Study Companion: Programme F.4 — Graphs

You have drawn graphs before: choose some values of x, work out y, plot the points, join them up. This programme asks what that procedure actually produces. An equation such as y=x^2 has infinitely many solutions — every pair (x,y) that makes it true — and its graph is the picture of all of them at once. Once the solutions are a shape, questions about them become questions about the shape: where a line joining plotted points can be trusted and where it lies; how a spreadsheet draws a graph from one formula; which points satisfy an inequality; how far one number is from another. The one idea underneath is that a graph is the set of solutions, and every rule in this programme is a statement about where that set lies.

The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in six units instead, one for each idea: the graph as every solution, the places where joining the dots goes wrong, the spreadsheet as a plotting machine, inequalities as regions, the modulus as a distance, and the solving of modulus inequalities.

How to use this companion

Each unit is an argument: a real situation and a question, the attempt a sensible person would make first, the picture that settles it, a check with numbers, and only then the rule. Read it once front to back. The Narration spine at the end of each unit summarises what the video shows.

Unit 01: A Graph Is Every Solution at Once

The problem

Drop a ball from a balcony and film it. Starting from rest, with g taken as 10 m s^{-2}, it has fallen

after t seconds. A slow camera taking two pictures a second catches it at t=0,\ 0.5,\ 1 s, fallen 0,\ 1.25,\ 5 m. Where was the ball at t=0.75 s, between two pictures? And what exactly is "the graph of d=5t^2" — three photographs, or something more?

First attempt

Plot the three pairs and join them with a ruler. At t=0.75 s the ruler line from (0.5,\ 1.25) to (1,\ 5) is halfway between them, at \frac{1.25+5}{2}=3.125 m. The equation says 5\times0.75^2=2.8125 m. The ruler is out by 0.3125 m, about 31 cm.

A straight segment between two photographs assumes the ball fell at a steady speed between them. A falling ball speeds up, so the ruler line is a guess about the gap, and here it is a wrong one.

The picture

Each photograph is an ordered pair (t,d): the time is chosen, and the distance is what the equation makes it. Put the chosen quantity on the horizontal axis and the calculated one on the vertical axis. The two axes measure different things — seconds and metres — so there is no reason for their scales to match, and they do not.

Now take pictures more often. At four a second there is a photo every 0.25 s, and the ruler line has less gap to guess across. At ten a second, less still; at twenty, less again. The ruler lines bend closer and closer to one curve, and the dots crowd onto it. No finite set of photographs is that curve, but every set lies on it, and the closer the photos, the less the join between them matters.

That curve is the graph of the equation: every pair (t,\ 5t^2) at once, infinitely many solutions drawn as one line. In practice nobody plots infinitely many points. We plot enough of them, and join them with a smooth curve rather than a ruler.

How fast does the ruler's error shrink? Take any two photos a time \delta t apart, at a and a+\delta t. Halfway between them the ruler gives the average of the two distances, while the equation gives 5\left(a+\frac{\delta t}2\right)^2. The difference is

The a has cancelled, so the error is the same everywhere on this curve, and it goes as the square of the spacing: halve the gap between photos and the ruler's error falls to a quarter.

Check it with numbers

Two pictures a second, \delta t=0.5: \frac54\times0.25=0.3125 m, the 3.125-2.8125 found above.

Four a second, \delta t=0.25: at t=0.625, halfway between the photos at 0.5 and 0.75, the ruler gives \frac{1.25+2.8125}{2}=2.03125 m and the equation 5\times0.625^2=1.953125 m, an error of 0.078 m =\frac54\times0.0625 — a quarter of the first.

Ten a second, \delta t=0.1: \frac54\times0.01=0.0125 m. Twenty a second, \delta t=0.05: \frac54\times0.0025=0.003 m, three millimetres.

The rule it gives

Equations and graphing form. An equation in two variables, such as x-y=1, is a conditional equation: it is true only for some pairs of values. Transposed to make y the subject, y=x-1, it is in the form "y= an expression in x". Then x is the independent variable, chosen freely, and y is the dependent variable, restricted to the value the right-hand side gives.

Ordered pairs. Choosing x=2 in y=x^2 gives the pair (2,\ 4). It is ordered because the first number is always the independent variable and the second the dependent one; (4,\ 2) is a different point.

Cartesian axes. Two perpendicular numbered lines crossing at their common zero form the Cartesian coordinate frame. The horizontal axis is always the independent-variable axis (x-axis); the vertical is the dependent-variable axis (y-axis). The scales need not be the same; choose them to use the paper well.

The graph of an equation is the continuous line on which every ordered pair lies. Plot a collection of points, then join them with a smooth curve. Some graphs have names: y=x^2 is a parabola, and y=x+1 is a straight line.

Domain restrictions. Transposing can restrict x as well as y. A crank pin on a wheel of radius 5 cm, measured from the axle, sits at (x,y) with x^2+y^2=25. Solving for y:

For each x there are two values, the pin above the axle and the pin below it. And 25-x^2 must not be negative, so -5\le x\le5: the pin never gets more than 5 cm to either side.

The book's own example is the circle x^2+y^2=1, giving y=\pm\sqrt{1-x^2} for -1\le x\le1. A note on notation: the book writes \sqrt x for the positive square root and x^{1/2} for both roots. Here, as in F.1 Unit 07, both \sqrt x and x^{1/2} mean the non-negative root, and the second value is written with an explicit \pm.

Tables for any equation. For an equation that is not a polynomial, nesting is not available, so the book builds the table one term per row and adds the rows. It works for polynomials too, as below.

Worked example

Tabulate y=x^3-2x^2-x+2 for -2\le x\le3 in steps of 0.5, term by term:

The book prints y to 1 dp (-4.4, 1.9, 1.1, -0.6, 2.6). Plotted and joined smoothly, the curve meets the x-axis at x=-1, 1 and 2, and indeed x^3-2x^2-x+2=(x+1)(x-1)(x-2) — the factors of F.3 Unit 06, seen on a graph.

Where this shows up

A phone films at 30 frames a second, so \delta t=\frac1{30} s, and a smooth curve through the ball's positions in the frames is out by at most \frac54\left(\frac1{30}\right)^2=0.0014 m — under a millimetre and a half, less than the blur of the ball itself. That is why a motion-tracking app can join video frames into a trustworthy graph of position against time. Try it: film a dropped ball, step through the frames, and plot the fall. Keep only every fifteenth frame — two pictures a second — and the ruler's 31 cm error is plain to see; use every frame and it has vanished into the curve.

Narration spine. A ball falls beside the axes, and three photographs become three dots, each an ordered pair. A ruler joins them and misses the ball at three-quarters of a second by thirty-one centimetres. The photos come faster and faster, the dots crowd together and the ruler's error readout shrinks by quarters, until the dots merge into the smooth curve d=5t^2: every solution at once.

Unit 02: Joining the Dots Needs Care

The problem

Hold a magnifying glass of focal length 10 cm a height u cm above a page. The print seen through it is magnified

At 5 cm the letters look twice their size; at 8 cm, five times. Keep lifting the lens: what happens at 10 cm, and beyond?

First attempt

Tabulate m every 2 cm, as the book does for y=\frac1{1-x} in steps of 0.2 (the same function with x=\frac u{10}):

At u=10 the formula asks for \frac{10}{0}, so that row is skipped. Plot the rest and join them. The segment from (8,\ 5) to (12,\ -5) passes through (10,\ 0): the graph now says that at 10 cm the magnification is zero, and the print shrinks to nothing.

Try it with a real lens. At 10 cm the print does not vanish. It swells until one smear fills the lens.

The picture

Instead of trusting the join, look closer to 10:

As u creeps up to 10 the magnification grows without limit; just past 10 it is hugely negative, and a negative magnification means an inverted image. So the curve climbs alongside the vertical line u=10 on the left and returns from far below it on the right. It gets as close to the line as you like and never reaches it, because at u=10 there is no value to plot.

That line is a vertical asymptote. The ruler segment across it is not part of the graph: it joins two points that lie on different pieces of the curve, and it invents a value, m=0, at the one place where the formula has none.

Two more ways a table can hide the truth:

• A piecewise equation uses different formulas on different parts of the x-axis. y=x^2 for -5\le x<0 and y=x for x\ge0 is a parabola arm on the left and a straight line on the right. Here both formulas give 0 at x=0, so the pieces meet and the graph is one unbroken line with a change of shape.
• A discontinuity. A reversing switch sends +1 V until t=2 s and -1 V after: y=1 for x\le2, y=-1 for x>2. Samples taken every half second and joined by a ruler draw a slope from +1 V at t=2 s down to -1 V at t=2.5 s, through 0 V, but the signal is never 0 V. The graph is two horizontal lines with a gap. A dashed line may be drawn to guide the eye across the gap, but it is not part of the graph.

Check it with numbers

The ruler's midpoint: \frac{5+(-5)}{2}=0 at u=10. The truth near there: m(9.9)=\frac{10}{0.1}=100 and m(10.1)=\frac{10}{-0.1}=-100. Two points 0.2 cm apart have magnifications 200 apart; no straight line through the table captures that.

The rule it gives

Asymptotes. The book's definition: when a curve approaches a second curve (or line) arbitrarily closely without meeting or crossing it, the second is an asymptote to the first. A vertical asymptote sits where a denominator becomes zero. Not all asymptotes are vertical, or even straight.

Beyond the book, a sharper definition says what "approaches" means: x=a is a vertical asymptote when the values grow without bound as x approaches a; y=L is a horizontal asymptote when the values settle towards L as x grows very large. The magnifier has one of each. Far from the page, m(110)=\frac{10}{-100}=-0.1: the image is small and inverted, and m settles towards 0. A curve may cross a horizontal asymptote — y=\frac{x}{x^2+1} passes through 0 at the origin and still settles towards 0 far out on both sides — so "never crosses" holds only for vertical ones. And the curve y=x^2+\frac1x settles towards the parabola y=x^2 far from the origin: an asymptote that is not a line.

Piecewise equations. Which formula applies depends on where x lies. Not every equation has the single form "y= an expression in x".

Discontinuities. Not every equation gives a smooth unbroken curve. Where the graph has a gap, that gap is a discontinuity, and a line drawn across it is not part of the graph. Where the two pieces end at the same x, mark which piece owns the endpoint: a closed dot for the value that belongs to the graph, an open circle for the one that does not (a convention beyond the book). For the switch, y=1 at x=2 itself, so the upper line ends in a closed dot and the lower line starts with an open circle.

Beyond the book, discontinuities come in three kinds. An infinite one, as at an asymptote. A jump, as at the switch. And a hole: y=\frac{x^2-1}{x-1} equals x+1 everywhere except at x=1, where it has no value, so its graph is the line y=x+1 with the single point (1,\ 2) missing.

The procedure. Before joining plotted points, ask two questions. Where does the formula have no value — a zero denominator, the square root of a negative number? And where does a piecewise formula change? Never join across either; sample more densely near them.

Worked example

Plot y=4-x^2 for x\le1 and y=x+1 for x>1, for -2\le x\le3.

At x=1 the first formula applies and gives 3 (a closed dot). Just past 1 the second formula gives values close to 2 (an open circle at (1,\ 2)). So the graph falls from 3 to 2 at x=1: a jump discontinuity of 1, and the points at 1 and 1.5 must not be joined.

Where this shows up

This is an experiment you can do at a desk. Hold a magnifying glass over a page and look through it from arm's length as you lift it. The letters grow — twice their size at 5 cm, five times at 8 — smear into a blur at about 10 cm, and then reappear upside down, five times their size at 12 cm and shrinking as you go on lifting. A 2 mm letter at u=9 cm looks 20 mm tall. The blur is the asymptote, and the flip is the change of sign across it. The joined table, with its zero at 10 cm, predicts something no one has ever seen through a lens.

Narration spine. The magnifier's table is plotted and joined with a ruler, and the join crosses zero at 10 cm — struck out. A probe creeps towards 10 and the readout climbs through 10, 100, 1000; on the far side it is hugely negative. The false segment is removed, the dashed asymptote drawn, and the true curve's two pieces appear. A letter beside the graph grows and then flips as the lens passes the focal length. Last, a reversing switch: the ruler's slope through zero is replaced by two lines and a dashed guide that is not part of the graph.

Unit 03: A Spreadsheet Is a Function Plotter

The problem

Tabulating the magnifier of Unit 02 every 2 cm from u=1 to u=21 cm means eleven calculations of \frac{10}{10-u}. A lens of a different focal length means eleven more. A spreadsheet does the whole table from one formula typed once. How can one formula give eleven different answers — and how do you change the lens without retyping anything?

First attempt

Type each calculation into its own cell of column B: =10/(10-1), =10/(10-3), =10/(10-5), and so on. It works, but it is eleven formulas typed by hand, and changing to an 8 cm lens means editing both 10s in every one of them: twenty-two edits. Copying the first formula down the column is no help either — every copy computes \frac{10}{10-1}, and the column fills with 1.111.

The picture

A spreadsheet is a grid of cells. Each has an address, its column letter followed by its row number: A1, B7, P123.

Put the lens heights in column A: type 1 in A1, then use Fill → Series with step value 2 to fill A1:A11 with 1,3,5,\ldots,21. In B1 type

=10/(10-A1)

and read A1 inside it not as "the cell A1" but as an arrow: "the cell one column to my left, on my row". This is a relative address. Now copy B1 down to B11 by dragging the small square at its bottom-right corner. Each copy keeps the same arrow, so B2 works out \frac{10}{10-\text{A2}}, B7 works out \frac{10}{10-\text{A7}}, and one formula gives eleven answers.

The lens needs the opposite kind of arrow. Put the focal length in D1 and write

=$D$1/($D$1-A1)

The dollar signs pin the address. $D$1 is an absolute address: its arrow lands on the actual cell D1 wherever the formula is copied, while A1 still slides down the rows.

D1 holds the focal length, 10: every row points at that one cell, while A1, A2, … each point one column to the left.

Select A1:B11 and insert an X-Y Scatter chart, "Scatter with Smooth Lines". Column A becomes the horizontal coordinate, column B the vertical. Now type 8 into D1. Every cell in column B recalculates, and the chart redraws itself for the new lens.

The chart also repeats Unit 02's mistake. It joins (9,\ 10) to (11,\ -10) with a line straight through the asymptote. The book's fix is to insert an empty row between the two rows that straddle the asymptote: the chart will not draw a line across an empty row. The fix follows the formula, not the spreadsheet: with 8 in D1 the asymptote moves to u=8, between rows 4 and 5, and the empty row must move with it.

Check it with numbers

With 10 in D1: B1 =\frac{10}{9}=1.111, B4 =\frac{10}{3}=3.333, B5 =10, B6 =-10, B11 =\frac{10}{-11}=-0.909.

With 8 in D1: B1 =\frac87=1.143, B4 =\frac81=8, B5 =\frac8{-1}=-8, B11 =\frac8{-13}=-0.615. A number changed in one cell has changed all eleven results.

The rule it gives

What the book teaches, in its order:

• Rows and columns. Columns are lettered from A, rows numbered from 1; P123 is column P (the 16th letter), row 123. The active cell is the one outlined by the cursor; move it with the arrow keys or the mouse.
• Entries. Any cell holds text or a number: type it, then press Enter.
• Formulas begin with =. Use * for multiplication and ^ for a power. A formula updates itself: =3*C15 in C16 changes whenever C15 does.
• Clearing. Select a cell and press backspace; for a block, highlight it and cut it (or Edit → Clear → All).
• Constructing a graph. First value in A1; highlight the column; Fill → Series → Columns with the step value. Formula in B1; copy it down by dragging the fill handle. Highlight both columns; Insert → X-Y Scatter → "Scatter with Smooth Lines". In short: x-values in the first column, y-values in the second, and the Scatter chart.
• Displays. Chart Design → Quick Layout; delete the "Series 1" legend; type axis titles; drag the handles to resize.
• A new equation. Overtype B1 and copy it down again. The graph updates itself.
• Stray lines. Smooth lines are drawn across asymptotes and jumps. Insert an empty row (Insert → Rows) between the two rows that straddle the break.
• Relative and absolute addresses. A1 means "the cell in this relative position" and changes when copied; $A$1 means that actual cell and never changes. The formula bar (View → Formula Bar) shows the relative addresses changing down the column while the absolute ones stay fixed.

Reading roots from a plot. A spreadsheet graph shows at once where a curve meets the x-axis. The book's four examples: y=x^2-5x+6=(x-2)(x-3) meets it at x=2 and 3; y=x^2-6x+9=(x-3)^2 touches it only at x=3; y=x^2-x+1 never reaches it (its lowest value is 0.75, at x=0.5), matching the fact that it does not factorize; and y=x^3-6x^2+11x-6=(x-1)(x-2)(x-3) meets it at 1, 2 and 3. A plot suggests where the roots are; the factor theorem of F.3 Unit 04 confirms them, and F.6 finds them when they are not whole numbers.

Beyond the book: use the X-Y Scatter chart, not a Line chart. A Line chart spaces the rows evenly along the horizontal axis whatever numbers are in column A, so a table with uneven steps comes out distorted. Only the scatter chart uses column A as coordinates.

Worked example

The book's first spreadsheet graph, y=(x-2)^3 for -1\le x\le5:

1. Type -1 in A1. Highlight A1:A21; Fill → Series → Columns, step value 0.3. Column A now runs -1,-0.7,-0.4,\ldots,5.
2. In B1 type =(A1-2)^3. It shows -27.
3. Drag the fill handle of B1 down to B21. Now B2 holds =(A2-2)^3 =(-2.7)^3=-19.683, and B11 holds =(A11-2)^3 =0, since A11 is 2.
4. Highlight A1:B21; Insert → X-Y Scatter → "Scatter with Smooth Lines".

The curve rises through the axis once, at x=2, where it flattens as it crosses — the triple root of (x-2)^3.

Where this shows up

This is how engineers run a sensitivity study: put every constant of a model in its own absolute cell, then change one and watch the whole table and chart respond. Here, swapping the 10 cm lens for a stronger 8 cm one moves the blur from 10 cm to 8 cm above the page, and a lens held at u=7 cm magnifies 8 times instead of 3.333. Laboratory sheets are built the same way, with a calibration constant or a material property in one pinned cell and a formula copied down a column of readings.

Narration spine. A grid appears; column A fills with 1 to 21. The formula in B1 sends one arrow left to A1 and one across to the pinned D1. Copied down, the left arrow slides row by row while the pinned arrow stays on D1, and the answers fill in. The points fly into a scatter chart, whose smooth line crosses the asymptote. D1 changes from 10 to 8: every answer and every point moves together. An empty row is inserted, and the false line disappears.

Unit 04: Inequalities Are Regions

The problem

A stone arch bridge has an opening 10 m wide at road level and 5 m high in the middle. Measuring x across from the centre line, its curve is

A lorry 2.5 m wide and 4.8 m tall drives down the middle. Does it fit?

First attempt

The arch is 5 m high and the lorry is 4.8 m tall, so it clears by 20 cm. That compares the wrong heights. The lorry's roof corners are not on the centre line; they are 1.25 m either side of it, and there the arch is only

The corners strike the arch 11.25 cm down from their tops.

The picture

Take any point (x,y) in the opening. Stand at its x and look straight up: the vertical line through x meets the arch exactly once, at the height f(x)=5-\frac{x^2}{5}. The point is under the arch if and only if its height is less than that, at the same x:

Now slide that vertical line across the opening from x=-5 to x=5. At every position, the part of the line below the curve is made of points with y<f(x), and the part above of points with y>f(x). Swept across, the parts below fill out the whole space under the arch.

So the graph of an inequality is not a line but an area: a whole region of the plane. The curve y=f(x) is the boundary that separates the region y<f(x) below it from the region y>f(x) above it. That is the book's rule, and the vertical line is why it works: every vertical line meets the graph of y=f(x) once, so "below" is well defined at every x.

Check it with numbers

The 4.8 m lorry's corner is the point (1.25,\ 4.8). There f(1.25)=4.6875, and 4.8>4.6875: the corner is above the curve, outside the region, and strikes.

A 4.5 m lorry of the same width: 4.5<4.6875, so its corners are inside the region, with 4.6875-4.5=0.1875 m, about 19 cm, to spare. The widest 4.5 m lorry that fits down the middle needs 5-\frac{w^2}{5}=4.5 at its corners, so its half-width is w=\sqrt{2.5}=1.581 m: 3.16 m wide in all.

The rule it gives

Less than or greater than. 3<5 and -2>-4. The inequality y>x is satisfied, for each x, by infinitely many y, so its graph is a region, not a line.

Regions.

• The points below the curve y=f(x) satisfy y<f(x).
• The points above it satisfy y>f(x).
• y\ge f(x) means y>f(x) or y=f(x): the region on or above the curve; y\le f(x) is on or below it.

Make y the subject first. The reading "above" or "below" only works once the inequality has the form y>\ldots or y<\ldots. Adding and subtracting on both sides, or multiplying and dividing both sides by a positive number, keeps the direction: 2x+3y>6 becomes 3y>6-2x, then y>2-\frac23x, the region above the line with gradient -\frac23 through (0,\ 2). Multiplying or dividing by a negative number reverses it (proved in Unit 06): 2x-y<4 becomes -y<4-2x, and dividing by -1 gives y>2x-4, the region above the line.

Beyond the book, three conventions make regions easier to draw and read:

• Draw the boundary dashed when it is excluded (<, >) and solid when it is included (\le, \ge).
• A test point confirms the side: substitute any point not on the boundary, and if the inequality holds there, its side is the region. The origin is the usual choice when the boundary misses it.
• Not every region is "above" or "below" a single y=f(x). x^2+y^2\le25 is the inside of the crank circle of Unit 01, the points within 5 of the origin; a vertical line meets that boundary twice, so use a test point: (0,\ 0) gives 0\le25, true, so the region is the disc.

Worked example

Describe the region y\le x^2-3x+2.

The boundary is the parabola y=x^2-3x+2=(x-1)(x-2), which meets the x-axis at 1 and 2 and is drawn solid, since \le includes it. The region is on or below it.

Test (0,\ 0): 0\le2, true, so the origin is in the region. Test (1.5,\ 0): the curve there is at 1.5^2-4.5+2=-0.25, and 0\le-0.25 is false, so (1.5,\ 0) is outside. Between x=1 and x=2 the parabola dips below the axis, and the region dips with it.

Where this shows up

A driver at speed v m/s needs a reaction distance of about 0.7v (a 0.7 s reaction) plus a braking distance of \frac{v^2}{14} (braking at 7 m s^{-2}) to stop:

A clear road of y metres ahead is safe when y\ge s(v): the region on or above the curve. At 20 m/s (72 km/h), s=14+28.6=42.6 m; at 30 m/s (108 km/h), s=21+64.3=85.3 m. The two-second rule gives a gap of y=2v, which is 60 m at 30 m/s: the point (30,\ 60) lies below the curve, too short to stop for something stationary. The line y=2v crosses the curve where 2v=0.7v+\frac{v^2}{14}, at v=18.2 m/s (65.5 km/h); above that speed, two seconds is not a stopping distance. And the arch is an observation anyone can make: an arch bridge's height is greatest only on its centre line, which is why a high lorry drives through the middle of the road.

Narration spine. The arch is drawn and a 4.8 m lorry arrives; "4.8<5, it fits" is written and crossed, because its corners hit the arch at 1.25 m. A vertical probe sweeps across the opening, splitting at the curve into a part below and a part above, and the part below leaves a shaded trail until the whole region under the arch is filled: y<5-\frac{x^2}{5}. The lorry's height drops to 4.5 m, and its corners turn from red to inside the region.

Unit 05: Modulus Is Distance

The problem

A train runs through a station without stopping, at a steady 20 m/s. At t=0 it is 700 m before the station. A display in the carriage shows the distance to the station. What does it show at t=50 s?

First attempt

The distance still to go is 700-20t metres. At t=20 s: 300 m. At t=35 s: 0 — the train is in the station. At t=50 s: 700-1000=-300 m.

A distance of -300 m means nothing on a display. The train is 300 m past the station. The formula 700-20t is not a distance; it is a signed position — positive before the station, negative after.

The picture

Draw the track as a number line with the station at 0. The train is at position p=20t-700: negative before the station, positive after. Its distance from the station is the length of the stretch of track between them, and a length is never negative:

• if p\ge0, the distance is p;
• if p<0, the distance is -p, which is positive.

That quantity — a number's distance from zero, whichever side it is on — is its absolute value or modulus, written |p|. So the display shows |20t-700|, which is the same as |700-20t|: the distance does not care which way you measure.

On a graph against time, the naive line 700-20t runs down to zero at t=35 and on into negative values. The modulus keeps the part above the axis and reflects the part below it upwards, because -p is the mirror image of p in the axis. The result is a V with its point at t=35.

Check it with numbers

At t=50: |700-1000|=|-300|=300 m. At t=20: |700-400|=300 m. The two times are 15 s either side of t=35, and the V gives them the same height, as a distance must: 15 s at 20 m/s is 300 m on either side. At t=0: 700 m; at t=60: |700-1200|=500 m.

The rule it gives

Absolute value or modulus. On the number line, a number's distance from zero is its absolute value, also called its modulus: |-5|=5, |3|=3, |-2.41|=2.41, |13.6|=13.6. The rule, read "y equals mod x":

So |6|=6 and |-4|=-(-4)=4. For a negative number, multiplying by -1 gives its modulus.

The modulus of an expression is piecewise in the same way, switching where the expression is zero:

Check: |x+5| at x=-7 is |-2|=2, and the second line gives -(-7+5)=2. In general |x-a| is the distance between x and a on the number line.

Graphs. y=|x| is two straight lines forming a V with its point at the origin, the arms rising with gradients -1 and 1. y=|x-a| is the same V with its point at x=a.

The family y=|a(x+b)|+c (with a>0). The point of the V is where x+b=0, at x=-b, and its height there is c. So increasing b moves the graph left, increasing c moves it up, and increasing a makes both arms steeper, with gradients \pm a. The train's distance is |700-20t|=|20(t-35)|: a=20, b=-35, c=0, a V with its point at t=35 and arms of gradient \pm20 m/s.

Worked example

Write y=|2x-6|+1 without the modulus sign, and give the point of its V.

2x-6\ge0 when x\ge3, so

The point is at (3,\ 1), and the arms have gradients 2 and -2: in the family's form, 2|x-3|+1 has a=2, b=-3, c=1. Check two points the same distance from 3: at x=1, |{-4}|+1=5 and 7-2=5; at x=5, |4|+1=5 and 10-5=5.

Where this shows up

A machinist measures two shafts meant to be 25 mm across: 24.97 mm and 25.03 mm. Their signed errors are -0.03 and +0.03 mm, and comparing those as numbers would call the first "smaller". The quality chart records the size of the error, |L-25|, which is 0.03 mm for both: equally good. The same V is in the full-wave rectifier inside a phone charger, which turns an input of -12 V into an output of 12 V, and in a map app counting down the distance to a place you drive through — down to zero, then up again. Watch the display on a train as it runs through a station: it traces the V.

Narration spine. A train moves along a track towards a station at zero, and the formula 700-20t is plotted as it goes. Past the station the formula turns negative — crossed out. A brace on the track shows the real distance, always positive. The negative part of the line reflects up into a V with its point at t=35. Last, the family y=|a(x+b)|+c on a small graph: b slides the V left, c lifts it, a steepens its arms.

Unit 06: Solving Modulus Inequalities, and Why Negatives Flip Them

The problem

The station of Unit 05 has Wi-Fi that reaches 300 m either side of it. A passenger on the train is in range while

For which times t?

First attempt

A distance less than 300 means the signed position lies between -300 and 300:

Subtract 700 from all three parts: -1000<-20t<-400. Divide all three parts by -20:

No number is both greater than 50 and less than 20, so the passenger never has Wi-Fi. But at t=30 the train is |700-600|=100 m from the station, well within range. The last step has broken something.

The picture

Think of multiplying by a number as doing something to the whole number line, as in F.1 Unit 02. Take two numbers in order, 3<5, and multiply both by k, sliding k down from 1 to -1.

While k is positive the two points shrink towards 0 but keep their order. At k=0 they meet. As k goes negative they pass through each other and come out on the other side of zero in reverse order: -3 is now to the right of -5, so -3>-5. Multiplying by any negative number is a flip of the line followed by a stretch, and the flip reverses every order. So when an inequality is multiplied or divided by a negative number, its direction must be switched.

The graph shows the same answer without any algebra. The line y=300 cuts the V of Unit 05 where |700-20t|=300, at t=20 and t=50. Between those times the V is below the line; outside them it is above.

Check it with numbers

Dividing -1000<-20t<-400 by -20 with the switch gives 50>t>20, that is 20<t<50 — the two times where the line cuts the V. At the endpoints the train is exactly 300 m away, which is not less than 300, so they are excluded. Inside, t=30 gives 100 m: in range. Outside, t=10 gives |700-200|=500 m: out of range.

The book's own counterexample shows why the switch is forced. x=1 satisfies x<2. Multiplying by -1 without switching gives -1<-2, which is false; with the switch, -1>-2, which is true.

The rule it gives

Operations on an inequality, including a double one: add or subtract the same number across all parts freely; multiply or divide all parts by a positive number freely; when multiplying or dividing by a negative number, switch every inequality sign.

The rules for a>0 (the book's list):

Rules (a) and (c) are the horizontal line cutting the V of y=|x| at \pm a: below the line is the interval between, above it the two arms outside. Rules (e) and (f) are the switch.

Greater-than inequalities give two separate pieces, and each must be solved on its own. When is the passenger out of range? |700-20t|>300 means 700-20t>300 or 700-20t<-300. The first gives -20t>-400, so t<20; the second gives -20t<-1000, so t>50. The answer is t<20 or t>50 — never one double inequality, since no t is both.

With \le or \ge the endpoints are included, and the same steps apply.

Beyond the book, the distance reading does all of this at a glance: |x-c|<r says "x is within r of c", the interval from c-r to c+r, and |x-c|>r says "x is further than r from c". Because |700-20t|=20|t-35|, the Wi-Fi condition is |t-35|<15: within 15 s of t=35, from 20 to 50.

Worked example

Solve |7-2x|<9.

So -1<x<8. Check x=0: |7|=7<9, inside. Check x=8: |7-16|=9, not less than 9, so 8 is excluded.

A second route avoids the switch and gives the same answer, because a distance is the same measured either way: |7-2x|=|2x-7|. Then -9<2x-7<9, so -2<2x<16 and -1<x<8.

Where this shows up

A shaft is turned down on a lathe from 25.4 mm, each pass removing 0.02 mm, so after n passes it measures L=25.4-0.02n mm. It must be within 0.05 mm of 25 mm:

Subtract 0.4: -0.45\le-0.02n\le-0.35. Divide by -0.02 and switch: 22.5\ge n\ge17.5. So the shaft is in tolerance after 18, 19, 20, 21 or 22 passes. Check with calipers: after 18 passes it is 25.04 mm and after 22 passes 24.96 mm, both inside; after 17 it is still 25.06 mm, and after 23 it has gone to 24.94 mm. Forget the switch and the working says 22.5\le n\le17.5: no number of passes would ever be in tolerance.

Narration spine. The line y=300 cuts the train's V at t=20 and t=50. The algebra is written out and divides by -20 to reach 50<t<20, which is crossed out. Two dots at 3 and 5 on a number line are multiplied by k as it slides from 1 to -1: they close in, meet at 0 and pass through each other, landing at -3>-5. Back on the graph the endpoints 50 and 20 swap into order, and the interval 20<t<50 lights up under the line, with a check at t=30.