When B is zero or negative
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.
- 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.
the distance is at most zero, so the inside must be zero
Only the center has zero distance; every other position has positive distance.
- |x − 6| ≤ 0
- x − 6 = 0
- x = 6
- {6}
- |x − 6| > 0 means x ≠ 6
- (−∞, 6) ∪ (6, ∞)
- {x : x ≠ 6}
A leash of length zero keeps you at its post. Being a positive distance from the post means standing anywhere else.
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.
For |x − 2| > −1, try the smallest distance at x = 2: |0| = 0 > −1. Larger distances also pass, so every real input works.
|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.
| Comparison | Negative B | B = 0 |
|---|---|---|
| |A| < B | No solution | No solution |
| |A| ≤ B | No solution | Solve A = 0 |
| |A| > B | Every allowed input | All allowed inputs except A = 0 |
| |A| ≥ B | Every allowed input | Every 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.
- Comparing negative numbers. 0 > −1, so any number zero or greater is also greater than −1.
Solve |x − 2| ≤ −1 and |x − 2| > −1.
- No distance is negative.
- A greater than sign can pass even when a negative equality cannot.
- The first has no solution.Every absolute value is at least 0, and 0 is larger than −1.
- The second holds for every real x.Even the smallest possible distance 0 exceeds −1.
- For |x − 2| ≤ −1: no solution
- Set: ∅
- For |x − 2| > −1: every real x
- Interval: (−∞, ∞)
- 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.
- Nonzero. Nonzero means any number except 0. Its absolute value is positive.
Solve |x − 2| > 0.
- ≠ means not equal to.
- The excluded center separates two open rays.
- 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.
- Check |2 − 2| = 0.The proposed excluded input really has zero distance and fails the question.
- Keep every x ≠ 2, meaning x < 2 or x > 2.Every other real number has a nonzero inside and therefore positive absolute value.
- 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.
- One member set. {3} lists the single produced output 3; it does not mean an interval.
zero times the distance plus k
Every allowed input produces the same output k.
- f(x) = 0|x − h| + k = k
- Domain: (−∞, ∞)
- Range: {k}
A flattened pair of ramps becomes one level road.
For f(x) = 0|x − 7| + 3, find the domain, range, and zeros. What changes if the outside addition 3 is replaced by 0?
- Zero times any distance is zero.
- A constant graph reaches output zero only if its constant height is zero.
- Reduce f(x) = 0 × |x − 7| + 3 = 3.Every distance is multiplied by zero, leaving only the outside addition.
- The domain is every real number and the range is {3}.Every input is permitted, and every one produces the same single output 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.
- 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.
- For f(x) = 3: domain (−∞, ∞)
- Range: {3}
- Zeros: none
- For g(x) = 0: domain (−∞, ∞)
- Range: {0}
- Zeros: every real input
- 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 = . Percent amount = × center.
Arithmetic: divide by − by multiplying by −2. Fractions must have matching bottoms before addition or subtraction.
- 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.
A V has corner (2, 1) and rises 3 units for each unit you move away. What rule belongs on your practice sheet?
- Distance from 2 is |x − 2|.
- A height increase of 3 per unit means multiply that distance by 3.
- 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.
- 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.
- Formula: f(x) = 3|x − 2| + 1.
- Corner: (2, 1).
- Use this reference after the lessons, then cover it and rebuild each line for closed-book practice.
- Get the absolute value alone. In an inequality, negative multiplication or division flips the sign. In an equation, equality remains equality.
- Inspect B, the number on the other side, after that isolation.
- If B is negative, equality or a smaller distance is impossible. A greater distance accepts every allowed input.
- If B is zero, decide whether the sign requires equality, excludes zero distance, or allows every nonnegative distance.
- 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.
- 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.
Solving any absolute value problem
- Isolate the bars and preserve the equation or comparison.
- Inspect the sign of B, the number on the other side.
- Choose the equation, between, outside, zero, or negative case from the decision map.
- Solve for every allowed input, then substitute into the original question.
- Write the requested notation, keeping individual answers as a set and intervals as intervals.
Solve −2|x − 6| + 4 ≥ 4, and then solve −2|x − 6| + 4 > 4.
- Dividing by negative two reverses either comparison.
- At most zero allows distance zero.
- Below zero allows no distance.
- For the first condition, subtract 4: −2|x − 6| ≥ 0.This undoes the outside addition before inspecting the isolated distance.
- Divide by −2: |x − 6| ≤ 0.The negative divisor changes the at least comparison into an at most comparison.
- 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.
- 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.
- For the second condition, subtract 4: −2|x − 6| > 0.The outside addition is removed in the same way, but the comparison remains strict.
- Divide by −2: |x − 6| < 0.Negative division changes greater than into less than.
- There is no solution to the second condition.No distance can be below zero, so no input can make the original output exceed four.
- For output at least 4: x = 6
- Set for that answer: {6}
- For output greater than 4: no solution
- Set for that answer: ∅
Solve |x| < −3.
- The bars are already alone.
- No distance lies below zero.
- |x| ≥ 0 for every real x.Distance cannot be negative.
- There is no solution.A value at least 0 cannot be smaller than −3.
- No solution
- Set: ∅
Solve |2x − 1| ≥ −3.
- Zero is already greater than negative three.
- The inside expression is defined for every real input.
- |2x − 1| ≥ 0 > −3.All values of the bars are nonnegative.
- Every real x satisfies the condition.The linear expression is defined at every real input.
Solve |2x − 1| ≤ 0.
- The distance cannot be negative, so at most zero means exactly zero.
- Add one, then divide by two.
- Set 2x − 1 = 0.A nonnegative distance at most zero must equal zero. This equation locates the required center input.
- Add 1: 2x = 1. Divide by 2: x = .Undoing both inside operations isolates the one input with zero distance.
- Substitute: |2 × − 1| = |1 − 1| = 0.The original comparison accepts this zero distance, confirming the answer.
- x =
- Set: {}
Solve −|3x + 2| + 1 < 3.
- Subtract one before removing the negative half.
- Every distance is greater than negative four.
- Subtract 1: −|3x + 2| < 2.Undo the outside addition before judging the limit.
- Multiply by −2: |3x + 2| > −4.The negative multiplier cancels the coefficient and flips the less than comparison into a greater than comparison.
- Every real input works.All distances are at least zero, which exceeds −4.
- 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 (−∞, ∞).