Picture your distance from a bus stop as you walk toward it and keep going. The stop is 5 blocks ahead, and you walk 1 block each minute. Your distance after t minutes is |t − 5|. It decreases until minute 5, then increases. Draw time across the page and distance upward, and you get two straight ramps meeting in a V. That meeting point is the corner point, or vertex. The plain y = |x| is the parent function, the shape before any changes. Moving its corner shifts the V. Multiplying its distance heights changes its steepness. Taking their opposites makes both arms go down.
The solid V has corner (3, 4); the dashed V is the unchanged parent function.
Reminder
Slope. Slope = rise ÷ run. A rise of 2 for 1 step right gives slope 2; a fall of 2 gives slope −2.
Coordinates and graph shadows. (x, y) records horizontal position first, height second. Domain is the shadow on the x-axis; range is the shadow on the y-axis.
Substitution. Keep a negative input together: at x = −2, 2x + 2 means 2(−2) + 2 = −4 + 2 = −2.
Why it works. Start with the tables below. Subtracting 3 puts zero distance under x = 3. Multiplying those distances by 2 doubles their heights. Adding 4 raises every height by 4. Thus 2|x − 3| + 4 has corner (3, 4). The parent rule draws y = −x on the left of zero and y = x on the right. For a|x − h| + k, zero distance occurs at h and leaves height k. Each unit farther from h changes height by a. A negative a makes that change downward.
RuleFor f(x) = a|x − h| + k with a ≠ 0: a is the number multiplying the distance, h is its horizontal center, and k is the outside added height. Vertex: (h, k). Left slope: −a; right slope: a. It opens up if a > 0 and down if a < 0. Domain: all real numbers. Range: y ≥ k if a > 0; y ≤ k if a < 0.
The same idea, five ways
Say it
Say a times the absolute value of x minus h, plus k.
Write it
For a ≠ 0, the formula draws a V whose corner is (h, k). Positive a makes the arms rise away from it; negative a makes them fall.
In math
f(x) = a|x − h| + k
f(x) = 2|x − 3| + 4
Vertex: (3, 4)
Domain: (−∞, ∞)
Range: [4, ∞)
{y : y ≥ 4}
Like
Place two ramps at a meeting point, then choose how much each step changes their height.
See itThe corner is where the distance part is zero.
The same idea, other ways
As two ramps
One straight arm goes toward the corner from the left; the other goes away to the right. Both are straight because each unit of horizontal distance changes the height by the same amount.
The parent function joins two straight arms at zero.
As three table stages
The zero moves under input 3 first. Then double the distance row. Then raise it by 4. Follow the column under 1: distance 2, doubled distance 4, final height 8.
First stage: subtracting 3 locates the zero distance under 3.
As a distance machine
For 2|x − 3| + 4, subtract 3, measure the distance, multiply by 2, then add 4. At x = 4 the machine gives 2 × |1| + 4 = 6.
The distance is doubled before the height 4 is added.
Why the corner must be there
At the center h, the distance is zero. Multiplying that zero gives zero, and adding k gives k. One unit either side has the same distance 1, so both heights are k + a. This explains the matching arms.
f(h) = a × 0 + k = k
f(h − 1) = a + k
f(h + 1) = a + k
The corner and matching one-step heights follow from distance.
.1Parent function
A toolkit function is a basic shape you reuse. The parent function is the unchanged y = |x|: two matching ramps meeting at the origin, the point (0, 0). Moving one unit farther from zero adds one unit to the height.
f(x) = |x|.
Vertex: (0, 0). Slopes: −1 on the left and 1 on the right.
Domain: (−∞, ∞). Range: [0, ∞).
The parent V has its lowest point at the origin.
Worked exampleThree parent points
Find the outputs at x = −1, 0, 1.
The output under each input supplies its plotted height.The table columns place the corner and one point on each arm.
What it asks. Find the heights corresponding to inputs −1, 0, and 1.
Plan. Measure each input's distance from zero and read the table columns.
Opposite inputs have matching distances.
The column under 0 has output 0.
f(−1) = |−1| = 1 and f(1) = |1| = 1.Both positions are one unit from zero.
f(0) = |0| = 0.Home has distance zero.
Answer
The output table gives the three plotted points.
Vertex: (0, 0)
Check The outputs at −1 and 1 match because they are equally far from zero. This is symmetry: matching sides across the vertical line x = 0.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The parent V has negative outputs on its left arm.
Negative inputs describe position, while outputs describe distance.
✓ Instead: For x = −1, the output is |−1| = 1.
Tips and tricks
Keep the original parent shape in mind before describing a shift or stretch.
.2Horizontal shift
Changing the home address moves the V sideways. Find the address that makes the inside zero.
|x − h| shifts right h when h > 0.
|x + c| shifts left c when c > 0.
Adding 2 inside moves the corner to x = −2.
Worked exampleA left shift
Where is the vertex of |x + 2|?
The inside addition puts the corner at negative 2.
What it asks. Find the horizontal position and height where this V turns.
Plan. Set the inside to zero, solve for the input, and substitute it.
x + 2 is x minus negative 2.
No number outside raises or lowers the corner.
Set x + 2 = 0 and subtract 2 to find x = −2.This locates the input where the distance is zero and the V changes direction.
|−2 + 2| = |0| = 0, so the vertex is (−2, 0).Substitution finds the height at the turning input.
Answer
Vertex: (−2, 0).
Check Inputs −3 and −1 both give 1 and surround −2 equally.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: |x + 2| shifts the V right 2.
At positive 2 the inside is 4, while at −2 it is zero.
✓ Instead: The corner shifts left to (−2, 0).
Tips and tricks
Find the inside zero instead of guessing from its sign.
.3Vertical shift
Adding the same height to every point raises the entire road. Subtracting lowers it.
|x| + k shifts up k if k > 0.
The added number is also the vertex height.
Subtracting 3 lowers every output by three units.
Worked exampleA downward shift
Find the vertex of |x| − 3.
The subtraction lowers the corner by three units.
What it asks. Find where the lowered V turns.
Plan. Use the parent corner input 0 and compute its new height.
The inside still becomes zero at x = 0.
Subtract 3 after taking absolute value.
At x = 0, |0| − 3 = −3.Zero remains the position of minimum distance, meaning the smallest possible distance. The outside subtraction changes the height, rather than the input where the distance is zero.
The vertex is (0, −3).Every height of the parent V was lowered by 3.
Answer
Vertex: (0, −3).
Check At x = 1 the output is −2, exactly three below the parent output 1.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: |x| − 3 moves the corner left 3.
The subtraction acts on the completed output rather than changing the input inside.
✓ Instead: It moves the corner down to (0, −3).
Tips and tricks
An outside addition or subtraction changes every output by the same amount.
.4Vertical stretch and compression
Multiplying distance heights changes the ramp's steepness. A factor above 1 makes the V narrower; a factor between 0 and 1 makes it wider.
|a| > 1 gives a vertical stretch.
0 < |a| < 1 gives a vertical compression.
The 2 in 2|x − 3| + 4 doubles the distance part alone: at x = 4, 2 × 1 + 4 = 6, rather than 2 × (1 + 4) = 10.
At the same horizontal distance, the factor 2 V rises more than the factor one-half V.
Worked exampleSame input, different steepness
At x = 2, compare 2|x| with 12|x|.
At input 2, the steeper V reaches 4 and the wider V reaches 1.
What it asks. Find and compare two outputs at the same input 2.
Plan. Take the same distance |2|, then apply each multiplier separately.
Check The parent output 2 lies between 1 and 4, as the factors predict.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The outside factor doubles an added height shift too.
In 2|x − 3| + 4, multiplication comes before adding 4.
✓ Instead: At x = 4 the output is 6, with distance height 2 and added height 4.
Tips and tricks
Compare heights above or below the corner, rather than multiplying the final corner height.
.5Reflection
A vertical reflection flips the distance heights upside down across the horizontal x-axis. A minus sign in front of the bars takes the opposite of each height. After that flip, adding 3 raises every point by 3. The order matters because flipping after a lift would also flip the added height.
a < 0 reflects the distance part across the x-axis before the vertical shift.
For −|x − h| + k, the vertex is a maximum, the greatest output.
Vertically reflected means upside down across the x-axis, rather than a sideways reflection.
The distance heights are reflected across the x-axis and then raised 3.
Worked exampleLeft two, upside down, then up three
Shift |x| left 2, reflect vertically, and shift up 3.
The resulting V has its highest point at (−2, 3).
What it asks. Write the formula obtained by performing the three changes in the stated order.
Plan. Move the distance center, take opposite distance heights, then add the upward shift.
A left shift 2 makes the center −2.
A vertical reflection puts a minus sign in front of the bars.
Left 2 gives |x + 2|.The inside becomes zero at −2.
Reflection gives −|x + 2|, and raising 3 gives −|x + 2| + 3.Taking the opposite turns the distance heights downward. Adding 3 afterward raises every resulting height.
Answer
f(x) = −|x + 2| + 3.
Check f(−2) = 3 and f(−3) = f(−1) = 2, so the corner is at (−2, 3) and opens down.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: −(|x + 2| + 3) performs the requested reflection followed by an upward shift 3.
The outside minus also reverses the added 3, producing a downward shift instead.
✓ Instead: Reflect first, then raise: −|x + 2| + 3, with corner (−2, 3). The wrong order gives −|x + 2| − 3 with corner (−2, −3).
Tips and tricks
Keep the upward shift outside the reflected bars when the instruction says reflect, then shift up.
.6Horizontal scaling
A number multiplying the whole inside distance can be moved outside as its positive size. Multiplying by −2 reverses a trip and doubles its length; the bars keep the doubled length. The product rule allows this for multiplication. It does not allow splitting an addition or subtraction into separate absolute values. Here b is the number multiplying the entire inside, and h is still its center.
For real numbers u and v, |uv| = |u| × |v|: multiplication multiplies lengths while the bars remove direction.
|b(x − h)| = |b| × |x − h|.
For b ≠ 0, horizontal widths are multiplied by 1 ÷ |b|. If |b| > 1 this compresses them; if 0 < |b| < 1 this stretches them.
An inside negative reverses left and right; the absolute value graph has matching sides, so the shape is unchanged.
Doubling the inside makes a V rise twice as fast as the parent V.
Reminder
Factoring a number. 2x − 6 = 2(x − 3), since 2 × x = 2x and 2 × (−3) = −6. Multiply back to check.
Worked exampleEquivalent scalings
Rewrite |−2(x − 3)| − 2 with the multiplying number outside the bars. Also compare |2x − 6| − 2 with 2|x − 3| − 2.
The inside-factor formula gives the same output as the next table.Factoring and the product rule preserve every column's output.All three equal formulas trace this one V.
What it asks. Show which inside multiplication can be moved outside, and verify two equal formulas with their output table.
Plan. Use the length product rule, keep the outside subtraction, and factor 2x − 6 before applying the same rule.
The positive size of −2 is 2.
Products may move outside; sums cannot be split.
|−2(x − 3)| = |−2| × |x − 3| = 2|x − 3|.Multiplying a trip by −2 doubles its length and reverses direction. The bars retain the length 2 times the original distance.
Thus |−2(x − 3)| − 2 = 2|x − 3| − 2.The subtraction outside the bars stays after the distance multiplication.
Factor 2x − 6 = 2(x − 3). Then |2x − 6| − 2 = 2|x − 3| − 2 too.Distributing the 2 checks the factoring, and the product rule removes its positive factor.
In the column under 1, both formulas give 2; under 3, both give −2. Every remaining column matches too.Factoring and the product rule preserve the result for every input, and these columns check its numerical effect.
Answer
|−2(x − 3)| − 2 = 2|x − 3| − 2.
Check At x = 1, |−2(1 − 3)| − 2 = |4| − 2 = 2 and 2|1 − 3| − 2 = 2 × 2 − 2 = 2. The table also confirms the factored positive-inside formula has the same heights.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Because products can be split, |−2 + 3| = |−2| + |3|.
The product rule describes multiplication. Adding two numbers can cancel their directions, changing the resulting distance.
✓ Instead: |−2 + 3| = 1, while |−2| + |3| = 5. For a product, |(−2)(3)| = 6 = |−2| × |3|.
Tips and tricks
Factor multiplication before moving a number outside. Keep sums together.
.7Build the transformed output table
Treat the formula as a short measuring job. The first row finds distance from the new center. The next row doubles each distance. The last row adds the new corner height. Follow the same input column through these stages so you can see why the corner moved sideways and upward.
For inputs 1 through 5, subtracting 3 centers the zero distance at input 3.
Multiplying the distance heights by 2 changes steepness.
Adding 4 to each result raises the entire graph.
Stage one puts zero distance under input 3.
Worked exampleThree stages of 2|x − 3| + 4
Build the distance, doubled distance, and final output tables for inputs 1 through 5.
First measure distance from 3.Then double those distance heights.Finally add 4 to every height.
What it asks. See exactly what each operation changes in the table.
Plan. Evaluate the distance first, double its outputs, and add 4 last.
The zero distance is under 3.
The doubling acts before the addition.
At x = 1 and 5, |x − 3| = 2. At x = 2 and 4 it is 1. At x = 3 it is 0.These are the inputs' distances from 3, shown in the first table.
Double every distance output to fill the second table.The outside multiplier 2 acts on the bars alone.
Add 4 to every doubled output to fill the third table.The outside addition lifts each height equally.
Answer
The final table has corner height 4 under input 3.
Check In the column under 1, the three stages give 2, then 4, then 8. Direct substitution gives 2|1 − 3| + 4 = 8 too.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Adding 4 moves the corner input from 3 to 7.
The addition changes output heights after the distance is measured.
✓ Instead: The corner input stays 3, while its output rises from 0 to 4.
Tips and tricks
Follow the same column through all three stages.
.8Range on a limited input stretch
A viewing window is the stretch of inputs a calculator screen shows. Imagine looking at only part of a long road through a window. The full V may rise forever, while the visible part reaches only certain heights. To find the range for a stated input stretch, check both ends and the corner if it is inside that stretch.
A viewing window [a, b] shows inputs from a to b. A stated restricted domain uses only those inputs.
For an upward V, the lowest output is at the corner if the corner lies inside the stretch; otherwise it is at the end nearest the corner.
For an upward V, the highest output over a closed limited stretch is at the end farther from the corner.
For a downward V, greatest and smallest switch roles.
On inputs 3 through 6, the V reaches heights from 0 through 4.
Worked exampleA range inside a viewing window
Find the range of f(x) = 2|x − 5| when 3 ≤ x ≤ 6.
The corner column shows the minimum, and the left end shows the maximum.Only the stated input stretch is used to find these output limits.
What it asks. Find every output reached while inputs stay in the stated stretch.
Plan. Check the left end, the included corner, and the right end, then compare their heights.
The corner input 5 lies between 3 and 6.
The greatest distance from 5 occurs at the left end 3.
Set x − 5 = 0 to find the turning input x = 5. It lies in [3, 6], and f(5) = 2|0| = 0.This checks whether the lowest point of the full V belongs to the allowed stretch.
f(3) = 2|3 − 5| = 4; f(6) = 2|6 − 5| = 2.Each straight arm has its other extreme at the permitted endpoint.
The smallest output is 0 and the largest is 4, so write [0, 4].The connected arms reach every height between those two, and both extreme points are included.
Answer
Domain in this question: [3, 6]
Range: [0, 4]
Output statement: 0 ≤ y ≤ 4
Set-builder: {y : 0 ≤ y ≤ 4}
Check The distance from 5 over this stretch ranges from 0 to 2. Multiplying those distances by 2 gives exactly the output limits 0 and 4.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: For 2|x − 5| on 3 ≤ x ≤ 6, the range is [0, ∞).
That is the range over all real inputs. The limited stretch stops at distances 2 and 1 from its center.
✓ Instead: Its largest shown output is 4 and its smallest is 0, so the range is [0, 4].
Tips and tricks
Check the corner before choosing only endpoint outputs. A V can turn between them.
.9A distance graph for passing a bus stop
A V can tell a time story. Start 5 blocks before a bus stop and walk 1 block each minute. Before you pass the stop, the distance shrinks. After you pass it, distance grows. The corner tells you the moment you are at the stop.
For t ≥ 0 minutes, d(t) = |t − 5| blocks.
The corner is (5, 0), meaning at minute 5 the distance is zero.
Time in this story starts at zero, so only t ≥ 0 is part of the model.
Distance falls toward minute 5 and rises after you pass the stop.
Worked exampleFive times in the walking model
Find the distances at times 0, 3, 5, 7, and 10 minutes for d(t) = |t − 5|.
The column under 5 identifies when you reach the stop.The five measured times follow a V-shaped distance pattern.
What it asks. Find your distance from the stop at each stated time.
Plan. Substitute the time into the distance formula and record the results in a table.
Times equally spaced from 5 give the same distance.
Minute 5 gives the corner output zero.
At t = 0 and 10, the differences from 5 are −5 and 5, so both distances are 5.The bars keep distance without the direction of the difference.
At t = 3 and 7, the differences are −2 and 2, so both distances are 2.These times are equally far from the moment you pass the stop.
At t = 5, d(5) = |5 − 5| = 0.At the passing time you stand at the stop itself.
Answer
The table gives distance in blocks for each time in minutes.
Corner: (5, 0)
Check The distances in the columns under 3 and 7 match because both times are two minutes from passing the stop.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: After minute 5, the distance becomes negative.
Passing the stop changes which side you are on, while distance stays nonnegative.
The corner's input is the moment you reach the center; its output is the distance there.
Strategy: step by step
Set the entire inside equal to 0 and solve. This finds the input where its distance is zero, the corner's x-coordinate.
Put that input into the full formula to find the corner's y-coordinate, then plot the corner.
Choose one and two units on each side. Substitute all four inputs into the full formula, especially when a number multiplies x inside the bars.
Draw an input and output table, plot its five points, connect each side with a straight arm, and extend the arms.
Read domain as the graph's reach across the x-axis and range as its reach up the y-axis. For a limited input stretch, check both ends and the corner if it lies inside.
Strategy
Graph a V, including a number multiplying x inside
1
Does the outside multiplying number equal zero?
YesEvery output equals the outside added number. The graph is a flat line, rather than a V.
NoContinue to find the corner.
↓
2
Is the outside multiplying number positive?
YesThe arms rise away from the corner.
NoThe arms fall away from the corner.
↓
3
Does the inside have a number multiplying x?
YesUse the whole inside in every substitution. Its change per step affects arm steepness.
NoOne step from the corner gives distance 1.
↓
4
Is the input limited to a stated stretch?
YesKeep only that stretch. Evaluate both ends and the corner when included to find the output limits.
NoThe domain is all real numbers, and the arms keep extending.
Make the inside zero to locate the corner input.
Substitute that input to find the corner height.
Evaluate four inputs surrounding the corner and record all five outputs.
Plot and connect the arms. Read the input and output reach.
Worked exampleMove and stretch the parent graph
Graph f(x) = 2|x − 3| + 4. Give its vertex, arm slopes, domain, and range.
The column under 3 gives the corner height; matching distances give matching outputs.The solid V meets at (3, 4) and rises two per step away from the center.
What it asks. Draw the V and describe its turning point, the steepness of each arm, and all allowed inputs and outputs.
Plan. Find the corner by making the inside zero. Calculate four surrounding points, then use their heights to read the arms and range.
Only the distance part is doubled. Add 4 afterward.
The column under 3 contains the corner height 4.
Set x − 3 = 0 and add 3 to find x = 3. Then f(3) = 2|3 − 3| + 4 = 4, so the vertex is (3, 4).Zero distance marks where the left and right instructions meet. Substituting that input finds the corner height.
At x = 2 and x = 4, |x − 3| = 1, so f(x) = 2 × 1 + 4 = 6.Positions equally far from the center have equal outputs.
At x = 1 and x = 5, |x − 3| = 2, so f(x) = 2 × 2 + 4 = 8.The two-step distance gives a doubled distance height of 4, followed by the upward shift 4.
Use the five columns of the output table to plot the points. Connect each arm with a straight line.Each side uses a straight-line instruction, and the shared corner connects them.
From (2, 6) to (3, 4), go 1 right and 2 down, giving slope −2. From (3, 4) to (4, 6), go 1 right and 2 up, giving slope 2.Slope is change in height divided by change across. These changes are −2 ÷ 1 and 2 ÷ 1.
The domain is (−∞, ∞), and the range is [4, ∞).The formula accepts every real input. The lowest height is 4 and the arms rise without an upper limit.
Answer
Vertex: (3, 4).
Left slope: −2.
Right slope: 2.
Domain: (−∞, ∞).
Range: [4, ∞).
Check Starting from |x|, shift right 3, double the distance heights, and move up 4. The parent vertex and the points one unit away become (3, 4), (2, 6), and (4, 6), matching the substitutions.
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 basic V
Graph y = |x| using five points.
Five columns give the parent V's plotting points.The plotted points lie on two straight arms.
What it asks. Draw the V and find its turning point and output heights.
Plan. Make the inside zero to find the corner input, substitute to find its height, and plot the surrounding table columns.
The corner occurs at zero inside distance.
Each table column gives one plotted point.
At x = 0, y = |0| = 0.Zero has no distance from itself, so this is the corner.
At inputs −1 and 1, the distance is 1. At inputs −2 and 2, the distance is 2.Each pair is equally far from zero.
Read the five table columns, plot their inputs and output heights, and connect two straight arms through the corner.The piecewise rules are straight lines on their respective sides.
Answer
Vertex: (0, 0)
Domain: (−∞, ∞)
Range: [0, ∞)
Check The left arm lies on y = −x and the right arm on y = x.
Rung 2Rung 2: move the home address
Graph y = |x + 4|.
The middle column locates the shifted corner.The unchanged ramp heights meet at x = −4.
What it asks. Draw the V and find its turning point and output heights.
Plan. Make the inside zero to find the corner input, substitute to find its height, and plot the surrounding table columns.
The corner occurs at zero inside distance.
Each table column gives one plotted point.
Set x + 4 = 0 and subtract 4 to get x = −4. Then y = |−4 + 4| = 0.The inside zero locates the corner input; substituting finds its height.
At x = −5 and −3, y = 1. At x = −6 and −2, y = 2.These pairs are one and two units from the center −4.
Use all five table columns to plot the points, then extend two straight arms from the corner.Only the center moved; each extra unit of distance still adds one to the height.
Answer
Vertex: (−4, 0)
Domain: (−∞, ∞)
Range: [0, ∞)
Check The graph is the parent V moved four units left.
Rung 3Rung 3: stretch and shift
Graph y = 3|x − 1| − 2.
The distance part is multiplied by 3 before the downward shift.The V rises three units for each step away from its corner.
What it asks. Draw the V and find its turning point and output heights.
Plan. Make the inside zero to find the corner input, substitute to find its height, and plot the surrounding table columns.
The corner occurs at zero inside distance.
Each table column gives one plotted point.
Set x − 1 = 0 and add 1 to get x = 1. Then y = 3|1 − 1| − 2 = −2.This finds and verifies the corner (1, −2).
One unit away, y = 3 × 1 − 2 = 1. Two units away, y = 3 × 2 − 2 = 4.The factor 3 triples each distance height before subtracting 2.
Plot all five table columns. The left arm drops 3 per step right and the right arm rises 3 per step right.Distance shrinks toward the corner and grows away from it, so the slopes are −3 and 3.
Answer
Vertex: (1, −2)
Left slope: −3
Right slope: 3
Domain: (−∞, ∞)
Range: [−2, ∞)
Check At x = 3, direct substitution gives 3|2| − 2 = 4, consistent with another three-unit rise.
Rung 4Rung 4: reflect and compress
Graph y = −12|x + 3| + 5. Use points two units from the vertex.
Two-unit input steps avoid decimal output heights.Each two-unit move away from −3 lowers the output by 1.
What it asks. Draw the V and find its turning point and output heights.
Plan. Make the inside zero to find the corner input, substitute to find its height, and plot the surrounding table columns.
The corner occurs at zero inside distance.
Each table column gives one plotted point.
Set x + 3 = 0 and subtract 3 to get x = −3. Then y = −12|0| + 5 = 5.The inside zero finds the turning input, and substitution gives corner height 5.
At x = −5 and −1, distance is 2, so y = −12 × 2 + 5 = 4.The negative half multiplier turns two units of distance into a one-unit fall.
At x = −7 and 1, distance is 4, so y = −12 × 4 + 5 = 3.Four units of distance produce a two-unit fall from the corner.
Plot the five table columns and connect the downward arms. The domain is all real numbers and the range is (−∞, 5].Every nonzero distance subtracts from the greatest height 5, and the arms keep falling.
Answer
Vertex: (−3, 5)
Left slope: 12
Right slope: −12
Domain: (−∞, ∞)
Range: (−∞, 5]
Check The rise from (−5, 4) to (−3, 5) is 1 over a run of 2, so the left slope is 12, the opposite of a.
Rung 5Rung 5: a number multiplying x inside
Graph y = |2x + 2| − 1 using five points.
The whole inside determines these five heights.An inside factor of 2 makes the V rise two per step.
What it asks. Draw a V whose inside changes by two each time x changes by one.
Plan. Find the turning input, evaluate two inputs on each side, and connect the plotted table points.
Use the entire inside 2x + 2.
The inside multiplying number affects steepness.
Set 2x + 2 = 0. Subtract 2 to get 2x = −2, then divide by 2 to get x = −1.This finds the input where the inside distance is zero and the V turns.
At x = −1, y = |2(−1) + 2| − 1 = |0| − 1 = −1.Substitution checks the turning input and locates its height.
At x = −2 and 0, the inside is −2 and 2, so y = 2 − 1 = 1.These are one unit each side of the corner, and the inside changes by 2 per input unit.
At x = −3 and 1, the inside is −4 and 4, so y = 4 − 1 = 3.Two input steps each way give inside distance 4.
Plot all five table columns and connect straight arms. The left slope is −2 and right slope is 2.Each step toward the corner lowers height by 2; each step away raises it by 2.
Answer
Vertex: (−1, −1)
Left slope: −2
Right slope: 2
Domain: (−∞, ∞)
Range: [−1, ∞)
Check 2x + 2 = 2(x + 1), so the product rule rewrites y as 2|x + 1| − 1. Its corner and steepness match the graph.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: |x − 3| + 4 has corner (−3, 4).
At x = −3, the inside is −6 rather than zero. It becomes zero at x = 3.
✓ Instead: x − 3 = 0 gives x = 3, and the corner is (3, 4).
✗ Not this: 2|x − 3| + 4 at x = 4 equals 2 × (1 + 4) = 10.
The factor 2 multiplies the bars alone. The height 4 is added afterward.
✓ Instead: 2|4 − 3| + 4 = 2 × 1 + 4 = 6.
✗ Not this: |x − 9| = |x| − 9.
A subtraction cannot be moved out of the bars. At x = 1, the two sides would be 8 and −8.
Use five points: the corner, then one and two steps each side.
Inside lies, outside tells the truth is a memory cue for shifts. Find the input that makes the inside zero to verify its direction: x − 3 becomes zero at positive 3.
Domain is the horizontal reach. Range is the vertical reach. A plotted five-point table locates the V; its arms continue beyond the table unless the domain is limited.
Trap. Reading x − 3 as a left shift. Find where the inside is zero: x − 3 = 0 gives x = 3, so the corner moved right. The memory device is 'inside lies, outside tells the truth'; the zero calculation explains it.
