Reads absolute value as distance rather than as sign removal, which turns equations into 'which points are this far from the centre' and inequalities into 'inside' or 'outside'. Graphs the V-shaped parent and every transformation of it from the previous section, and connects the graph to the solution sets of the corresponding equations and inequalities.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 1 — Functions
§1.6 Absolute Value Functions, pp. 137-150
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 137-150 — the pages these objectives are drawn from
Warm-up
The usual description is 'it makes things positive'. That description is almost right and gets a whole class of problems wrong.
Discussion prompt
Simplify the absolute value of x, given no information about x. Is the answer x?
Hint: What if x is negative?
Answer:
It is not x, and it is not always x with the sign removed either. If x is negative, the absolute value of x is the negative of x — which is a positive number, because negating a negative gives a positive.
So the honest answer is: it is x when x is at or above zero, and negative x when x is below zero. That is a piecewise definition, and it is why absolute value cannot be simplified without knowing the sign of what is inside.
The description that works everywhere is distance from zero. Distance is never negative, and it does not require knowing the sign in advance, which is what makes it the useful reading.
Concept
The absolute value of a number is its distance from zero on the number line. More usefully, the absolute value of a difference is the distance between the two numbers.
absolute value — The distance of a number from zero on the number line, written with vertical bars. Equivalently, the piecewise rule that returns the number itself when it is nonnegative and its negative otherwise. Outputs are never negative.
\[ |x-a| = \text{the distance between } x \text{ and } a \]
The second reading is the one that pays. Almost every absolute value equation and inequality in this section, and every one in the chapters that follow, is a question about which points lie a given distance from a given centre. Framed that way, they need no case analysis at all.
Figure (svg): A number line showing that the absolute value of a number is its distance from zero, with two points equidistant from the origin marked and both labelled with the same distance
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 137-139
Section
Section 1
Concept
The formal definition is piecewise: return the input if it is nonnegative, and its negative otherwise. The useful description is distance from zero, and the two agree everywhere.
\[ |x| = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases} \]
The phrase 'negative x' in the definition trips people up, because it looks as though it produces a negative output. It does not: when x is below zero, negating it gives a positive number. The minus sign in the definition is doing the work of removing a sign, not adding one.
Figure (svg): A number line showing that the absolute value of a number is its distance from zero, with two points equidistant from the origin marked and both labelled with the same distance
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 137-140
Picture it
The difference inside the bars names the centre, and the value is the distance from it.
Figure (svg): A number line showing that the absolute value of x minus a is the distance between x and a, with a marked as the centre and two points at equal distances on either side
Two points lie at any given positive distance from a centre, one on each side. That single observation explains why absolute value equations have two solutions.
Worked example
Translating the notation into a sentence about distance is most of the work.
\[ \text{Describe } |x-7| = 2 \text{ in words, and solve it.} \]
Identify the centre
Why: It is the number being subtracted inside.
\[ \text{centre } 7 \]
State the condition in words
Why: The distance from x to 7 is 2.
\[ \text{distance } 2\text{ from } 7 \]
Find the point 2 to the right
Why: Add the distance to the centre.
\[ x = 9 \]
Find the point 2 to the left
Why: Subtract the distance from the centre.
\[ x = 5 \]
Figure (svg): A number line showing that the absolute value of x minus a is the distance between x and a, with a marked as the centre and two points at equal distances on either side
\[ x = 5 \quad \text{and} \quad x = 9 \]
Verify: substitute both back
Why: At 9 the inside is 2 and its absolute value is 2. At 5 the inside is negative 2 and its absolute value is also 2. Both work. Note that no case analysis was needed: the centre plus and minus the distance gives both answers directly.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 139-140
Translation
Each expression names a centre.
Match the pairs
Why: The sign inside flips when read as a centre, exactly as a shift does, because the standard form is a subtraction. The third is the special case with centre zero, which is the version most people meet first and which hides the general pattern.
Worked example
When the inside is written with a plus, the centre is negative.
\[ \text{Find the centre of } |x+3|. \]
Rewrite as a difference
Why: The distance reading needs a subtraction.
\[ | x - (-3) | \]
Read the centre
Why: It is the number being subtracted.
\[ \text{centre } -3 \]
Check with an input
Why: The expression should be zero at the centre.
\[ \text{at } x = -3,\text{ gives } 0 \]
State the meaning
Why: The expression is a distance from negative 3.
\[ \text{distance from } -3 \]
Figure (svg): The solution to Worked example a sign hidden inside shown as a ladder of expressions, one row per legal move
\[ |x+3| = |x-(-3)|: \text{ the distance from } x \text{ to } -3 \]
Verify: confirm with a second input
Why: At x equal to 0 the expression gives 3, and 0 is indeed 3 units from negative 3. The rewriting step is the same sign convention as the shifts in §1.5, and for the same reason: the standard form uses a subtraction so that the number in it can be read directly.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 140-141
Trap
\[ |x-5| = x-5 \quad \text{for all } x \]
Remove the bars and keep what was inside
Why: The bars are treated as brackets that can simply be dropped.
The expression is then used as though it were the linear rule x minus 5.
This holds only when x is at or above 5. Below 5 the inside is negative, and the absolute value is its negative, which is 5 minus x.
Test at x equal to 2: the inside is negative 3 and the absolute value is 3, while the expression x minus 5 gives negative 3. They differ by a sign.
Bars cannot be dropped without knowing the sign of what is inside. That is what makes absolute value genuinely piecewise, and why its graph has a corner rather than being a single straight line.
Prediction
An absolute value expression is set equal to a negative number.
Predict first
How many solutions does the equation have?
Correct: None, because a distance cannot be negative.
Why: An absolute value is a distance and distances are never negative, so no input can produce a negative output. Recognising this immediately saves a page of algebra, and it is a case worth checking for before starting: an equation set equal to zero has exactly one solution, to a positive number two, and to a negative number none.
Sorting
The right-hand side decides before any algebra is done.
Sort into buckets
Sort each equation by its number of solutions.
Socratic
Both descriptions of absolute value are correct.
Discussion prompt
Why does reading it as a distance make the equations easier than reading it as the piecewise rule?
Hint: How many cases does each reading require you to handle?
Answer:
The piecewise reading forces a case analysis every time: assume the inside is nonnegative and solve, then assume it is negative and solve again, then check both answers against their assumptions. That is four steps and two chances to slip.
The distance reading gives both answers at once: the centre plus the distance, and the centre minus it. No assumptions are made and nothing needs checking against them.
The case analysis is still what justifies the shortcut, and it is worth being able to do. But once the shortcut is understood, it is faster and much less error-prone — and it is the reading that generalises to the inequalities in the last section of this lesson.
Section
Section 2
Concept
Because the definition is two linear rules, the graph is two straight rays. They meet at the input where the inside expression is zero, and that meeting point is the corner.
Finding the corner is always the same one-line calculation: set the inside equal to zero and solve. Everything else about the graph follows from it, because the arms are straight and their steepness comes from the coefficient outside.
Figure (svg): The parent absolute value graph as a V with its corner at the origin, alongside the piecewise definition that produces it, showing the two straight rays meeting at the vertex
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 141-145
Picture it
The piecewise definition on the left produces the two rays on the right.
Figure (svg): The parent absolute value graph as a V with its corner at the origin, alongside the piecewise definition that produces it, showing the two straight rays meeting at the vertex
Each piece contributes one ray. They meet at the origin because that is where the inside expression, which is just x here, equals zero.
Worked example
One equation gives the input; one substitution gives the height.
\[ \text{Find the corner of } f(x)=|2x-6|+1. \]
Set the inside equal to zero
Why: The corner is where the absolute value bottoms out.
\[ 2 x - 6 = 0 \]
Solve for the input
Why: Divide by 2.
\[ x = 3 \]
Substitute to find the height
Why: The absolute value contributes zero there.
\[ f(3) = 0 + 1 \]
State the corner
Why: Both coordinates together.
\[ \text{corner at } (3, 1) \]
Figure (svg): The solution to Worked example locate the corner shown as a ladder of expressions, one row per legal move
\[ \text{corner } (3,\,1) \]
Verify: check the height is the lowest
Why: Since the absolute value part is never negative and the outside adds 1, the smallest possible output is 1, attained exactly when the inside vanishes. So the corner is also the minimum, which is true whenever the coefficient outside the bars is positive. If it were negative, the corner would be the maximum instead.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 142-143
Faded example
Locate the corner of the rule that takes the absolute value of 3x plus 12, then subtracts 2.
Fill in the blanks
3x + 12 = 0 \;\Longrightarrow\; x = -4, \qquad \text-2 = ___
Why: Solving the inside for zero gives x equal to negative 4. At that input the absolute value contributes 0, so the height is whatever is added outside, which is negative 2. The corner is at the point with coordinates negative 4 and negative 2, and since the coefficient outside is positive, it is the graph's minimum.
Worked example
Every transformation from §1.5, applied to one parent.
\[ \text{Graph } g(x)=-2|x-3|+4. \]
Read the inside
Why: Minus 3 shifts right 3.
\[ \text{right } 3 \]
Read the outside multiplier
Why: The 2 makes the arms twice as steep.
\[ \text{steeper by } 2 \]
Read the sign
Why: The minus flips the V upside down.
Read the outside constant
Why: Plus 4 raises it.
\[ \text{corner at } (3, 4) \]
Figure (svg): A transformed absolute value graph shown against the parent, with the corner moved and the arms flipped and steepened, and each transformation labelled
\[ \text{corner } (3,4), \text{ opening downward, arm slopes } \pm 2 \]
Verify: test a point on an arm
Why: At x equal to 4 the rule gives negative 2 times 1 plus 4, which is 2. The corner is at height 4 and moving one unit right has dropped the graph by 2, which matches an arm slope of negative 2. The reflection makes the corner a maximum rather than a minimum, so the range is everything at or below 4.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 144-145
Error analysis
A student locates the corner of the rule with x plus 5 inside the bars.
Annotate
On: \( f(x)=|x+5| \;\Longrightarrow\; \text{corner at } x = 5 \)
Never read the corner off the constant. Set the inside equal to zero and solve — it is one line and it is right regardless of how the expression is written.
Prediction
An absolute value rule has a negative number multiplying the bars.
Predict first
What does that do to the graph?
Correct: It opens downward, and the corner is a maximum.
Why: The multiplier is outside, so it acts on the output: a negative one flips the graph over the horizontal axis, turning the V upside down. The corner is then the highest point rather than the lowest. Note that reflecting an absolute value graph left to right changes nothing visible, because the V is already symmetric.
Sorting
Only the sign of the coefficient outside the bars matters.
Sort into buckets
Sort each rule.
Explain it to yourself
The absolute value graph has a sharp corner rather than a smooth turn.
Discussion prompt
Explain why the graph has a corner, using the piecewise definition.
Hint: What are the slopes of the two pieces?
Answer:
Each piece is a straight line, one with slope 1 and one with slope negative 1 for the parent. Straight lines do not bend, so all the direction change has to happen at the single point where the pieces meet.
At that point the graph jumps from one slope to the other without passing through the values in between, which is what a corner is. A parabola turns smoothly because its slope changes gradually; the V has nothing gradual about it.
This becomes important in Chapter 12. A corner is a point where a function is continuous but has no single well-defined steepness, so it is continuous but not differentiable — and the absolute value at zero is the standard example of exactly that.
Section
Section 3
Concept
An absolute value equation is solved by isolating the absolute value expression and then setting the inside equal to both the positive and the negative of the right-hand side.
\[ |A| = c \;\Longleftrightarrow\; A = c \;\text{ or }\; A = -c, \quad c \ge 0 \]
Isolating first is not optional. An equation with the bars still multiplied by something has a different right-hand side once the multiplication is undone, and splitting before that produces two equations with the wrong constant in them.
Figure (svg): Three number lines showing the solution sets of an absolute value equation, a less-than inequality giving a bounded interval, and a greater-than inequality giving two rays
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 145-148
Picture it
The top line is the equation; the two below are what happens when the equals sign becomes an inequality.
Figure (svg): Three number lines showing the solution sets of an absolute value equation, a less-than inequality giving a bounded interval, and a greater-than inequality giving two rays
The equation picks out the two boundary points. The inequalities pick out everything on one side of them or the other, which is the subject of the next section.
Worked example
The isolation step is the one that is skipped.
\[ \text{Solve } 3|x-2|+4 = 19. \]
Subtract the constant
Why: Get the multiplied bars alone.
\[ 3 | x - 2 | = 15 \]
Divide by the multiplier
Why: Now the bars are isolated.
\[ | x - 2 | = 5 \]
Split into two equations
Why: The inside is 5 or negative 5.
\[ x - 2 = 5\text{ or } -5 \]
Solve each
Why: Add 2 to both.
\[ x = 7\text{ or } x = -3 \]
Figure (svg): The solution to Worked example isolate, then split shown as a ladder of expressions, one row per legal move
\[ x = 7 \quad \text{and} \quad x = -3 \]
Verify: substitute both back into the ORIGINAL
Why: At 7: the inside is 5, its absolute value is 5, three times that is 15, plus 4 is 19. At negative 3: the inside is negative 5, absolute value 5, and the rest follows identically. Both check. Note that splitting before dividing by 3 would have given the inside as 15 and produced two wrong answers.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 146-147
Ranking
For solving a general absolute value equation.
Put in order
Why: Isolating comes first because everything else depends on the bars standing alone. Checking the sign of the right-hand side comes next, since a negative value ends the problem immediately and saves the remaining work. Only then is it correct to split, and checking the answers in the original equation closes it out.
Worked example
Spotting this early saves all the work.
\[ \text{Solve } |2x+1|+7 = 3. \]
Isolate the bars
Why: Subtract 7 from both sides.
\[ | 2 x + 1 | = -4 \]
Read the right-hand side
Why: It is negative.
\[ \text{right side is } -4 \]
Apply the distance argument
Why: A distance cannot be negative.
State the conclusion
Why: There is nothing to solve.
Figure (svg): The solution to Worked example an equation with no solutions shown as a ladder of expressions, one row per legal move
\[ \text{No solution: an absolute value is never negative.} \]
Verify: confirm from the graph
Why: The graph of the left-hand side is a V whose corner sits at height 7, so its lowest output is 7 and it never reaches 3. A horizontal line at height 3 passes entirely below the graph. Recognising the impossibility at the isolation step is faster, but the graph confirms it independently.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 147-148
Trap
\[ 3|x-2| = 15 \;\Longrightarrow\; 3(x-2) = 15 \;\text{ or }\; 3(x-2) = -15 \]
Split into two cases while the 3 is still attached
Why: The bars are removed and the two sign cases are written down immediately.
Solving gives x equal to 7 or x equal to negative 3, which happen to be correct here.
The answers are right this time and the method is not. Multiplying the whole equation happens to be equivalent to multiplying inside the bars only because 3 is positive.
With a negative multiplier the method breaks: the negative cannot be moved inside the bars, since the absolute value would absorb it. Splitting first then gives the wrong two equations.
Isolate the absolute value completely before splitting. It costs one line, it always works, and it also puts you in position to notice a negative right-hand side and stop.
Prediction
An absolute value equation is isolated and the right-hand side comes out as zero.
Predict first
How many solutions are there?
Correct: One, where the inside equals zero.
Why: Only one point is at distance zero from the centre, namely the centre itself. Splitting into two equations gives the inside equal to 0 and the inside equal to negative 0, which are the same equation, so the two solutions coincide. This is the case where the graph's corner sits exactly on the horizontal line.
Faded example
Solve the equation setting twice the absolute value of x plus 1 equal to 10.
Fill in the blanks
|x+1| = 5 \;\Longrightarrow\; x+1 = 5 \text-6 x+1 = -5 \;\Longrightarrow\; x = 4 \text___ x = ___
Why: Dividing by 2 isolates the bars and gives 5 on the right. The two cases then give 4 and negative 6. Reading it as a distance confirms both: the centre is negative 1 and the distance is 5, so the answers are negative 1 plus 5 and negative 1 minus 5.
Counterexample
A classmate claims every absolute value equation has exactly two solutions.
Discussion prompt
Give two equations that break the claim, in two different ways.
Hint: Think about what the right-hand side can be.
Answer:
One solution: set the absolute value of x minus 4 equal to 0. Only the centre is at distance zero from itself, so x equal to 4 is the only answer.
No solutions: set the absolute value of x minus 4 equal to negative 1. A distance is never negative, so nothing satisfies it.
So the count is decided entirely by the right-hand side once the bars are isolated: negative gives none, zero gives one, positive gives two. Checking that number before doing the algebra is the habit worth building, because two of the three cases require no algebra at all.
Section
Section 4
Concept
A less-than inequality asks which points are closer to the centre than a given distance, giving one interval. A greater-than inequality asks which are further, giving two rays.
\[ |A| < c \;\Longleftrightarrow\; -c < A < c; \qquad |A| > c \;\Longleftrightarrow\; A < -c \text{ or } A > c \]
The mnemonics 'less than is and' and 'greater than is or' are commonly taught and commonly misremembered. The distance reading makes them unnecessary: close to a centre is obviously one stretch around it, and far from a centre is obviously the two pieces at either end.
Figure (svg): Three number lines showing the solution sets of an absolute value equation, a less-than inequality giving a bounded interval, and a greater-than inequality giving two rays
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 148-150
Picture it
Same centre, same distance, three different questions.
Figure (svg): Three number lines showing the solution sets of an absolute value equation, a less-than inequality giving a bounded interval, and a greater-than inequality giving two rays
The equation's two answers are the boundary points of both inequalities. Everything between them is the less-than set, and everything outside is the greater-than set.
Worked example
Closer than a given distance, so one interval.
\[ \text{Solve } |x-3| < 5. \]
Read it as a distance statement
Why: The distance from x to 3 is under 5.
\[ \text{closer than } 5\text{ to } 3 \]
Find the two boundary points
Why: The centre plus and minus the distance.
\[ -2\text{ and } 8 \]
Take everything between them
Why: Closer means inside.
\[ -2 < x < 8 \]
Write it as an interval
Why: Both ends are excluded, since the inequality is strict.
\[ (-2, 8) \]
Figure (svg): The solution to Worked example a less-than inequality shown as a ladder of expressions, one row per legal move
\[ (-2,\,8) \]
Verify: test one point inside and one outside
Why: At x equal to 3 the distance is 0, which is under 5, so it should be in the set — and it is, since 3 lies between negative 2 and 8. At x equal to 10 the distance is 7, which is not under 5, and 10 is indeed outside the interval. Two tests, one from each side, confirm the direction.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 148-149
Discrimination
The direction of the inequality decides the shape of the answer.
Sort into buckets
Sort each inequality by its solution set.
Worked example
Further than a given distance, so two rays.
\[ \text{Solve } |2x+1| \ge 7. \]
Split into the two cases
Why: Further from the centre means beyond either boundary.
\[ 2 x + 1 \ge 7\text{ or } \le - 7 \]
Solve the first
Why: Subtract 1 and halve.
\[ x \ge 3 \]
Solve the second
Why: Subtract 1 and halve.
\[ x \le - 4 \]
Join with a union
Why: Two separate rays.
\[ (-\infty, -4) U [3, \infty] \]
Figure (svg): The solution to Worked example a greater-than inequality shown as a ladder of expressions, one row per legal move
\[ (-\infty,-4] \cup [3,\infty) \]
Verify: check a point in the gap
Why: At x equal to 0 the expression gives 1, which is not at or above 7, so 0 should be excluded — and it is, sitting between negative 4 and 3. The brackets are square because the inequality allowed equality, and the boundary points 3 and negative 4 are exactly the solutions of the corresponding equation.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 149-150
Error analysis
A student solves an inequality asking for a distance above 5 from the centre 3.
Annotate
On: \( |x-3| > 5 \;\Longrightarrow\; -2 > x > 8 \)
A chained inequality can only describe a single interval. Whenever the answer is two rays, it must be written as a union or with the word or, never chained.
Prediction
An absolute value expression is required to be greater than a negative number.
Predict first
What is the solution set?
Correct: Every real number.
Why: An absolute value is never negative, so it is automatically greater than any negative number, whatever the input. The inequality is satisfied everywhere. This is the mirror image of the equation case: a negative right-hand side makes an equation impossible and makes a greater-than inequality universally true.
Faded example
Solve the inequality requiring the absolute value of x minus 6 to be at most 2.
Fill in the blanks
-2 \le x-6 \le 2 \;\Longrightarrow\; 4 \le x \le 8
Why: Adding 6 throughout gives 4 and 8. As a distance statement: the points at most 2 units from the centre 6 run from 4 to 8, both included because the inequality allows equality. Chaining is legitimate here precisely because the answer is a single interval.
Real world
Manufacturing tolerances are absolute value inequalities, and they are written that way in specifications.
Discussion prompt
A part must be 50 millimetres long, to within 0.2 millimetres. Write that as an absolute value inequality and say which kind it is.
Hint: Which is the centre, and which the allowed distance?
Answer:
The centre is the target, 50, and the allowed distance is the tolerance, 0.2. The requirement is that the absolute value of the length minus 50 is at most 0.2.
It is a less-than statement, so the solution set is a single interval: lengths from 49.8 to 50.2 inclusive. That interval is exactly what a quality inspector checks against.
The greater-than version is the rejection criterion: a part is rejected when the distance from target exceeds the tolerance, and its solution set is the two rays of parts that are too short or too long. Reading the specification as a distance is not a mathematical convenience here — it is how the engineering statement is meant.
Section
Section 5
Concept
Solving an absolute value equation is asking where the V meets a horizontal line. Solving an inequality is asking where it lies below or above that line.
This picture makes the three cases inevitable rather than memorised. A horizontal line and an upward V can meet in two points, one point, or none, and those are exactly the three possible solution counts of the equation.
Figure (svg): Three number lines showing the solution sets of an absolute value equation, a less-than inequality giving a bounded interval, and a greater-than inequality giving two rays
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 145-150
Picture it
The corner's height decides which horizontal lines can be met at all.
Figure (svg): A transformed absolute value graph shown against the parent, with the corner moved and the arms flipped and steepened, and each transformation labelled
For this downward V with corner at height 4, lines above 4 miss entirely, the line at 4 touches once, and every line below 4 meets it twice — the same three cases with the inequalities reversed.
Worked example
Same question, answered by looking rather than by algebra.
\[ \text{Using the graph of } y=|x-3|, \text{ solve } |x-3| < 5 \text{ graphically.} \]
Draw the horizontal line at the right-hand value
Why: The line at height 5.
\[ \text{line } y = 5 \]
Find where the V meets it
Why: Two crossings, symmetric about the corner.
\[ \text{at } x = -2\text{ and } x = 8 \]
Decide which side the inequality wants
Why: Less than means below the line.
\[ \text{below } y = 5 \]
Read off the inputs where that holds
Why: Between the two crossings.
\[ -2 < x < 8 \]
Figure (svg): The solution to Worked example read the solutions off a graph shown as a ladder of expressions, one row per legal move
\[ (-2,\,8) \]
Verify: compare with the algebra
Why: The algebraic solution in the earlier section gave exactly this interval. The graph adds something the algebra does not: it makes visible that the answer had to be a single interval, because the region of a V below a horizontal line is always the connected stretch around the corner.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 148-149
Matching
For an upward-opening V with its corner at height 2.
Match the pairs
Why: Everything is decided by the line's height relative to the corner. Above the corner gives two crossings, at the corner gives one, below gives none. The last item shows the transition is continuous: just above the corner the two solutions exist but sit almost on top of each other, and they merge exactly at the corner.
Worked example
The corner's height decides everything before any solving happens.
\[ \text{Without solving, say how many solutions } |x+2|+6 = 4 \text{ has.} \]
Find the corner's height
Why: The absolute value contributes zero at the corner.
\[ \text{lowest output is } 6 \]
Compare with the right-hand side
Why: The line sits at height 4.
\[ 4\text{ is below } 6 \]
Interpret geometrically
Why: The line passes under the whole V.
Conclude
Why: No crossings means no solutions.
Figure (svg): The solution to Worked example use the corner to predict the case shown as a ladder of expressions, one row per legal move
\[ \text{No solutions: the minimum output is } 6 > 4. \]
Verify: confirm algebraically
Why: Isolating gives the absolute value equal to negative 2, which is impossible. The two arguments agree, and the graphical one arrives faster: comparing the right-hand side against the corner's height settles the case in one step, without isolating anything.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 147-148
Trap
\[ |x-1|+3 = 1 \;\Longrightarrow\; x-1 = -2 \text{ or } 2 \;\Longrightarrow\; x = -1 \text{ or } 3 \]
Move the 3 across and split immediately
Why: The bars are isolated to give negative 2, and the usual two cases are written down.
Two answers are reported, and neither is checked in the original equation.
The isolated equation reads: absolute value equals negative 2, which is impossible. The splitting step was applied to a right-hand side that should have stopped the problem.
Testing either answer confirms it. At x equal to 3 the original gives the absolute value of 2, plus 3, which is 5 — not 1.
Check the sign of the right-hand side after isolating, and check every answer in the original. The V's corner sits at height 3, so nothing on the graph ever reaches height 1, and the geometry says so before the algebra does.
Prediction
A downward-opening V has its corner at height 7.
Predict first
For which right-hand values does the corresponding equation have two solutions?
Correct: Any value below 7.
Why: A downward V has 7 as its maximum and falls away on both sides, so it attains every height below 7 twice, height 7 once at the corner, and nothing above. This is the mirror image of the upward case, and thinking in terms of the corner and the direction of opening handles both without separate rules.
Two truths and a lie
Two of these are true of absolute value graphs and one is false.
Eliminate the wrong options
One of these claims is wrong.
Survives elimination: B
Why: B is the false claim. A horizontal line meets a V twice, once, or not at all, depending on whether it sits above, on, or below the corner. Those three cases are exactly the three possible solution counts of an absolute value equation.
Edge cases
The two solutions of an absolute value equation merge as the right-hand side approaches the corner's height.
Discussion prompt
What happens to the solution set of the corresponding less-than inequality as that happens, and at the moment it happens?
Hint: The interval between the two crossings is shrinking. What does it shrink to?
Answer:
As the line descends towards the corner, the two crossings move together and the interval between them shrinks towards a single point.
At the corner's height exactly, the strict less-than inequality has no solutions at all: the graph is below the line nowhere, since it touches the line at one point and is above it everywhere else. The interval has shrunk to nothing.
The non-strict version behaves differently at that instant: it holds at exactly the corner, giving a one-point solution set. That difference between strict and non-strict, invisible everywhere else, becomes the entire answer at the boundary — which is a good illustration of why the distinction is kept.
Comparison
Fill the blanks from memory. All three are the same picture with a different question asked of it.
Comparison matrix
| equals c | less than c | greater than c | |
|---|---|---|---|
| distance reading | exactly c from the centre | closer than c to the centre | further than c from the centre |
| solution set | two points | one bounded interval | two rays, joined by a union |
| on the graph | where the V meets the line | where the V is below the line | where the V is above the line |
| if c is negative | no solutions | no solutions | every real number |
| written as | two equations | a chained inequality | two separate statements, never chained |
The fourth row is the one worth checking first: two of those three cases need no algebra at all once the bars are isolated.
Pattern
The same five steps handle equations and both directions of inequality.
Step 2 is the cheapest step in the list and the one that most often decides the answer. An isolated absolute value set equal to a negative number, or required to exceed one, is finished without further work.
OpenStax Algebra and Trigonometry 2e, §3.6 Absolute Value Functions §3.6
Check
Set the inside to zero.
Check your understanding
Where is the corner of the graph of the rule taking the absolute value of x plus 4, then subtracting 5?
Answer: A
Why: The inside x plus 4 vanishes at x equal to negative 4, and there the absolute value contributes 0, so the height is negative 5. The corner is at the point with coordinates negative 4 and negative 5.
Check
Isolate first, then look at the right-hand side.
Check your understanding
How many solutions does the equation setting the absolute value of x minus 3, plus 8, equal to 5 have?
Answer: A
Why: Isolating gives the absolute value equal to negative 3, and a distance is never negative. Geometrically the V's corner sits at height 8, so the graph never descends to 5 and a horizontal line there misses it entirely.
Check
Which side of the boundary?
Check your understanding
What is the solution set of the inequality requiring the absolute value of x minus 2 to be greater than 6?
Answer: A
Why: Greater than means further from the centre than 6. The centre is 2, so the boundary points are negative 4 and 8, and the solution is everything beyond them on either side — two rays joined by a union.
Real world
Error bars, tolerances and confidence intervals are all absolute value inequalities.
Discussion prompt
A poll reports 46 percent support with a margin of error of 3 points. Write that as an absolute value statement and say what the corresponding greater-than statement would mean.
Hint: What is the centre, and what is the distance?
Answer:
The centre is the reported figure, 46, and the margin is the distance. The claim is that the absolute value of the true support minus 46 is at most 3 — a less-than statement, whose solution set is the interval from 43 to 49.
The greater-than version would describe the values the poll is claiming to rule out: true support further than 3 points from 46, meaning below 43 or above 49. Those are the two rays, and they are exactly the outcomes that would make the poll wrong.
The same structure appears in every tolerance and every confidence interval. Reading 'plus or minus' as a distance rather than as two separate arithmetic operations is what makes the connection visible — and it is why this section spends its effort on the distance reading rather than on the piecewise definition.
Commit first
State your confidence along with your answer.
Predict first
What is the solution set of the inequality requiring an absolute value to be greater than negative 4?
Correct: Every real number.
Why: An absolute value output is never negative, so it exceeds any negative number no matter what the input is. The inequality is satisfied everywhere. Compare with the same right-hand side in an equation or a less-than inequality, where the answer is instead no solutions — the sign of the right-hand side is decisive, but which way it decides depends on the direction of the statement.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why a less-than absolute value inequality gives one interval while a greater-than one gives two rays, without using the words 'and' or 'or' as a memory aid.
Hint: Ask them to stand on a number line at the centre.
Answer:
Read the absolute value as a distance from a centre. The points closer to a centre than a given distance form one connected stretch around it — you cannot be near a point without being in the neighbourhood of it.
The points further from that centre are everything beyond the boundary in each direction, and those two regions are separated by the whole neighbourhood in between. So there are necessarily two of them, and they cannot be joined.
A good explanation makes the shapes feel inevitable rather than rule-governed. Near is one region; far is two. Once that is clear, the mnemonics are unnecessary — and unlike the mnemonics, the reasoning survives being asked about a case the student has not seen before.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The first is the one that makes the other three easier, so it is often worth revisiting even when the difficulty appeared elsewhere. The fourth is where marks are most often lost, usually by writing two rays as a chained inequality.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw one V with its corner clearly marked, and draw three horizontal lines across it: one above the corner, one exactly at it, and one below. Beside each line, write how many solutions the corresponding equation has. Then shade, in two different colours, the parts of the horizontal axis that solve the less-than and the greater-than inequalities for the top line.
If your shading shows one connected stretch for less-than and two disconnected rays for greater-than, you have the fact that the mnemonics are trying to encode.
Recap
Five things, and the first one is what makes the other four quick.
| if you remember one thing | it should be this |
|---|---|
| about the meaning | absolute value is distance, and distance is never negative |
| about the graph | the corner is where the inside equals zero |
| about equations | isolate first, then check the sign of what is left |
| about inequalities | near the centre is one region; far from it is two |
Section 1.7 closes the chapter by asking when a function can be run backwards, which needs the one-to-one test from Section 1.1 and the composition machinery from Section 1.4 at the same time.
OpenStax, Precalculus, §1.6 Absolute Value Functions §1.6, pp. 137-150 — everything on these slides traces back here
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