Math 8 · 8.F.A.2 · 8.F.A.3

Constant rate of change and linear tables

A table lists a few inputs of a function and their outputs. How can you tell whether one steady rule made every row, and how do you find that rule from the rows alone?

You need

Average rate of changeGrade 9

Unlocks

Choosing a function family for dataAlgebra 1 Reading a linear function from its graphMath 8 Comparing linear functions given different waysMath 8

Where you are 5 in Linear functions

Functions13

Linear functions5

Constant rate of change and linear tables

Reading a linear function from its graph Comparing linear functions given different ways Linear and nonlinear functions Linear functions in context

Why this exists

A press’s sheet counter, the points on a loyalty card, a chair’s height on a ski lift: each is a function, one output for each input. Often all anyone has is a table, a handful of rows written down at odd moments. Did one steady rule produce every row? And if it did, what is that rule, so that outputs nobody wrote down can be found?

Rate of change gave the number that says how fast an output moves, and slope from a table gave the test for whether a table’s points lie on one line. This lesson puts that test into the language of functions. A function whose rate never changes is a linear function; it always has a rule y = mx + b; and the rows of its table are enough to write that rule, even when the row for an input of 0 is missing.

Definition

A function has a constant rate of change when the rate

\frac{\text{change in output}}{\text{change in input}}

comes out the same number m whichever two inputs you pick.

A linear function is a function whose rule can be written

y = mx + b

with x the input, y the output, and m and b fixed numbers. The number m is the rate of change, and b is the initial value: the output when the input is 0.

A table of values is linear when it could have come from a linear function, that is, when the rate of change between any two of its rows is the same number.

Intuition

Take this table.

x	1	2	4	5
y	7	10	16	19

From one row to the next, the input moves 1, then 2, then 1, and the output moves 3, then 6, then 3. The output steps are not equal, yet each one is exactly three times its input step. The rate is 3 every time.

Draw each step as a right triangle, across then up (Figure 1). The triangles have different sizes but one shape, so every step leaves in the same direction and the points line up. That is the picture of slope triangles, and it is why a constant rate and a straight graph go together.

Figure 1Uneven steps, one direction. Every step goes up three times as far as it goes across, so the points line up, and walking back one step reaches the initial value 4.

Now walk backwards one step, from x = 1 to x = 0. The input falls by 1, so the output falls by 3, from 7 to 4. No row shows it, but 4 is the initial value, and the rule is y = 3x + 4. Every row obeys it: 3 \times 4 + 4 = 16.

Results

Constant rate means linear. A function has a constant rate of change m exactly when its rule is y = mx + b. Its graph is then a straight line with Slope m, crossing the y-axis at (0, b).

The table test. Put the rows in order of input and find the rate from each row to the next, each output change divided by its own input change. The table is linear exactly when these rates are all the same number m. When the inputs go up in equal steps, every rate has the same divisor, so it is enough to see that the outputs change by equal steps.

The initial value from any row. If a linear function has rate m and (x_1, y_1) is one of its rows, then

b = y_1 - m x_1

Why it is true

A rule y = mx + b has a constant rate. For any two inputs x_1 and x_2,

\begin{aligned} \frac{(m x_2 + b) - (m x_1 + b)}{x_2 - x_1} &= \frac{m(x_2 - x_1)}{x_2 - x_1} \\ &= m \end{aligned}

The b is in both outputs, so it cancels, and the answer does not depend on the pair chosen.

A constant rate gives such a rule. Fix one row (x_1, y_1). For any other input x with output y, the rate between the two is m, so \displaystyle \frac{y - y_1}{x - x_1} = m. Multiplying by x - x_1 and adding y_1,

y = mx + (y_1 - m x_1)

which is y = mx + b with b = y_1 - m x_1. Putting x = 0 gives y = b, so this b is the initial value, whether or not 0 is one of the table’s inputs.

The table test. Why equal rates between neighboring rows are enough, and why the points then lie on one straight line, is shown with slope triangles in slope from a table.

Worked examples

Example 1

A press counter read every five minutes

Kolya reads the sheet counter on a printing press every five minutes.

time (minutes)	counter (sheets)
0	1200
5	1500
10	1800
15	2100
20	2400

Is the counter reading a linear function of time? If it is, write its rule, and find the reading at 45 minutes if the press keeps running the same way.

Solution

The input is the time t in minutes and the output is the reading c. The inputs go up in equal steps of 5, so the output steps can be compared directly: 1500 - 1200 = 300, and the other three are 300 as well. The table is linear, with

m = \frac{300 \text{ sheets}}{5 \text{ minutes}} = 60 \text{ sheets per minute}

The row for t = 0 is in the table, so b = 1200, the sheets already on the counter when Kolya started watching. The rule is

c = 60t + 1200

At 45 minutes the reading is 60 \times 45 + 1200 = 3900 sheets.

Check with a row the working did not use: 60 \times 15 + 1200 = 2100, as in the table.

Example 2

A loyalty card with uneven steps

Nanami’s coffee shop loyalty card gives points for stamps. A few rows from the card’s history:

stamps	2	3	6	10
points	70	85	130	190

Show that the points are a linear function of the stamps, and write the rule.

Solution

The stamp counts jump by 1, 3 and 4, so the point steps 15, 45 and 60 cannot be compared as they stand. Divide each by its own input step:

\begin{aligned} \frac{85 - 70}{3 - 2} &= \frac{15}{1} = 15 \\ \frac{130 - 85}{6 - 3} &= \frac{45}{3} = 15 \\ \frac{190 - 130}{10 - 6} &= \frac{60}{4} = 15 \end{aligned}

Every neighboring rate is 15, so the table is linear with m = 15 points per stamp. There is no row for 0 stamps. Walk back from 2 stamps to 0: two stamps fewer means 2 \times 15 points fewer, so

b = 70 - 15 \times 2 = 40

The rule is p = 15s + 40, for s stamps and p points. The 40 is what the card holds before its first stamp, a welcome balance.

Check: 15 \times 10 + 40 = 190, the last row.

Example 3

A ski lift and its top station

A ski lift carries Leocadia’s chair up a mountain at a steady speed. Her elevation is noted a few times, counting minutes from when the chair left the bottom station.

time (minutes)	elevation (m)
4	1310
7	1430
9	1510
13	1670

(a) Show that the table is linear and write the rule. (b) What does the initial value mean here? (c) The top station is at 1830 m. When does the chair reach it?

Solution

(a) The neighboring rates are \displaystyle \frac{120}{3} = 40, \displaystyle \frac{80}{2} = 40 and \displaystyle \frac{160}{4} = 40, so the table is linear with m = 40 meters per minute. From the row (4, 1310),

b = 1310 - 40 \times 4 = 1150

and the rule is e = 40t + 1150.

(b) At t = 0 the chair is leaving the bottom station, so the bottom station stands at 1150 m.

(c) The chair is at the top when 40t + 1150 = 1830. Subtracting 1150 from both sides gives 40t = 680, and dividing by 40 gives t = 17 minutes. That answer reaches past the last row: it trusts the lift to keep the same steady climb all the way up.

Example 4

A feeder tray that empties

Perpetua fills the feeder tray of a printing press and starts the press, which draws sheets at a steady speed. She counts the sheets left at these times since the start:

time (minutes)	sheets left
1.5	4100
4	3600
6.5	3100
9	2600

Write the rule, find how many sheets she loaded, and find when the tray runs empty.

Solution

The times go up in equal steps of 2.5 minutes and the counts fall by 500 each step, so the table is linear, with

m = \frac{-500}{2.5} = -200 \text{ sheets per minute}

The rate is negative because the output falls as the input grows. From the row (1.5, 4100), b = 4100 - (-200)(1.5), which is 4100 + 300 = 4400: going back in time adds sheets. The rule is

n = -200t + 4400

She loaded 4400 sheets, the count at t = 0. The tray is empty when -200t + 4400 = 0, so 200t = 4400 and t = 22 minutes. After that the rule would give a negative number of sheets: it describes the press only from t = 0 to t = 22.

Example 5

Filling in a table

The table below belongs to a linear function. Find its rule and the missing entries p and q.

x	-3	-1	2	6	q
y	14	10	p	-4	-10

Solution

The first two rows are complete, so they give the rate:

m = \frac{10 - 14}{-1 - (-3)} = \frac{-4}{2} = -2

From the row (-3, 14), b = 14 - (-2)(-3) = 14 - 6 = 8, and the rule is y = -2x + 8.

Then p = -2(2) + 8, so p = 4. For q, solve -2q + 8 = -10: subtracting 8 gives -2q = -18, so q = 9.

Check with the row the working did not use: -2(6) + 8 = -4. Had it come out different, no linear function could have made the table.

Example 6

A lift over a ridge

Oisin rides a chairlift whose cable crosses a ridge. His height above the bottom station is recorded every two minutes.

time (minutes)	height (m)
0	0
2	90
4	180
6	240
8	300

Is the height a linear function of time?

Solution

The inputs go up in equal steps of 2, and the height steps are 90, 90, 60 and 60. They are not all equal, so the table is not linear: no single rule y = mx + b fits all five rows.

The rows still have a pattern. Up to minute 4 the chair climbs \displaystyle \frac{90}{2} = 45 m per minute, and after it \displaystyle \frac{60}{2} = 30 m per minute: two straight pieces with a bend at the ridge (Figure 2). Had the chair kept climbing 45 m per minute, it would have been at 45 \times 8 = 360 m at minute 8, on the dashed line, not at 300.

Figure 2Two straight pieces with different rates, so no single line fits all five rows. The dashed line is where a steady 45 m per minute would have gone.

Boundaries

A table shows only its rows. Passing the test proves that the listed points lie on one line. It cannot prove that the quantity changes steadily between the rows or after the last one; that comes from the situation, such as a press running at a steady speed.

The rule holds where the situation does. The feeder rule stops at 22 minutes and the lift rule at the top station. Outside that stretch the rule still gives numbers, but they describe nothing.

A constant rate is not a constant ratio. The rate compares changes. The output divided by its own input, \displaystyle \frac{y}{x}, is constant only when b = 0, which is a proportional relationship. For y = 3x + 4, the row (1, 7) gives \displaystyle \frac{7}{1} = 7, and the rate is still 3.

Where this goes next

Reading a linear function from its graph takes m and b off a picture instead of a table, and slope-intercept form works with the rule y = mx + b directly. Linear and nonlinear functions carries the test of this lesson to graphs and equations, and comparing linear functions sets a table beside a graph or a rule. Linear functions in context says what m and b mean for the quantities in a story.

An arithmetic sequence is a linear function whose inputs are the counting numbers. Later, choosing a function family asks of a table of data whether equal steps in the input give equal differences, the mark of this lesson, or something else, such as equal ratios.

Exercises

Is the table linear?

Decide whether each table could come from a linear function. If it could, give the rate of change.

Inputs x = 0, 1, 2, 3, 4 give outputs y = 7, 11, 15, 19, 23.
Inputs x = 0, 2, 4, 6 give outputs y = 1, 4, 9, 16.
Inputs x = 1, 4, 5, 9 give outputs y = 2, 11, 14, 26.
Inputs x = 2, 3, 5, 6 give outputs y = 10, 8, 6, 4.

Writing the rule

Each table comes from a linear function. Write its rule.

Inputs x = 0, 3, 6, 9 give outputs y = -5, 1, 7, 13.
Inputs x = 2, 4, 6, 8 give outputs y = 9, 8, 7, 6.
Inputs x = -4, -1, 5, 7 give outputs y = -9, -3, 9, 13.
Inputs x = 10, 15, 25, 40 give outputs y = 34, 31, 25, 16.
A linear table has rows (1, 4), (3, a), (c, 19) and (8, 25). Find the rule, a and c.
A linear function’s table has only two rows, (4, 21) and (10, 48). Write the rule.
A linear table has rate 3 and contains the row (2, 5). Which of (0, -1), (5, 14) and (-1, -3) could also be rows of it?

Tables from situations

Nanami times another printing press. Its counter reads 1530, 2050, 2570 and 3350 sheets at 6, 10, 14 and 20 minutes. Show that the reading is a linear function of time, write the rule, say what its initial value means, and find when the counter reads 4650.
Oisin’s loyalty card shows 95, 145, 220 and 320 points after 3, 5, 8 and 12 stamps. Write the rule, say what the initial value means, and find the fewest stamps that bring the card to at least 500 points.
On the way down, a ski lift chair is 616, 490, 406 and 238 m above the bottom station at 2, 5, 7 and 11 minutes after leaving the top. Write the rule for its height, and find when it reaches the bottom station, in minutes and seconds.
A press warming up has printed 0, 20, 60, 120 and 200 sheets after 0, 1, 2, 3 and 4 minutes. Is the count a linear function of time? Say what the table shows about the press.

Reasoning

Can a table with only two rows fail the test for a linear table? What does a two-row table tell you about whether a function is linear?
Explain, by walking from a row to the input 0, why b = y_1 - m x_1 gives the initial value even when 0 is not one of the table’s inputs.
Malachy has a table with inputs 1, 2, 3, 4 and outputs 8, 13, 18, 23. He says: “The rate is 5, so the rule is y = 5x.” What is wrong, and what is the rule?

Answers

Linear, rate 4. The inputs go up by 1 and the outputs by 4 every time (the rule is y = 4x + 7).
Not linear. The inputs go up in equal steps of 2, but the outputs go up by 3, 5 and 7.
Linear, rate 3: \displaystyle \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3, \displaystyle \frac{14 - 11}{5 - 4} = 3 and \displaystyle \frac{26 - 14}{9 - 5} = \frac{12}{4} = 3.
Not linear. The outputs fall by 2 each row, but the input steps are 1, 2, 1, so the rates are -2, \displaystyle \frac{-2}{2} = -1 and -2.
y = 2x - 5. The rate is \displaystyle \frac{6}{3} = 2 every step, and the row for x = 0 gives b = -5.
\displaystyle y = -\frac{1}{2}x + 10. The rate is \displaystyle \frac{-1}{2} every step, and \displaystyle b = 9 - (-\frac{1}{2})(2) = 10.
y = 2x - 1. The rates are \displaystyle \frac{6}{3}, \displaystyle \frac{12}{6} and \displaystyle \frac{4}{2}, all 2, and b = -9 - 2(-4) = -1.
y = -0.6x + 40. The rates are \displaystyle \frac{-3}{5}, \displaystyle \frac{-6}{10} and \displaystyle \frac{-9}{15}, all -0.6, and b = 34 - (-0.6)(10) = 40.
y = 3x + 1, a = 10 and c = 6. The complete rows give \displaystyle m = \frac{25 - 4}{8 - 1} = 3 and b = 4 - 3 = 1. Then a = 3(3) + 1 = 10, and 3c + 1 = 19 gives c = 6.
y = 4.5x + 3. The rate is \displaystyle \frac{48 - 21}{10 - 4} = \frac{27}{6} = 4.5, and b = 21 - 4.5 \times 4 = 3.
(0, -1) and (5, 14), but not (-1, -3). The rule is y = 3x - 1, since b = 5 - 3 \times 2 = -1, and it gives -4 at x = -1.
c = 130t + 750; it reads 4650 at 30 minutes. The rates are \displaystyle \frac{520}{4}, \displaystyle \frac{520}{4} and \displaystyle \frac{780}{6}, all 130 sheets per minute, and b = 1530 - 130 \times 6 = 750: the counter already showed 750 when timing began. Then 130t + 750 = 4650 gives 130t = 3900 and t = 30.
p = 25s + 20, and 20 stamps. The rates are \displaystyle \frac{50}{2}, \displaystyle \frac{75}{3} and \displaystyle \frac{100}{4}, all 25, and b = 95 - 25 \times 3 = 20, the points on the card before any stamp. For 500 points, 25s + 20 = 500 gives s = 19.2; 19 stamps give only 495, and 20 stamps give 520.
h = -42t + 700, reaching the bottom after 16 minutes 40 seconds. The rates are \displaystyle \frac{-126}{3}, \displaystyle \frac{-84}{2} and \displaystyle \frac{-168}{4}, all -42 m per minute, and b = 616 + 42 \times 2 = 700. Then -42t + 700 = 0 gives \displaystyle t = \frac{700}{42} = 16\frac{2}{3} minutes.
Not linear. The minutes go up by 1, but the count goes up by 20, 40, 60 and 80: each minute the press prints 20 sheets more than the minute before, so it is speeding up.
No. Two rows give one rate, with nothing to compare it to, and any two points lie on a line. So a two-row table fits a linear function but also many nonlinear ones; only more rows or the situation can show the function is linear.
Going from the input x_1 to the input 0 changes the input by 0 - x_1 = -x_1. At a constant rate m the output changes by m times that, -m x_1, so the output at 0 is y_1 - m x_1. No row at 0 is needed, only the rate and one row.
The rule is y = 5x + 3. The rate is 5, but y = 5x would give 5 at x = 1, not 8: the rule needs its initial value too, b = 8 - 5 \times 1 = 3. Only a table with b = 0 has the rule y = mx.