Quarry School

When B is zero or negative

Explain it like I am five

Before choosing between or outside, compare the requested distance with zero. Think of a ruler that begins at zero and has no negative lengths. A length smaller than a negative number is impossible. Every length is greater than a negative number, including a length of zero. Zero itself needs care. A distance at most zero forces you to stay at the center. A distance greater than zero allows every place except the center. A distance at least zero allows every place. These cases follow from what the ruler can measure. Inspect B, the number on the other side, after clearing operations outside the bars, because that arithmetic can change it.

2345678910−5−4−3−2−112345range(6, 4)
The graph reaches height 4 at one input and never rises above it.
Reminder
  • Letters in the rules. A names the whole inside, and B names the number on the other side. In |2x − 1| ≤ 0, A is 2x − 1 and B is 0.
  • Nonnegative. Nonnegative means zero or positive. Every absolute value has that property.
  • Equality versus inequality. Dividing −2x = 6 by −2 gives x = −3; equality stays equality. Dividing −2x < 6 gives x > −3.
  • One answer and no answers. {6} is the set containing the number 6. ∅ is the empty set, with no numbers in it.
Why it works. Every inside expression A has |A| ≥ 0 wherever A is defined. If B < 0, every such distance is larger than B, so smaller than or equal to B is impossible. At B = 0, equality holds exactly when A = 0; every other distance is positive. These facts cover all four inequality signs. They also explain the equation cases: negative B is impossible, and zero B requires A = 0. Our inside lines are defined for all real inputs, so a condition that is always true gives the whole real line, (−∞, ∞).
RuleIf B < 0: |A| = B, |A| < B, and |A| ≤ B have no solution; |A| > B and |A| ≥ B hold for every allowed input. If B = 0: |A| = 0 and |A| ≤ 0 mean A = 0; |A| < 0 has none; |A| > 0 means A ≠ 0; |A| ≥ 0 holds for every allowed input.
The same idea, five ways
Say it

the distance is at most zero, so the inside must be zero

Write it

Only the center has zero distance; every other position has positive distance.

In math
  • |x − 6| ≤ 0
  • x − 6 = 0
  • x = 6
  • {6}
  • |x − 6| > 0 means x ≠ 6
  • (−∞, 6) ∪ (6, ∞)
  • {x : x ≠ 6}
Like

A leash of length zero keeps you at its post. Being a positive distance from the post means standing anywhere else.

See it
24681024(6, 0)
Only input 6 gives zero distance; every other point is above the axis.
The same idea, other ways
As a ruler

The first possible ruler length is zero. A negative comparison number sits below every measurable length. So a smaller distance is impossible and a greater distance is automatic.

−22−224
Every distance height lies above the line y = −2.
With the smallest possible distance

For |x − 2| > −1, try the smallest distance at x = 2: |0| = 0 > −1. Larger distances also pass, so every real input works.

At zero

|x − 2| ≤ 0 forces x = 2. |x − 2| > 0 permits every x except 2. |x − 2| < 0 permits none. |x − 2| ≥ 0 permits all real x.

ComparisonNegative BB = 0
|A| < BNo solutionNo solution
|A| ≤ BNo solutionSolve A = 0
|A| > BEvery allowed inputAll allowed inputs except A = 0
|A| ≥ BEvery allowed inputEvery allowed input
.1B negative

Every length sits at zero or above. A negative B lies below the whole ruler. Equality and a smaller distance cannot reach it. A greater distance is larger than it even at zero.

  • |A| < B or |A| ≤ B with B < 0: no solution.
  • |A| > B or |A| ≥ B with B < 0: all allowed inputs.
  • |A| = B with B < 0 also has no solution, because it requires a negative distance.
−2246−2246
The distance graph is always above the negative comparison line.
Reminder
  • Comparing negative numbers. 0 > −1, so any number zero or greater is also greater than −1.
Worked exampleOpposite comparisons with negative B

Solve |x − 2| ≤ −1 and |x − 2| > −1.

−2246−2246
The distance graph is always above the negative comparison line.
(−∞, ∞)
The greater than comparison permits the whole real line.
What it asks. First find inputs giving distance at most negative one. Then find inputs giving distance greater than negative one.
Plan. Compare the smallest possible distance, zero, with negative one.
  1. The first has no solution.Every absolute value is at least 0, and 0 is larger than −1.
  2. The second holds for every real x.Even the smallest possible distance 0 exceeds −1.
Answer
  • For |x − 2| ≤ −1: no solution
  • Set: ∅
  • For |x − 2| > −1: every real x
  • Interval: (−∞, ∞)
Check At x = 2 the distance is its smallest value, 0. It fails the first condition and satisfies the second; larger distances behave the same way.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Use the equation's no solution rule for |x − 2| > −1.
The question asks for a greater distance, and even distance zero is greater than −1.
✓ Instead: Every real input satisfies this inequality.
Tips and tricks
  • Say the comparison aloud: a distance greater than a negative number is always possible.
.2B zero

A zero length leash keeps you at its post. A requirement for positive distance sends you anywhere except the post. At a zero B, inspect whether equality is required, forbidden, or allowed along with every positive distance.

  • |A| = 0 and |A| ≤ 0 both mean A = 0.
  • |A| > 0 means A ≠ 0. The symbol ≠ means 'not equal to'.
  • |A| < 0 is impossible. |A| ≥ 0 accepts every input where A is defined.
−224624(2, 0)
Only the labeled corner has output zero; every other point has positive height.
Reminder
  • Nonzero. Nonzero means any number except 0. Its absolute value is positive.
Worked exampleAll positions except the center

Solve |x − 2| > 0.

2(−∞, 2) ∪ (2, ∞)
A hollow dot at 2 excludes only that center input.
What it asks. Find all positions whose distance from two is positive.
Plan. Find the single zero distance input and leave it out of the real line.
  1. Find the zero distance input from x − 2 = 0: add 2 to obtain x = 2.Positive distance excludes exactly the inputs whose inside equals zero, so solving locates the one input to remove.
  2. Check |2 − 2| = 0.The proposed excluded input really has zero distance and fails the question.
  3. Keep every x ≠ 2, meaning x < 2 or x > 2.Every other real number has a nonzero inside and therefore positive absolute value.
Answer
(−∞, 2) ∪ (2, ∞).
Check At 2 the distance is 0 and fails. Every other input has strictly positive distance.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: |x − 2| > 0 accepts x = 2.
The center has distance zero, and the strict greater than sign rejects zero.
✓ Instead: Exclude 2 and keep (−∞, 2) ∪ (2, ∞).
Tips and tricks
  • Solve the inside equal to zero to locate the center. Then decide whether this question keeps or removes it.
.3When the outside factor is zero

If you flatten both ramps completely, they stop making a V. In a|x − h| + k, choosing a = 0 removes the distance part. Every input then gives the same height k. This is a constant function, a flat horizontal line. The choice of h cannot create a corner because zero times any distance is still zero.

  • If a = 0, f(x) = 0|x − h| + k = k, a constant function with no unique corner.
  • Domain: (−∞, ∞). Range: {k}, a set containing the one output k.
  • If k ≠ 0 there are no zeros. If k = 0, every real input is a zero.
−4−22424
The constant output 3 makes a horizontal line with no x-intercepts.
Reminder
  • One member set. {3} lists the single produced output 3; it does not mean an interval.
The same idea, five ways
Say it

zero times the distance plus k

Write it

Every allowed input produces the same output k.

In math
  • f(x) = 0|x − h| + k = k
  • Domain: (−∞, ∞)
  • Range: {k}
Like

A flattened pair of ramps becomes one level road.

See it
−4−22424
Every input produces the single height 3.
Worked exampleA flattened V

For f(x) = 0|x − 7| + 3, find the domain, range, and zeros. What changes if the outside addition 3 is replaced by 0?

−4−22424
The nonzero constant has no horizontal intercepts.
−4−22424
The zero constant lies on the horizontal axis at every input.
What it asks. Describe the permitted inputs, produced outputs, and output zero inputs when the distance term is multiplied by zero.
Plan. Reduce each function to its constant output before finding the requested sets.
  1. Reduce f(x) = 0 × |x − 7| + 3 = 3.Every distance is multiplied by zero, leaving only the outside addition.
  2. The domain is every real number and the range is {3}.Every input is permitted, and every one produces the same single output 3.
  3. There are no zeros because solving f(x) = 0 would require 3 = 0.Setting the output to zero asks for horizontal-axis crossings, but the height never equals zero.
  4. Replacing the addition by zero gives g(x) = 0|x − 7| = 0. Its domain is every real input, its range is {0}, and every real input is a zero.The flat graph now lies on the horizontal axis, and substituting any input gives output zero.
Answer
  • For f(x) = 3: domain (−∞, ∞)
  • Range: {3}
  • Zeros: none
  • For g(x) = 0: domain (−∞, ∞)
  • Range: {0}
  • Zeros: every real input
Check f(0) = 3 and f(7) = 3 confirm the constant height above zero. g(0) = 0 and g(7) = 0 confirm that removing the outside addition puts the whole graph at output zero.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The zero outside factor still makes a V with corner (7, 3).
Every point has the same height 3. There is no change of direction and no unique corner.
✓ Instead: Treat it as the constant function y = 3.
Tips and tricks
  • The usual zero, one, or two x-intercepts rule describes a genuine V, with a ≠ 0. Check a before applying that count.
.4Study priorities and a printable reference after the lessons

Pack for this section the way you pack for a short trip. Keep the few ideas you need constantly in your pocket. Learn how to rebuild the details from those ideas. Keep a compact reference for checking notation during practice. The three lists below separate those jobs. For a closed-book exam, use the reference while studying, then cover it and explain each line. You do not need to memorize every worked answer or every transformed V. You can rebuild them from distance, the corner, and the two straight arms.

  • know cold
    Absolute value is nonnegative distance. Memory device: distance has size, direction has sign. This explains |−6| = 6 and |0| = 0.
    Isolate before splitting. Memory device: alone, then two roads. This keeps outside arithmetic from entering the two cases.
    A positive number B gives two inside values, a number B equal to 0 gives one, and a negative number B gives none. Memory device: two, one, none. This follows from the two sides of home.
    Less than means between with and; greater than means outside with or, for a positive limit. Memory device: less thAND, greatOR. This remembers whether one region or two regions work.
    Multiplying or dividing an inequality by a negative reverses its sign. Memory device: a negative turns the street around. This remembers that left and right trade places.
    A bracket includes an endpoint; a parenthesis excludes it. Memory device: a bracket grabs. This remembers whether equality is allowed.
  • understand, then rebuild it when needed
    Rebuild |x| = x for x ≥ 0 and |x| = −x for x < 0 from distance, because the negative address needs its direction removed.
    Rebuild the corner and arm slopes of a|x − h| + k from x = h and a one-unit step, because zero inside the bars marks the turn.
    Rebuild shifts, reflection, horizontal scaling, domain, and range from the parent V, because these describe movements or changes in its two arms.
    Rebuild an equation from a corner and a second point by substitution, because the point supplies the missing stretch factor a.
    Rebuild zeros and positive or negative regions from boundaries and a graph or test point, because outputs can change sign only through a zero.
    Do not memorize each finished graph or individual answer. These can be rebuilt from the picture and a few arithmetic steps.
  • put on the cheat sheet
    Keep the V formula, the equation cases, the between and outside cases, and endpoint notation together, because these are the short lines you compare during practice.
    Keep the zero and negative limit cases beside them, because the positive-limit shortcuts do not cover those cases.
    Keep slope, percent tolerance, and the reciprocal of a negative fraction visible, because arithmetic slips can spoil an otherwise correct solution.
    Use the printable reference below as one page. For a closed-book exam, practice rebuilding it from memory rather than bringing it.
  • A small set of pictures explains the whole reference. Distance explains the cases. A corner and one step explain the V. Number-line order explains inequalities and endpoints. The compact sheet records the results so you can compare your reasoning during practice.

    One-page cheat sheet: absolute value functions
    Meaning: |x| is distance from 0. |x − c| is distance from c. Distance is always ≥ 0.
    Piecewise: |x| = x for x ≥ 0; |x| = −x for x < 0.
    Graph: f(x) = a|x − h| + k, a ≠ 0. Corner (h, k). Right arm slope a; left arm slope −a. Up if a > 0; down if a < 0. Domain (−∞, ∞). Range [k, ∞) if a > 0; (−∞, k] if a < 0.
    Scaling: |b(x − h)| = |b|·|x − h|. A nonzero inside factor b changes widths by 1 ÷ |b|.
    Equation: isolate |A| = B. For B > 0, A = B or A = −B. For B = 0, A = 0. For B < 0, no solution. Check the original.
    For B > 0: |A| < B means −B < A < B. |A| ≤ B means −B ≤ A ≤ B.
    For B > 0: |A| > B means A < −B or A > B. |A| ≥ B means A ≤ −B or A ≥ B.
    For B < 0: |A| < B and |A| ≤ B have no solution; |A| > B and |A| ≥ B hold for every allowed input.
    For B = 0: |A| < 0 has no solution; |A| ≤ 0 means A = 0; |A| > 0 means A ≠ 0; |A| ≥ 0 holds for every allowed input.
    Inequalities: add or subtract everywhere. Multiply or divide by a negative and reverse every sign.
    Intervals: (p, q) excludes both ends; [p, q] includes both. Mix symbols as needed. Infinity always takes a parenthesis. ∪ means union, combining allowed regions.
    Intercepts: set output y = 0 for horizontal intercepts; set input x = 0 for the vertical intercept.
    Slope = changeinychangeinx. Percent amount = percent100 × center.
    Arithmetic: divide by −12 by multiplying by −2. Fractions must have matching bottoms before addition or subtraction.
Distance: size without direction
Alone, then two roads
Less thAND, greatOR
A bracket grabs the endpoint
Keep a few memory devices ready, then rebuild the details from the pictures.
Reminder
  • Reference letters. A is the whole inside calculation; B is the isolated number on the other side. The V uses h for the corner's horizontal coordinate, k for its height, and a for the outside multiplier.
Worked exampleRebuild a formula instead of memorizing the drawing

A V has corner (2, 1) and rises 3 units for each unit you move away. What rule belongs on your practice sheet?

24246810cornerone right
The corner and one rise rebuild the whole V.
What it asks. Use the corner and steepness to rebuild the formula, then check one point.
Plan. Put h = 2 and k = 1 into the V formula, use a = 3, and substitute the corner and one nearby input.
  1. f(x) = 3|x − 2| + 1.The distance is zero at x = 2; multiplying by 3 sets the rise and adding 1 sets the corner height.
  2. f(2) = 3|0| + 1 = 1 and f(3) = 3|1| + 1 = 4.These checks recover the corner and a three-unit rise one unit to its right.
Answer
  • Formula: f(x) = 3|x − 2| + 1.
  • Corner: (2, 1).
Check At x = 1 the distance is also 1 and the output is 4, so both arms have the same steepness.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Memorize every completed graph and every worked answer.
That does not tell you how to handle new numbers on the exam.
✓ Instead: Know distance, isolation, and endpoint choices; rebuild each drawing from its corner and steepness.
Tips and tricks
  • Use this reference after the lessons, then cover it and rebuild each line for closed-book practice.
Strategy: step by step
  1. Get the absolute value alone. In an inequality, negative multiplication or division flips the sign. In an equation, equality remains equality.
  2. Inspect B, the number on the other side, after that isolation.
  3. If B is negative, equality or a smaller distance is impossible. A greater distance accepts every allowed input.
  4. If B is zero, decide whether the sign requires equality, excludes zero distance, or allows every nonnegative distance.
  5. For a required zero distance, solve A = 0 to find the center input. Substitute it back to confirm distance zero. For positive distance, exclude exactly the inputs where A = 0.
  6. If B is positive, use two equations for equality, between for a smaller distance, or outside for a greater distance. Check the answer in the original problem.
Strategy
Solving any absolute value problem
1
Are the bars alone on one side?
YesInspect B, the number on the other side.
NoUndo outside addition or subtraction, then multiplication or division. Negative multiplication or division flips an inequality. Equality remains equality.
↓
2
Is B negative?
YesEquality and a smaller distance have no solution, ∅. A greater distance gives every allowed input, here (−∞, ∞).
NoCompare B with zero.
↓
3
Is B zero?
YesFor equality or at most zero, solve A = 0. Strictly below zero has no solution. Strictly above zero excludes A = 0. At least zero accepts every allowed input.
NoB is positive. Go on.
↓
4
Is the sign equality?
YesSolve A = B or A = −B, check both, and report every input. For an inside line with x multiplied by a nonzero number, there are two distinct answers.
NoChoose between or outside.
↓
5
Is the sign less than, with or without equality?
YesUse −B < A < B, retaining equality when allowed. Change all three parts together, then write one interval.
NoUse A < −B or A > B, retaining equality when allowed. Solve each alternative and join the intervals with ∪.
  1. Isolate the bars and preserve the equation or comparison.
  2. Inspect the sign of B, the number on the other side.
  3. Choose the equation, between, outside, zero, or negative case from the decision map.
  4. Solve for every allowed input, then substitute into the original question.
  5. Write the requested notation, keeping individual answers as a set and intervals as intervals.
Worked exampleA negative coefficient hides a zero limit

Solve −2|x − 6| + 4 ≥ 4, and then solve −2|x − 6| + 4 > 4.

246810−4−224range(6, 4)
The graph reaches height 4 at one input and never rises above it.
What it asks. For the same function, first find inputs with output at least four. Then find inputs with output strictly greater than four.
Plan. Undo the outside addition and negative multiplier in each comparison, then inspect whether the distance can be at most zero or below zero.
  1. For the first condition, subtract 4: −2|x − 6| ≥ 0.This undoes the outside addition before inspecting the isolated distance.
  2. Divide by −2: |x − 6| ≤ 0.The negative divisor changes the at least comparison into an at most comparison.
  3. Set x − 6 = 0 and add 6: x = 6.A nonnegative distance at most zero must equal zero, so this finds the only center input.
  4. Substitute x = 6 into the original: −2|6 − 6| + 4 = −2 × 0 + 4 = 4.The result confirms that this input passes the first inclusive comparison.
  5. For the second condition, subtract 4: −2|x − 6| > 0.The outside addition is removed in the same way, but the comparison remains strict.
  6. Divide by −2: |x − 6| < 0.Negative division changes greater than into less than.
  7. There is no solution to the second condition.No distance can be below zero, so no input can make the original output exceed four.
Answer
  • For output at least 4: x = 6
  • Set for that answer: {6}
  • For output greater than 4: no solution
  • Set for that answer: ∅
Check The greatest output of the downward V is 4, attained only at its vertex (6, 4). It can reach 4 but can never exceed 4.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: impossible negative length

Solve |x| < −3.

−4−224−4−224
The V is never below the negative comparison line.
What it asks. Find whether any distance from zero can be smaller than negative three.
Plan. Compare every possible distance with its minimum value zero.
  1. |x| ≥ 0 for every real x.Distance cannot be negative.
  2. There is no solution.A value at least 0 cannot be smaller than −3.
Answer
  • No solution
  • Set: ∅
Check Even the smallest output |0| = 0 is too large.
Rung 2Rung 2: every distance beats a negative limit

Solve |2x − 1| ≥ −3.

(−∞, ∞)
Every real input passes this comparison.
What it asks. Find all inputs whose distance expression is at least negative three.
Plan. Use nonnegative distance instead of splitting into two branches.
  1. |2x − 1| ≥ 0 > −3.All values of the bars are nonnegative.
  2. Every real x satisfies the condition.The linear expression is defined at every real input.
Answer
(−∞, ∞).
Check At x = 12, 2x − 1 = 1 − 1 = 0, the smallest possible distance. Since 0 ≥ −3, this input passes; every larger distance also passes.
Rung 3Rung 3: exactly zero distance

Solve |2x − 1| ≤ 0.

−22246[[1|2]]
The graph reaches height zero only at the exact input 12.
What it asks. Find inputs making the distance at most zero.
Plan. Force the inside to zero, solve for x, and substitute back.
  1. Set 2x − 1 = 0.A nonnegative distance at most zero must equal zero. This equation locates the required center input.
  2. Add 1: 2x = 1. Divide by 2: x = 12.Undoing both inside operations isolates the one input with zero distance.
  3. Substitute: |2 × 12 − 1| = |1 − 1| = 0.The original comparison accepts this zero distance, confirming the answer.
Answer
  • x = 12
  • Set: {12}
Check The corner is at x = 12. Every other input gives a nonzero inside and a positive distance, which is not at most zero.
Rung 4Rung 4: isolate a hidden negative limit

Solve −12|3x + 2| + 1 < 3.

−4−22−4−224
The function's greatest output is 1 at x = −23, below the comparison height 3.
(−∞, ∞)
The original inequality passes at every real input.
What it asks. Find inputs for which the whole expression is smaller than three.
Plan. Isolate first, flip after multiplying by negative two, then inspect the negative B.
  1. Subtract 1: −12|3x + 2| < 2.Undo the outside addition before judging the limit.
  2. Multiply by −2: |3x + 2| > −4.The negative multiplier cancels the coefficient and flips the less than comparison into a greater than comparison.
  3. Every real input works.All distances are at least zero, which exceeds −4.
Answer
(−∞, ∞).
Check The original expression subtracts a nonnegative distance from 1, so it is always at most 1 and therefore smaller than 3. At x = −23, 3x + 2 = −2 + 2 = 0 and the original output is 1, confirming that even its greatest output passes.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: |x − 2| > −1 has no solution because the other side is negative.
A greater than comparison is different from equality. Every distance is at least zero, and zero is already greater than −1.
✓ Instead: Every real x works, so write (−∞, ∞).
✗ Not this: |x| < 0 gives x = 0.
|0| = 0 equals zero rather than being smaller than zero.
✓ Instead: There is no solution, written ∅.
✗ Not this: |x − 6| ≤ 0 gives x = 6 or x = −6.
The only permitted distance is zero. At x = −6, |−6 − 6| = 12, which fails.
✓ Instead: Solve x − 6 = 0, giving the single answer x = 6 and Set: {6}.
Tips and tricks
  • Inspect B after isolating, not before. A negative outside multiplier can change its sign.
  • Use the smallest possible distance, zero, to reason about a negative B.
  • For one answer, write a set containing that one number, such as {6}. For no answers, write ∅. For every real input, write (−∞, ∞).
Trap. Using the formulas for B positive before checking the isolated limit. |x − 6| > −2 is true for every real x, so it does not need a pair of boundary equations.