two quotes.
- Plan E is graphed in Figure 4 (a). Plan F is “12 dollars to join, then 8 dollars a month.” Compare the two plans’ rates and initial values.
- An airship’s height H(t) above sea level, in meters, t hours after it is first logged, is H(1) = 2250, H(2.5) = 2025, H(3) = 1950 and H(4.5) = 1725. A second airship, logged from the same moment, has height G(t) = 180t + 750. Find the rate and initial value of H. Which starts higher, whose function grows faster, and whose height changes faster?
- A function k has k(-2) = 2, k(1) = 1, k(4) = 0 and k(10) = -2. Compare it with the function p graphed in Figure 4 (b): which has the greater rate of change, and which the greater initial value?
Part 1: three at once, and forms to unpack
- Rank these three functions by rate of change and by initial value: f given by 3x + y = 7, with y = f(x); g, which “starts at -1 and rises 4 for each 1 the input rises”; and h, whose table pairs the inputs 2, 4, 8 with the outputs 1, 2, 4.
- Under one streaming plan, with a joining fee and the same charge each month, a subscriber had paid 57 dollars in all after 3 months and 101 dollars after 7 months. Find the plan’s rate and initial value, and compare it with plan P(m) = 12m + 15.
- A third shop quotes 60 dollars for 100 posters, 110 for 200 and 150 for 300. Can this quote be compared with G(n) = 0.4n + 20 by rate and initial value? Explain.
Part 2: who is ahead, and when that changes
- For plans C and D in Exercise 1, which is cheaper for 3 months, and which for 12 months? After how many months do they cost the same, and how much is that?
- For the quotes T and Q in Exercise 2, for how many posters do they agree, and at what cost? Which is cheaper for 100 posters, and which for 300?
- For the airships in Exercise 4, assume both keep their rates for at least 6 hours. When are they at the same height, and what is it?
- For the airships in Exercise 4, write the gap H(t) - G(t) as a linear function, give its initial value and rate, and find the gap after 2 hours.
- One airship descends from 2000 meters at a steady 500 meters per hour and lands at an airfield at 1000 meters after 2 hours: A(t) = -500t + 2000 for 0 \le t \le 2. Another lifts off at the same moment from 600 meters and climbs at 100 meters per hour: B(t) = 100t + 600. Does the second reach the first one’s height before the first lands?
- One streaming plan costs 30 dollars to join and 7 dollars a month. A second costs 15 dollars to join. What monthly charge for the second plan makes the two cost the same after 5 months? Which plan is cheaper after that?
Reasoning
- Let f(x) = m_1x + b_1 and g(x) = m_2x + b_2 with b_1 < b_2 and m_1 > m_2. Explain why f overtakes g at a positive input, find that input, and explain why f stays ahead after it.
- Explain why two linear functions with the same rate and different initial values never take the same output, and say what their graphs look like.
- A quote with rate 0.25 dollars per poster is drawn on squares 100 posters across and 5 dollars up; a quote with rate 0.4 dollars per poster is drawn on squares 50 posters across and 25 dollars up. Which line looks steeper? Explain why comparing how steep two graphs look can mislead, and what to compare instead.
- A rising linear function starts below a falling one. Explain why they take the same output at exactly one input, and why that input is the starting gap divided by the sum of the sizes of the two rates.
Answers
- C: rate 13 dollars per month, initial value 5 dollars. D: D(m) = 9m + 25, rate 9 dollars per month, initial value 25 dollars. C grows faster, by 4 dollars a month; D starts higher, by 20 dollars.
- T(n) = 0.3n + 25: each quotient is \displaystyle \frac{18}{60} = \frac{27}{90} = 0.3, and b = 37 - 0.3(40) = 25. Q grows faster (0.35 > 0.3); T starts higher (25 > 12).
- From (0, 30) to (10, 90), E rises 60 dollars in 10 months: E(m) = 6m + 30. F(m) = 8m + 12. F grows faster; E starts higher.
- H(t) = -150t + 2400: each quotient is -150, and b = 2250 + 150 = 2400. H starts higher; G grows faster (180 > -150) and also changes faster (180 > 150).
- \displaystyle p(x) = \frac{2}{3}x - 2, from (0, -2) and (3, 0). \displaystyle k(x) = -\frac{1}{3}x + \frac{4}{3}, since each quotient is \displaystyle -\frac{1}{3} and \displaystyle k(0) = 1 + \frac{1}{3}. So p has the greater rate and k the greater initial value.
- f(x) = -3x + 7, g(x) = 4x - 1, \displaystyle h(x) = \frac{1}{2}x. By rate: g, h, f. By initial value: f (7), h (0), g (-1).
- Rate \displaystyle \frac{101 - 57}{7 - 3} = 11 dollars per month, initial value 57 - 11(3) = 24 dollars. P grows faster (12 > 11); this plan starts higher (24 > 15).
- No. The first 100 extra posters add 50 dollars and the next 100 add 40, so the quote is not linear and has no single rate. Only particular orders can be compared, such as 110 against G(200) = 100 dollars for 200 posters.
- For 3 months C is cheaper: C(3) = 44, D(3) = 52. For 12 months D is cheaper: C(12) = 161, D(12) = 133. The 20 dollar gap closes at 4 a month, so they agree after 5 months, at 70 dollars.
- The 13 dollar gap closes at 0.05 a poster: they agree at \displaystyle \frac{13}{0.05} = 260 posters, at 103 dollars. For 100 posters Q is cheaper (47 against 55); for 300, T is cheaper (115 against 117).
- After 5 hours, at 1650 meters: the gap 2400 - 750 = 1650 closes at 150 + 180 = 330 meters an hour.
- H(t) - G(t) = -330t + 1650: initial value 1650 meters, rate -330 meters per hour. After 2 hours the gap is 990 meters.
- No. The gap 1400 closes at 600 meters an hour, which would take \displaystyle \frac{7}{3} hours, but the first lands at t = 2. Then it is at 1000 meters and the second at 800.
- 10 dollars a month: the first costs 30 + 7(5) = 65, and 15 + 5r = 65 gives r = 10. After that the second costs 3 dollars more each month, so the first is cheaper.
- At input 0, f is behind by b_2 - b_1 > 0. Each unit, f gains m_1 - m_2 > 0, so the gap reaches 0 at \displaystyle x = \frac{b_2 - b_1}{m_1 - m_2}, positive as a quotient of positives. After that f(x) - g(x) = (m_1 - m_2)x - (b_2 - b_1) keeps increasing, since its rate is positive, so f stays ahead.
- With f(x) = mx + b_1 and g(x) = mx + b_2, f(x) - g(x) = b_1 - b_2 for every x, and it is not 0. So the outputs are never equal, and the graphs are parallel lines.
- The 0.25 line looks steeper: it climbs 25 dollars, 5 squares, per square across, while the 0.4 line climbs 20 dollars, under one square, per square across. How steep a line looks depends on what the squares stand for. Compare the rates themselves: 0.4 > 0.25.
- Let the rising one have rate m_1 > 0, the falling one rate m_2 < 0, and the starting gap be d > 0. Each unit the gap shrinks by m_1 + (-m_2), the sum of the sizes. The gap is linear with a nonzero rate, so it is 0 at exactly one input, \displaystyle \frac{d}{m_1 - m_2}, the starting gap divided by that sum.