A graph can tell you the recipe that made it. Think of inspecting two ramps at a building entrance. Their meeting point tells you where they begin and at what height. One other point tells you how steep they are. For a V, write a|x − h| + k using the corner's address (h, k). The missing number a is the stretch factor: height change per unit of distance from the corner. Read a point's horizontal position and height from the grid, and put both numbers into your recipe. This also recovers a distance model from measurements. A graph of distance from a bus stop can reveal where you pass the stop.
The point (1, 2) is two units sideways and four units above the corner, giving a = 2.
Reminder
Reading coordinates. Read the horizontal x position first and the vertical y height second. A point at x = 1, y = 2 is (1, 2).
Subtracting a negative center. x − (−1) = x + 1, so a corner at x = −1 uses |x + 1|.
Dividing by a fraction. 6 ÷ 32 = 6 × 23 = 4. The flipped fraction undoes multiplication by 32.
Why it works. At another point (p, q), |p − h| is its horizontal distance from the corner, and q − k is its height change. The recipe says q − k = a|p − h|, so dividing by that distance finds a. A point directly on the corner has distance zero and gives no steepness information. For an upward V, a horizontal cut D units above the corner meets each arm D ÷ a units from the center. The total width is 2D ÷ a. A narrower width at the same relative height therefore means a greater stretch factor.
RuleVertex form is f(x) = a|x − h| + k. From vertex (h, k) and another point (p, q) with p ≠ h: a = q − khorizontal distance, where horizontal distance = |p − h|. For an upward V, width at height D above the corner is 2D ÷ a, with D > 0. An x-intercept, or horizontal intercept, is a point (x, 0); its input is called a zero.
The same idea, five ways
Say it
Say find the formula from the corner and a point on the V.
Write it
The corner determines the two shifts; another point determines the stretch factor.
In math
f(x) = a|x − h| + k
q = a|p − h| + k
a = q − khorizontal distance
horizontal distance = |p − h|
f(x) = 2|x − 3| − 2
Like
Inspect a ramp's starting address and compare its rise with the distance walked.
See itThe corner and another point determine the formula of this V.
The same idea, other ways
Count the rise from the corner
Step 1 to the right of the corner. The change in height is a. From (3, −2) to (4, 0), the rise is 2, so the formula uses a = 2.
One horizontal unit from the corner reveals the stretch factor.
Put another point into the recipe
A point (1, 2) means input 1 must produce output 2. In a|x − 3| − 2, this gives 2 = 2a − 2. Add 2 and divide by 2 to obtain a = 2.
The given point supplies an equation for the unknown stretch factor.
Compare widths
At 4 above the corner, the parent V reaches 4 each side, giving width 8. This V reaches 2 each side, giving width 4. It reaches the same height in half the horizontal distance, so each unit must rise twice as much.
Height above corner: 4
Parent width: 8
Observed width: 4
Stretch factor: 8 ÷ 4 = 2
Compare widths at the same height above each corner.
.1Recover the bus-stop distance rule
A distance graph can be a set of measurements rather than a formula. When its corner says minute 5 and distance 0, that tells you when you passed the stop. If one minute later the distance is 1 block, the distance rises one block per minute away from that moment.
A graph with corner (5, 0) and height 1 at t = 6 gives d(t) = |t − 5|.
For this walking story, the time domain is t ≥ 0.
The corner and one-minute measurement recover the walking model.
Worked exampleA formula from distance measurements
The distance graph has corner (5, 0) and passes through (6, 1). Recover the distance formula for nonnegative times.
The recovered formula passes through both distance measurements.
What it asks. Write the distance rule from the turning time and pace shown by the graph.
Plan. Use the corner for the center, then divide the height change by the time difference.
Zero distance occurs at time 5.
One block per minute gives factor 1.
Write d(t) = a|t − 5|.The corner's time is 5 and its height is 0.
Insert (6, 1): 1 = a|6 − 5| = a × 1, so a = 1.The measured output is one block at a time one minute from the corner.
Write d(t) = |t − 5| for t ≥ 0.The model's measurements start at time zero.
Answer
d(t) = |t − 5|
Time domain: [0, ∞)
Check At t = 5, the distance is 0. At t = 6, it is 1. Both observed measurements fit.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: A corner at time 5 gives d(t) = |t + 5|.
That formula is zero at time −5, rather than at the observed time 5.
✓ Instead: Use d(t) = |t − 5|.
Tips and tricks
A recovered formula must reproduce the measured corner time and one other measurement.
Strategy: step by step
Read the corner from the grid: its horizontal position is h and its height is k.
Write f(x) = a|x − h| + k, leaving the stretch factor a unknown.
Read another clear point (p, q). Go to the x-axis to read p and to the y-axis to read q.
Subtract the corner height k from q and divide by the horizontal distance |p − h|. This finds the stretch factor a.
Put the resulting a into the formula and check it at a second point.
For a requested x-intercept, find where the drawing reaches height 0. Algebraically, set the output to 0 to find its input. A positive distance from h has one position on each side, h minus that distance and h plus it.
Strategy
Recover a V formula from a graph
1
Can you read the graph one unit right of the corner?
YesThe rise or fall from the corner is a.
NoUse another clear point and divide height change by horizontal distance.
↓
2
Is the selected point horizontally at the corner?
YesChoose a different point. Distance zero cannot determine steepness.
NoIts positive horizontal distance can be used in the division.
↓
3
Is the other point below the corner?
YesThe height change is negative, so a is negative and the V opens down.
NoA point above the corner gives positive a and upward arms.
Read the corner (h, k).
Write a|x − h| + k.
Read a different point and find its height change and horizontal distance from the corner.
Divide height change by distance to find a.
Check the formula at a second point.
Worked exampleRead a V from its picture
Write the equation of the V in the picture and find its x-intercepts.
Read the corner and another point from the grid before using the worked steps.
What it asks. Read the turning point and steepness from the grid, then identify where the graph has height zero.
Plan. Locate the corner, count a one-step rise, write the formula, check another point, and find ground-level positions.
The lowest point is the corner. Read horizontal position first and height second.
One step right from the corner reveals a when the rise is readable.
The lowest point reads (3, −2), so h = 3 and k = −2. Write f(x) = a|x − 3| − 2.The V turns where its distance part is zero; its grid address gives both shifts.
From the corner move 1 right to x = 4. The graph height is 0, which is 0 − (−2) = 2 above the corner. Thus a = 2 ÷ 1 = 2.The stretch factor measures height change per unit of distance from the corner.
Write f(x) = 2|x − 3| − 2.The corner and one-step height have supplied all three numbers in vertex form.
Check the point (1, 2): f(1) = 2|1 − 3| − 2 = 2 × 2 − 2 = 2.Another readable point must agree with the formula, including on the left arm.
Set the output to zero: 2|x − 3| − 2 = 0. Add 2, then divide by 2 to get |x − 3| = 1.A horizontal intercept has height zero. Undoing the outside operations identifies its required distance from 3.
A distance of 1 from 3 occurs at x = 3 − 1 = 2 or x = 3 + 1 = 4. Equivalently, x − 3 = −1 or x − 3 = 1.The distance picture places one input on each side of the center.
At height y = 2, the graph meets x = 1 and x = 5. Its width is 5 − 1 = 4, and that height is 2 − (−2) = 4 above the corner.For a second steepness check, compare widths at a fixed height above the corner. The parent V is 8 wide at height 4 above its corner, so this V is half as wide and twice as steep.
Answer
f(x) = 2|x − 3| − 2
Zeros: x = 2 and x = 4
x-intercepts: (2, 0) and (4, 0)
Check f(3) = 2|0| − 2 = −2 checks the corner. f(2) = f(4) = 2 × 1 − 2 = 0 checks both intercepts. Compared with |x|, the V moves right 3, stretches heights by 2, and moves down 2.
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: the stretch factor is given
An upward V has vertex (2, 1) and the same steepness as |x|. Write its formula.
The supplied steepness gives a rise of 1 per unit of distance.
What it asks. Find the formula of the V from its corner and the supplied information.
Plan. Use the corner to write vertex form, find the stretch factor from the other information, and check a point.
The corner supplies both shifts.
Divide the height change by horizontal distance, which is positive.
a = 1.The parent graph has one unit of rise per unit of distance.
Use h = 2 and k = 1 to write f(x) = |x − 2| + 1.These numbers place the corner at the stated address.
Answer
f(x) = |x − 2| + 1.
Check f(2) = 1 and f(3) = 2, so the corner and one-unit rise agree.
Rung 2Rung 2: one point reveals the stretch factor
Find a V with vertex (−1, 2) passing through (1, 8).
Two units from the corner raise the output by 6, so a = 3.
What it asks. Find the formula of the V from its corner and the supplied information.
Plan. Use the corner to write vertex form, find the stretch factor from the other information, and check a point.
The corner supplies both shifts.
Divide the height change by horizontal distance, which is positive.
f(x) = a|x + 1| + 2.The center is −1, and x − (−1) = x + 1. The corner height is 2.
Put (1, 8) into the formula: 8 = a|1 + 1| + 2 = 2a + 2.That point requires input 1 to give output 8, and the distance from −1 to 1 is 2.
Subtract 2 on both sides: 6 = 2a. Divide by 2 to get a = 3.Removing the corner height leaves the rise, and dividing by distance finds the stretch factor.
Answer
f(x) = 3|x + 1| + 2.
Check At x = 1, 3 × 2 + 2 = 8; at x = −1, the distance vanishes and the height is 2.
Rung 3Rung 3: the graph opens down
Find a V with vertex (4, 6) passing through (0, 4).
The given point is lower than the corner, so the multiplier is negative.
What it asks. Find the formula of the V from its corner and the supplied information.
Plan. Use the corner to write vertex form, find the stretch factor from the other information, and check a point.
The corner supplies both shifts.
Divide the height change by horizontal distance, which is positive.
f(x) = a|x − 4| + 6.The corner fixes center 4 and height 6.
Insert (0, 4): 4 = a|0 − 4| + 6 = 4a + 6.The given point lies four units horizontally from the corner.
Subtract 6: −2 = 4a. Divide by 4: a = −24 = −12.The height change is a fall of 2 over distance 4. Dividing those numbers finds the signed stretch factor.
Answer
f(x) = −12|x − 4| + 6.
Check At x = 0 the formula gives −2 + 6 = 4. A point away from the vertex is lower, so the negative outside multiplier makes sense.
Rung 4Rung 4: fractional coordinates
Find a V with vertex (12, −1) passing through (2, 5).
The height change 6 divided by distance three halves gives a = 4.
What it asks. Find the formula of the V from its corner and the supplied information.
Plan. Use the corner to write vertex form, find the stretch factor from the other information, and check a point.
The corner supplies both shifts.
Divide the height change by horizontal distance, which is positive.
f(x) = a|x − 12| − 1.The corner supplies center 12 and height −1.
Write 2 as 42, so |2 − 12| = |42 − 12| = 32.Matching fraction bottoms allows subtraction, and the positive result is the horizontal distance.
Insert (2, 5): 5 = a × 32 − 1. Add 1 to obtain 6 = a × 32.The point gives the output equation, and adding 1 removes the downward shift.
a = 6 ÷ 32 = 6 × 23 = 123 = 4.The reciprocal undoes multiplication by 32, isolating the stretch factor.
Answer
f(x) = 4|x − 12| − 1.
Check At x = 2 the output is 4 × 32 − 1 = 6 − 1 = 5. At the vertex it is −1.
Rung 5Rung 5: read a downward V from a picture
Read the downward V in the picture and write its formula.
Read the highest point and a one-step fall from this unlabeled graph.
What it asks. Recover the center, highest height, and signed stretch factor directly from the grid.
Plan. Read the corner and one step to its right. Check a point on the other arm.
The corner is the highest point for a downward V.
A fall gives a negative stretch factor.
The highest point reads (−2, 4), so write f(x) = a|x + 2| + 4.The corner gives h = −2, and x − (−2) = x + 2.
One step right reaches (−1, 1). The height change is 1 − 4 = −3 over distance 1, so a = −3.A downward change is negative, while horizontal distance is positive.
Write f(x) = −3|x + 2| + 4.The corner and one-step fall determine the V.
Check the left point (−3, 1): f(−3) = −3|−3 + 2| + 4 = −3 × 1 + 4 = 1.The recovered formula must fit the other arm too.
Answer
f(x) = −3|x + 2| + 4
Check At x = −2, output is 4. One step left or right lowers it to 1, matching the grid and its downward opening.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Use (2 − (−2)) ÷ (1 − 3) = −2 for the V through (1, 2) with corner (3, −2).
The denominator must be distance, which is positive, even for a point left of the corner. The wrong formula gives −2|1 − 3| − 2 = −6, missing height 2.
✓ Instead: a = 4 ÷ |1 − 3| = 4 ÷ 2 = 2.
Tips and tricks
From the corner, one step right gives a directly if that point is readable.
A second point checks the entire formula. Always choose a point away from the corner to find a.
For an upward V, compare widths at the same height above each corner, rather than at the same y-coordinate when corner heights differ.
Trap. Dividing by p − h instead of |p − h|. A point left of the corner has a negative coordinate difference, but its distance is positive. Keep the bars so a correctly records whether the V opens up or down.
