On a road map, an intersection is where two routes meet. On a graph, an x-intercept, also called a horizontal intercept, is where the graph meets the x-axis, the horizontal line at height zero. A zero is the input that makes that happen. A y-intercept, or vertical intercept, is where the graph meets the y-axis, the vertical line with horizontal position zero. To find zeros, ask for output zero and solve for the unknown input. To find the y-intercept, the input is already zero, so calculate its output. For distance from a bus stop, the zero tells you the time you reach the stop. That is why this input can matter.
The horizontal-axis points give zero inputs −2 and 32, and the vertical-axis point is (0, −6).
Reminder
A and B. In |4x + 1| = 7, A names the whole inside 4x + 1 and B is 7. Solve A = 7 or A = −7.
Coordinates. A point (x, y) lists input first and output second. (2, 0) is on the x-axis; (0, 2) is on the y-axis.
Negative substitution. For x = −2, write 4(−2) + 1 = −8 + 1 = −7 before taking absolute value.
Exact fractions.64 = 32 because dividing top and bottom by 2 keeps the same number.
Equation balance. Dividing −3u = −6 by −3 gives u = 2. Equality stays equality.
An upward V above zero has no x-intercepts.An upward V with corner height zero has one x-intercept.An upward V with corner below zero has two x-intercepts.A downward V with corner above zero has two x-intercepts.A downward V with corner below zero has no x-intercepts.
Why it works. For a|x − h| + k with a ≠ 0, zero height requires a|x − h| = −k, then |x − h| = −k ÷ a. Opposite signs for a and k make that distance positive, giving two crossings. If k = 0, the distance is zero, giving one. Matching nonzero signs make it negative, giving none. Every real input works in the formula, including zero, so there is always one y-intercept. A zero is a number, such as 2. Its x-intercept is the point (2, 0). Distinguishing the number from the point keeps your answer in the requested form.
RuleZeros: solve f(x) = 0 to find the inputs whose output is zero. Their x-intercepts are (x, 0). y-intercept: evaluate f(0) and write (0, f(0)). In a|x − h| + k with a ≠ 0: opposite signs for a and k give two x-intercepts; k = 0 gives one; the same nonzero sign gives none. The full V always has exactly one y-intercept.
The same idea, five ways
Say it
Say: a zero is an input that gives output zero. An x-intercept is its point on the horizontal axis. The y-intercept is the point at input zero.
Write it
Finding an x-intercept means finding where the graph reaches height zero; finding a y-intercept means finding its height directly over input zero.
In math
Zero: f(x) = 0
x-intercept: (x, 0)
y-intercept: (0, f(0))
For f(x) = |4x + 1| − 7, zero set {−2, 32}
x-intercepts: (−2, 0) and (32, 0)
y-intercept: (0, −6)
Like
Find where a road meets ground level, or where it crosses one fixed road running north and south. Those are different map questions.
See itThe horizontal-axis points give zero inputs −2 and 32, and the vertical-axis point is (0, −6).
The same idea, other ways
As ground level
Every point on the horizontal axis has height zero. A zero input is where your road reaches that height. A bus-stop distance |t − 5| has zero t = 5 because that is the minute you reach the stop.
The distance reaches zero at time 5, when you are at the stop.
As a table search
For f(x) = |4x + 1| − 7, look for zero in the output row. The column under −2 and the column under 32 have output 0, so those inputs are zeros. The column under 0 has output −6, giving the y-intercept (0, −6).
The two zero outputs point to zero inputs −2 and 32.
Without drawing
In 2|x − 5| − 6, a = 2 is positive and k = −6 is negative, so the lowest point is below zero and both arms rise through zero. In −2|x + 6| − 1, both a and k are negative, so its highest point is below zero and no arm reaches zero.
Opposite signs: two crossings
k = 0: one crossing
Same nonzero sign: no crossing
The corner height and opening direction determine how many horizontal-axis crossings exist.
Graph
Position relative to height 0
Number of x-intercepts
|x| + 2
Entirely above
0
|x|
Corner at zero
1
|x| − 3
Corner below, arms rise above
2
−|x| + 2
Corner above, arms drop below
2
−|x| − 1
Entirely below
0
.1Horizontal intercepts
Look for the inputs where the road reaches ground level. Each such input is a zero.
x-intercept and horizontal intercept mean the same point.
The point has coordinates (x, 0).
Both labeled points reach ground level y = 0.
Worked exampleTwo ground-level positions
Find the x-intercepts of |x| − 1.
Both labeled points reach ground level y = 0.
What it asks. Find the points where this V reaches height zero.
Plan. Set the output equal to zero, isolate the bars, solve both inside cases, and attach y-coordinate 0.
The input is unknown.
The intercept's output is fixed at zero.
Set the output equal to zero: |x| − 1 = 0.This asks which inputs place the graph on the horizontal axis.
Add 1 to both sides: |x| = 1.This isolates the distance and reveals positive B = 1.
Write x = −1 or x = 1, then check |−1| − 1 = 0 and |1| − 1 = 0.Both inside values have distance 1, and both inputs give zero in the original formula.
Write (−1, 0) and (1, 0).An intercept is a point, so include its output coordinate 0.
Answer
Zero inputs: x = −1 and x = 1
Zero set: {−1, 1}
x-intercepts: (−1, 0) and (1, 0)
Check Each gives 1 − 1 = 0.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Report x = 0 as the zero of |x| − 1.
At input 0 the output is |0| − 1 = −1, not zero.
✓ Instead: Solve |x| − 1 = 0 to get zero inputs −1 and 1. Their intercept points are (−1, 0) and (1, 0).
Tips and tricks
Write y-coordinate 0 in every x-intercept point.
.2Vertical intercept
Stand on the vertical axis, where the horizontal address is zero. Read the height of the graph there.
y-intercept and vertical intercept mean the same point.
The point has coordinates (0, f(0)).
The graph meets the vertical axis at height 3.
Worked exampleThe vertical-axis height
Find the y-intercept of |x − 2| + 1.
The graph meets the vertical axis at height 3.
What it asks. Find the graph point whose horizontal input coordinate is zero.
Plan. Put input zero into the original formula, compute the output, and write the two coordinates.
Every point on the y-axis has x = 0.
There is no equation to solve for x here; x is already given.
Use input 0: f(0) = |0 − 2| + 1.This finds the height over the vertical axis.
Compute 0 − 2 = −2, then |−2| = 2.Evaluate the entire inside before taking its distance.
Add 1: f(0) = 2 + 1 = 3. Write (0, 3).The output is 3 at input 0, and an intercept answer is a point.
Answer
y-intercept: (0, 3).
Check The vertex is (2, 1), and moving two units left raises the height by 2 to 3.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Set |x − 2| + 1 = 0 to find its y-intercept.
That searches for output zero and therefore asks for x-intercepts. It does not force input zero.
✓ Instead: Use x = 0: |0 − 2| + 1 = 3, so the y-intercept is (0, 3).
Tips and tricks
Write x = 0 before calculating a y-intercept so the two intercept tasks stay distinct.
.3Count x-intercepts without graphing
Imagine the corner of a V as a ramp's lowest or highest landing. Whether that landing lies above or below ground, and whether the ramps rise or fall away from it, determines how many times the road can reach ground level.
For a|x − h| + k with a ≠ 0, a sets opening direction and k sets corner height.
a and k with opposite signs give two x-intercepts. k = 0 gives one. Matching nonzero signs give none.
Algebra checks the same count: solving output 0 gives a distance −k ÷ a. Positive gives two, zero gives one, negative gives none.
Opposite signs, a = 2 and k = −6, place two crossings around a corner below ground.
Worked exampleA positive a and negative corner height
Without graphing, decide how many x-intercepts f(x) = 2|x − 5| − 6 has. Then find them.
The algebra finds the two zero-height points shown.
What it asks. Count the graph's zero-height points from its formula, then find their exact coordinates.
Plan. Compare a and k, then solve output zero and check the two inputs.
a = 2 and k = −6 have opposite signs.
The required distance after isolation is 3.
Read a = 2 > 0 and k = −6 < 0. There are two x-intercepts.The corner is below zero, and the two arms rise without stopping, so each arm crosses zero once.
Set the output to zero: 2|x − 5| − 6 = 0.This asks for the inputs where the graph reaches the horizontal axis.
Add 6 to both sides: 2|x − 5| = 6.This removes the outside subtraction while keeping the two sides equal.
Divide both sides by 2: |x − 5| = 3.This isolates the distance and reveals positive B = 3.
Write x − 5 = 3 or x − 5 = −3.Both whole inside values have absolute value 3.
Add 5 in each equation: x = 3 + 5 = 8 or x = −3 + 5 = 2.This finds the two zero inputs. Check them in the original formula before writing the intercept points.
Answer
Number of x-intercepts: 2
Zeros: x = 2 and x = 8
Zero set: {2, 8}
x-intercepts: (2, 0) and (8, 0)
Check f(2) = 2|2 − 5| − 6 = 2 × 3 − 6 = 0. f(8) = 2|8 − 5| − 6 = 0. Both inputs lie three units from center 5, as the distance equation requires.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: For −2|x + 6| − 1, assume a negative a guarantees two crossings.
Its corner height is already −1 and its arms drop from there, so every output is negative.
✓ Instead: Both a and k are negative, so there are no x-intercepts.
Tips and tricks
Inspect a and k together. Opening direction alone does not decide the crossing count.
Strategy: step by step
For zeros or x-intercepts, set the output f(x) equal to 0. This asks which inputs place the graph on the horizontal axis.
Isolate the bars. Inspect B, the number on the other side, to determine whether two, one, or no inside equations are possible.
Solve each possible equation to find the zero inputs. Substitute each input into the original formula to confirm output 0.
List zeros as numbers or an answer set. List x-intercepts as points with y-coordinate 0.
For the y-intercept, substitute x = 0 into the original formula and calculate the output. Write the point (0, f(0)).
Strategy
Zeros, intercepts, and how many crossings
1
Does the question ask for the y-intercept?
YesSubstitute x = 0, calculate f(0), and report (0, f(0)).
NoFor zeros or x-intercepts, set f(x) = 0 and isolate the bars.
↓
2
After isolation, is B negative?
YesNo zero inputs or x-intercepts: a distance cannot be negative.
NoCheck whether B is zero.
↓
3
Is B zero?
YesSet the inside equal to 0 to find the one zero input, then substitute it back.
NoB is positive. Solve the two whole-inside equations, check both inputs, and report zero values or intercept points as requested.
Read whether the question asks for inputs, points, or a count of crossings.
For zeros or horizontal intercepts, require output 0, isolate the bars, and solve the possible inside equations.
For the vertical intercept, use input 0 and evaluate the original formula.
For a crossing count without graphing, inspect the signs of a and k in a|x − h| + k, or inspect B after isolation.
Worked exampleFind zeros and intercepts
Find the zeros and both kinds of axis intercepts of f(x) = |4x + 1| − 7.
The horizontal-axis points give zero inputs −2 and 32, and the vertical-axis point is (0, −6).
What it asks. Zeros: the output is 0, so find every input that gives it. x-intercepts: write those zero-input points. y-intercept: the input is 0, so calculate the output.
Plan. Set the function equal to zero and solve the two inside equations. Then separately put input zero into the original function.
Output zero and input zero are different instructions.
Add 7 before splitting the inside.
Use parentheses around a negative input during substitution.
Set the output equal to 0: |4x + 1| − 7 = 0.This finds the inputs whose graph points lie on the horizontal axis.
Add 7 to both sides: |4x + 1| = 7.This removes the outside subtraction and reveals positive B = 7.
Write 4x + 1 = 7 or 4x + 1 = −7.The complete inside can be either of the two numbers with absolute value 7.
In the first equation subtract 1: 4x = 6. Divide by 4: x = 64 = 32.These steps find the input whose inside is 7. Dividing the fraction's top and bottom by 2 reduces it.
In the second equation subtract 1: 4x = −8. Divide by 4: x = −2.These steps find the input whose inside is −7.
Check both zero inputs in the original formula and write points (−2, 0) and (32, 0).Both substitutions give output 0, so both points are x-intercepts.
For the y-intercept use input 0: f(0) = |4 × 0 + 1| − 7 = |1| − 7 = 1 − 7 = −6.Points on the vertical axis have x-coordinate 0. This calculation finds the height there.
Answer
Zeros: x = −2 and x = 32
Zero set: {−2, 32}
x-intercepts: (−2, 0) and (32, 0)
y-intercept: (0, −6)
Check f(−2) = |4(−2) + 1| − 7 = |−8 + 1| − 7 = |−7| − 7 = 0. f(32) = |4 × 32 + 1| − 7 = |6 + 1| − 7 = 0. For the corner, 4x + 1 = 0 finds where the inside vanishes: subtract 1 and divide by 4 to get x = −14. Substituting gives f(−14) = |−1 + 1| − 7 = −7. The corner is below zero and the arms rise, confirming two crossings.
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: one touching point
Find all intercepts of f(x) = |x − 3|.
The corner is the one x-intercept, and input zero gives the separate y-intercept.
What it asks. Find every point where the V meets either coordinate axis.
Plan. Require output zero for the x-intercept; put in input zero for the y-intercept.
A zero-distance V has one horizontal touching point.
Input zero is a separate substitution.
Set f(x) = 0: |x − 3| = 0. Set the inside x − 3 = 0.This finds where the graph has height zero. Only inside zero has distance zero.
Add 3 to both sides: x = 3. Check f(3) = |3 − 3| = 0.This finds and verifies the one zero input.
For the y-intercept use input 0: f(0) = |0 − 3| = |−3| = 3.This calculates the height on the vertical axis.
Answer
Zero: x = 3
Zero set: {3}
x-intercept: (3, 0)
y-intercept: (0, 3)
Check The vertex lies at height zero, so there is exactly one horizontal touching point.
Rung 2Rung 2: two crossings
Find all intercepts of f(x) = |x + 1| − 2.
Two zero inputs give two x-intercepts, while input zero gives the y-intercept below the axis.
What it asks. Find all zero inputs, their horizontal-axis points, and the vertical-axis point.
Plan. Solve the zero-output equation with both inside signs, then compute the output at input zero.
Add 2 before splitting the bars.
Subtract 1 to solve each inside equation.
Set the output equal to zero: |x + 1| − 2 = 0. Add 2: |x + 1| = 2.This asks which inputs put the graph on the x-axis and isolates the required distance.
Write x + 1 = 2 or x + 1 = −2.The positive distance 2 has two possible whole inside values.
Subtract 1 in each equation: x = 2 − 1 = 1 or x = −2 − 1 = −3.This finds both zero inputs.
For the y-intercept use x = 0: f(0) = |0 + 1| − 2 = 1 − 2 = −1.This finds the graph's height on the vertical axis.
Answer
Zeros: x = −3 and x = 1
Zero set: {−3, 1}
x-intercepts: (−3, 0) and (1, 0)
y-intercept: (0, −1)
Check f(−3) = |(−3) + 1| − 2 = |−2| − 2 = 0 and f(1) = |2| − 2 = 0. The vertex (−1, −2) is below zero and the arms rise, agreeing with two crossings.
Rung 3Rung 3: no crossing
Find all intercepts of f(x) = 2|x − 1| + 3.
The entire V stays above zero and still meets the y-axis at (0, 5).
What it asks. Determine whether the V reaches output zero, and find its output at input zero.
Plan. Set output zero and inspect the distance after isolation. Compute f(0) separately.
A negative isolated distance cannot occur.
The y-intercept still exists even if there are no x-intercepts.
Set output zero: 2|x − 1| + 3 = 0. Subtract 3: 2|x − 1| = −3.This asks which inputs reach the horizontal axis and removes the outside addition.
Divide by 2: |x − 1| = −32.This isolates the distance so its possibility can be checked.
There are no zero inputs or x-intercepts: ∅.A distance cannot equal −32. Every original output is at least 3.
Use x = 0 for the y-intercept: f(0) = 2|0 − 1| + 3 = 2 × 1 + 3 = 5.This calculates the height over the vertical axis, where the input is zero.
Answer
Zeros: none; zero set ∅
x-intercepts: none
y-intercept: (0, 5)
Check The minimum height is 3, so the entire graph stays above zero.
Rung 4Rung 4: reflected V and fractions
Find all intercepts of f(x) = −3|2x + 1| + 6.
This downward V crosses zero at −32 and 12 and meets the y-axis at height 3.
What it asks. Find the two possible zero inputs of this downward V and the height at input zero.
Plan. Solve output zero, dividing by the negative outside multiplier, then compute the y-intercept separately.
Dividing this equation by −3 keeps equality.
Fractional zero inputs are exact.
Do the absolute value before multiplying by −3.
Set output zero: −3|2x + 1| + 6 = 0. Subtract 6: −3|2x + 1| = −6.This finds the inputs whose graph points reach the horizontal axis.
Divide both sides by −3: |2x + 1| = 2.This isolates the distance. Equality stays equality under negative division.
In the first equation subtract 1: 2x = 1. Divide by 2: x = 12.This finds the zero input on the right.
In the second equation subtract 1: 2x = −3. Divide by 2: x = −32.This finds the zero input on the left.
For the y-intercept use input 0: f(0) = −3|2 × 0 + 1| + 6 = −3 × 1 + 6 = 3.This calculates the height on the vertical axis. The outside minus acts after the distance is found.
Answer
Zeros: x = −32 and x = 12
Zero set: {−32, 12}
x-intercepts: (−32, 0) and (12, 0)
y-intercept: (0, 3)
Check At −32, 2x + 1 = −3 + 1 = −2. At 12, 2x + 1 = 1 + 1 = 2. Both original outputs are −3 × 2 + 6 = 0. The corner has height 6 and the V opens down, agreeing with two crossings.
Rung 5Rung 5: which constant removes the crossings?
For which real c does g(x) = −4|x + 6| + c have no x-intercepts?
Choosing c = −1 keeps the whole downward V below zero.Choosing c = 0 creates one zero at the corner.Choosing c = 4 creates two zero-height points.The permitted constants are every negative c, with zero excluded.
What it asks. Find every choice of the constant c that keeps the graph from reaching height zero for any input x.
Plan. Set the output equal to zero, isolate the distance, and find when the number on the other side is negative.
c is a constant choosing the corner height, rather than an input x to solve for.
The largest output is c because the graph opens down.
Dividing by positive 4 keeps the sign of c.
Require zero height: −4|x + 6| + c = 0.This tests whether any input can place the graph on the horizontal axis.
Subtract c: −4|x + 6| = −c. Divide by −4: |x + 6| = c4.This isolates the distance. Its sign decides whether crossing inputs exist.
No x-intercepts occur when c4 < 0, which is exactly c < 0.A negative distance has no solution. Since 4 is positive, c and c ÷ 4 have the same sign.
Check the other choices: c = 0 gives x + 6 = 0, so x = −6; c > 0 gives two inside equations and two crossings.This shows that no zero or positive c belongs to the requested no-crossing choices.
Answer
c < 0
Interval for c: (−∞, 0)
Set-builder notation: {c : c < 0}
Check For c = −1, the largest output is −1 at x = −6, so no input reaches zero. For c = 0, g(−6) = 0 gives one crossing. For c = 4, g(−7) = g(−5) = −4 × 1 + 4 = 0 gives two. These checks confirm why precisely the negative choices give none.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: For f(x) = |x − 2|, say the zero is (0, 2).
Input 0 gives output |0 − 2| = 2, so (0, 2) is the y-intercept. A zero is an input number where the output becomes zero.
✓ Instead: The zero is x = 2, set {2}. Its x-intercept is (2, 0).
✗ Not this: Assume every V has two x-intercepts because it has two arms.
For 2|x − 1| + 3, every output is at least 3. Neither arm reaches height zero.
✓ Instead: Set output 0 and isolate to get |x − 1| = −32. Negative B means no x-intercepts.
For a|x − h| + k with a ≠ 0, opposite signs for a and k mean two crossings. k = 0 means one. Matching nonzero signs mean none. This gives the count without graphing.
On an exam, answer the requested object. A zero is an x-value. An intercept is a point with two coordinates.
Trap. Substituting x = 0 when asked for zeros. That finds the y-intercept. For zeros set the output to 0 and leave x unknown.
