Quarry School

Build a line by changing the identity function

Explain it like I am five

Picture a flexible drawing of a ramp. You can make every height twice as large, flatten every height to half its size, turn the drawing upside down, or lift the whole drawing upward. Those changes are transformations. Start with the identity function, f(x) = x, whose output repeats its input. To make f(x) = mx + b, first multiply each height by m, then add b to that new height. Multiplying changes the slant. Adding lifts or lowers the entire line without changing the slant. When m is negative, multiplying also puts each height on the opposite side of the horizontal axis. The order matters because lifting before multiplying would multiply the lift too.

−4−3−2−1123456−6−4−22468(0, −3)(6, 0)
The solid line is the new graph; the dashed line is the earlier graph.
Reminder
  • Absolute-value bars. |−3| = 3, |3| = 3 and |0| = 0. Say absolute value of the number; the bars ask for distance from 0.
  • Order of operations. 12 × 2 − 3 = 1 − 3 = −2; multiply before subtracting.
  • Distribution. 2(x − 3) = 2x − 6 because 2 multiplies both terms.
  • Absolute value. |−12| = 12 measures size without sign.
  • Coordinate pairs. Changing (2, 3) to (2, 6) changes height only.
  • Multiplying signed numbers. −2 × (−2) = 4; two negative factors give a positive product.
Why it works. A point (x, x) on the identity graph becomes (x, mx) after its height is multiplied. It then becomes (x, mx + b) after the shift. That is exactly the requested formula. Reversing those operations gives m(x + b) = mx + mb, which usually has a different intercept. Absolute value means size without sign, so |m| controls steepness while the sign of m controls reflection. These descriptions use the same arithmetic that evaluates the function.
RuleRule: To transform y = x into y = mx + b, multiply outputs by m, including a reflection if m < 0, then shift outputs by b. Stretch if |m| > 1; compress if 0 < |m| < 1.
The same idea, five ways
Say it

Say: multiply the old height, then move it up or down.

Write it

The line y = mx + b is a transformed copy of the identity graph.

In math
  • f(x) = x
  • g(x) = mf(x) + b
  • g(x) = mx + b
  • (x, x) becomes (x, mx + b)
Like

Resize a ramp drawing's heights, turn it over if needed, then slide the drawing.

See it
−4−2246−6−4−22468(0, −3)(6, 0)
The solid line is flatter than the identity graph and crosses the y-axis three units lower.
The same idea, other ways
As three addresses

At input 2, the identity output is 2. Multiplication by 12 makes it 1. Subtracting 3 makes it −2. The final point is (2, −2).

2multiply by1/21subtract 3−2firstsecond
Compression occurs before the downward shift.
As a picture

The line y = 12x is flatter than y = x. Moving that whole flatter line down 3 gives y = 12x − 3; its slope remains 12.

−4−2246−6−4−22468(0, −3)(6, 0)
Both lines have the same slant; the solid one is three units lower.
As an order check

Input 0 is a fast check: multiply 0 by any m, then add b, and the output is b. If your graph starts somewhere else, the stated operations were not followed.

0multiply by 1/2,subtract 3−3inputoutput
The starting output must be the stated vertical shift.
.1Vertical stretch

Picture pulling a drawing taller while keeping its width fixed. A vertical stretch multiplies every height by the same positive factor greater than 1. Each nonzero height moves farther from the x-axis. A height below the axis stays below during this positive multiplication, but its distance from the axis grows.

  • Rule: Multiplication by a positive factor a > 1 stretches heights vertically.
  • Rule: In y = mx, |m| > 1 includes a vertical stretch; if m < 0, reflection also occurs.
  • Rule: |m| = 1 preserves height size. Multiplication by 0 sends all heights to 0, after which adding b gives y = b.
−4−2246−6−4−22468(0, 0)(−0, 0)
The solid line is the new graph; the dashed line is the earlier graph.
Reminder
  • Absolute value. |−3| = 3 is the distance of −3 from 0, with no direction sign.
The same idea, five ways
Say it

Say: multiply every height by the same size greater than 1.

Write it

A vertical stretch enlarges output distances from the x-axis while keeping inputs fixed.

In math
  • g(x) = af(x), a > 1
  • (x, y) becomes (x, ay)
  • |−2| = 2
Like

A copier changes the drawing's height while its width stays fixed.

See it
−4−2246−6−4−22468(0, 0)(−0, 0)
The solid line has twice the height at every input.
Worked exampleDouble the height

You need to transform y = x into y = 2x and show what happens to a point.

−4−2246−6−4−22468(0, 0)(−0, 0)positive height doublesnegative height doubles
The solid line is the new graph; the dashed line is the earlier graph.
  1. The old point at input 3 is (3, 3).The identity output equals its input.
  2. Multiply the height by 2: 2 × 3 = 6, so the new point is (3, 6).A vertical stretch changes the second coordinate only.
  3. At input −3 the old point is (−3, −3). Multiply its height by 2: (−3, −3) becomes (−3, −6).A positive stretch preserves which side of the axis the point occupies while doubling its distance from the axis.
Answer
  • y = 2x.
  • The point (3, 3) becomes (3, 6).
  • The point (−3, −3) becomes (−3, −6).
Check Substitute x = 3 into the new rule: y = 2 × 3 = 6. For the negative point, 2 × (−3) = −6; its distance grows from 3 to 6 while it stays below the axis.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: multiplying heights by 12 makes the identity line steeper.
Every nonzero height becomes half its size, so a fixed rightward move has a smaller climb.
✓ Instead: It is a vertical compression. The slope becomes 12.
Tips and tricks
  • Tip: A size above 1 spreads heights away from the axis; a size between 0 and 1 pulls them closer.
.2Vertical compression

Picture pressing a drawing shorter while leaving its width alone. A vertical compression multiplies each height by a positive factor between 0 and 1. Heights move toward the x-axis without crossing it. Half of a height of 2 is 1; half of a height of −6 is −3. Both points become closer to zero.

  • Rule: Multiplication by 0 < a < 1 compresses heights vertically.
  • Rule: In y = mx, 0 < |m| < 1 includes a compression; a negative m also reflects.
  • Rule: Factor 0 is not an ordinary compression: it collapses every output to 0.
−4−2246−4−2246(2, 1)
The solid line has half the height of the identity line at each input.
Reminder
  • Fraction multiplication. 12 × (−6) = −3, half of the signed amount.
The same idea, five ways
Say it

Say: take the same fractional share of every height.

Write it

A vertical compression shrinks output distances from the x-axis without changing inputs.

In math
  • g(x) = af(x), 0 < a < 1
  • (x, y) becomes (x, ay)
Like

Press a drawing shorter without changing its width.

See it
−4−2246−4−2246(2, 1)
The solid line has half the height of the identity line at each input.
Worked exampleHalve the heights

You need to transform y = x into y = 12x and show the effect on positive and negative heights.

−6−4−224−6−4−224(−6, −3)(2, 1)
The positive and negative heights both move halfway toward the x-axis.
  1. At input 2, the old height is 2. Multiply by 12: (2, 2) becomes (2, 1).The input stays fixed while the output becomes half its old size.
  2. At input −6, the old height is −6. Multiply by 12: (−6, −6) becomes (−6, −3).Positive multiplication preserves the sign, while reducing distance from zero.
  3. The new formula is y = 12x. Plug in both inputs: 12 × 2 = 1 and 12 × (−6) = −3.The transformed points must obey the new output rule.
Answer
  • New rule: y = 12x.
  • (2, 2) becomes (2, 1).
  • (−6, −6) becomes (−6, −3).
Check From (−6, −3) to (2, 1), rise = 4 and run = 8, so slope = 48 = 12, matching the compression factor.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: halving a negative height makes it more negative.
Half of −6 is −3, whose distance from zero is smaller.
✓ Instead: Positive fractional multiplication preserves the sign and compresses the height.
Tips and tricks
  • Tip: Between 0 and 1 means a share smaller than the whole, so distances shrink.
.3Reflection across the x-axis

Think of the horizontal axis as a mirror lying on a table. A vertical reflection moves a point the same distance to the opposite side of that mirror. The across coordinate stays fixed. Multiplication by a negative number includes this turn as well as any stretch or compression.

  • Rule: A vertical reflection sends (x, y) to (x, −y).
  • Rule: In y = mx + b, m < 0 reflects the identity graph before the shift.
−4−224−4−224reflected (2, −2)reflected (−2, 2)
Each solid-line height is the opposite of the dashed identity height at the same input; their distances from 0 match.
Reminder
  • Negative multiplication. −2 × (−3) = 6 because reversing a backward direction makes it forward.
The same idea, five ways
Say it

Say: keep the across position and reverse the height's sign.

Write it

A vertical reflection exchanges above and below the x-axis.

In math
  • (x, y) becomes (x, −y)
  • y = −f(x)
Like

A horizontal mirror puts a point equally far on the opposite side, exchanging above and below.

See it
−4−2246−6−4−22468(0, 0)(0, 0)
The heights have opposite signs at each unchanged input.
Worked exampleReflect without changing the size

You need to reflect y = x across the x-axis. Keep each input and reverse only its height.

−4−224−4−224reflected (2, −2)reflected (−2, 2)
The solid reflected line exchanges above and below at each fixed input.
  1. At input 2 the identity gives (2, 2). Keep x = 2 and change y = 2 to −2, making (2, −2).A reflection across the horizontal axis changes above to below at the same distance.
  2. At input −2 the old point is (−2, −2). Its reflection is (−2, 2).A point starting below the axis moves equally far above it; the input remains −2.
  3. The new rule is y = −x. Draw it through (0, 0) and (2, −2).Every old identity output x has been multiplied by −1, and two distinct points locate the resulting line.
Answer
  • Reflected rule: y = −x.
  • (2, 2) becomes (2, −2).
  • (−2, −2) becomes (−2, 2).
Check At x = −2, the new rule gives −(−2) = 2, agreeing with the reflected point. Reflecting it again gives the old point (−2, −2).
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: reflection changes (2, 3) into (−2, 3).
That reverses the horizontal coordinate, so it reflects across the y-axis instead.
✓ Instead: A vertical reflection across the x-axis sends (2, 3) to (2, −3).
Tips and tricks
  • Tip: The axis named in a reflection is the mirror. Across the x-axis, x stays fixed.
.4Vertical shift

Think of lifting the entire ramp drawing without bending it. A vertical shift adds the same amount to every height. The difference between any two heights stays the same, so the slope stays the same.

  • Rule: Adding b sends (x, y) to (x, y + b).
  • Rule: b > 0 shifts up; b < 0 shifts down; b = 0 makes no vertical shift.
−4−2246−6−4−22468(0, −3)(6, 0)
The solid line is the new graph; the dashed line is the earlier graph.
Reminder
  • Signed addition. 1 + (−3) = −2 means move three downward from 1.
The same idea, five ways
Say it

Say: move every height by the same signed amount.

Write it

A vertical shift changes the intercept while preserving slope.

In math
  • g(x) = f(x) + b
  • (x, y) becomes (x, y + b)
Like

Lift a straight board without tilting it.

See it
−4−2246−6−4−22468(0, −3)(3, 0)
The solid line has the same slope and every height is three lower.
Worked exampleShift the compressed line

You need to move y = 12x down 3. Also shift y = 2x − 5 up 4.

−4−2246−6−4−22468(0, −3)(6, 0)
The solid line is the new graph; the dashed line is the earlier graph.
−4−224−8−6−4−22468(0, −1)(0.5, 0)moved (1, 1)
The transformed points agree with the new rule.
  1. Subtract 3 from its output: y = 12x − 3.Down means a negative change to each height.
  2. At x = 2 the old height was 1; the new height is 1 − 3 = −2.The input stays 2 while every output shifts equally.
  3. For the upward case, add 4 outside the old rule: y = (2x − 5) + 4 = 2x − 1.The same addition acts on every output, leaving the coefficient of x unchanged.
  4. At input 1 the old height is 2 × 1 − 5 = −3. The moved height is −3 + 4 = 1.A shift may carry a point across the axis; it adds the same 4 regardless of the height's sign.
Answer
  • y = 12x − 3.
  • The point (2, 1) becomes (2, −2).
  • Upward case: y = 2x − 1.
  • The point (1, −3) becomes (1, 1).
Check The point at x = 0 moves from (0, 0) to (0, −3), confirming the new intercept. In the upward case, 2 × 1 − 1 = 1, and the unchanged slope 2 confirms that the shift did not tilt the line.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: subtracting 3 from y = 12x changes its slope to −52.
The subtraction is outside the multiplication and affects every height equally.
✓ Instead: The slope remains 12, and the intercept becomes −3.
Tips and tricks
  • Tip: Outside addition moves the whole graph, so it does not change rise over run.
.5A note on left and right shifts

You can slide a graph sideways too. When you move a point right h units, its new input address is the old input plus h. To use that new address in the old rule, subtract h first. This section's standard method needs vertical changes, but the source also mentions sideways changes.

  • Rule: g(x) = f(x − h) shifts the old graph right h units when h > 0; h < 0 shifts it left.
  • Rule: For a line, m(x − h) + b = mx + (b − mh), so a horizontal shift can also be described with a new vertical intercept.
−4−2246−6−4−22468(0, −2)(2, 0)
The solid line is the new graph; the dashed line is the earlier graph.
Reminder
  • Substitution. If f(u) = u, then f(x − 2) = x − 2; the whole new input replaces u.
The same idea, five ways
Say it

Say: move each address right, then undo that move when reading the old rule.

Write it

Subtracting a positive amount inside the input shifts right; adding a positive amount shifts left.

In math
  • g(x) = f(x − h)
  • (x, y) becomes (x + h, y)
Like

A relocated address is two blocks farther right, so subtract two to find its old address.

See it
−4−2246−6−4−22468(0, −2)(2, 0)
The identity point (0, 0) moves right to (2, 0).
Worked exampleRecognize a sideways shift

You need to describe g(x) = f(x − 2) when f(x) = x. Then move f(x) = 2x + 5 left 3 and find its new rule.

−4−2246−6−4−22468(0, −2)(2, 0)
The solid line is the new graph; the dashed line is the earlier graph.
−4−22246810121416(0, 11)(−5.5, 0)moved (−3, 5)moved (−2, 7)
The solid line is the old line shifted left 3; its slope remains 2.
  1. Replace the old input by x − 2: g(x) = x − 2.The identity rule outputs whichever input it receives.
  2. At the moved address x = 2, g(2) = 2 − 2 = 0.This plugs the new address back in to confirm the point moved from (0, 0) to (2, 0).
  3. Moving left 3 uses h = −3, so x − h = x − (−3) = x + 3. Write g(x) = f(x + 3).The new address must be increased by 3 to recover its old address.
  4. Replace the whole input in 2x + 5: g(x) = 2(x + 3) + 5.The factor 2 acts on the entire substituted input.
  5. Distribute: g(x) = 2x + 6 + 5 = 2x + 11.Both parts of x + 3 are multiplied before the outside addition.
  6. The old point at x = 0 is (0, 5). Moving it left 3 gives (−3, 5). Substitute: g(−3) = −6 + 11 = 5.Plugging the moved address into the new rule verifies the shift direction.
Answer
  • A shift right 2.
  • For the identity line this also equals a shift down 2.
  • Left-shift case: g(x) = 2x + 11.
  • The old point (0, 5) becomes (−3, 5).
Check Both descriptions give the same equation y = x − 2, so their graphs agree. Another old point is (1, 7). It moves to (−2, 7), and 2(−2) + 11 = 7 verifies that point too.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: f(x − 2) always shifts a graph left 2.
The new input 2 must become the old input 0 after subtracting 2.
✓ Instead: It shifts right 2.
Tips and tricks
  • Tip: Verify an old point at its moved address when an inside sign feels backward.
Strategy: step by step
  1. 1. Sketch the identity function y = x using (0, 0) and (1, 1).
  2. 2. Read |m|, the size of m without its sign, to decide whether heights stretch or compress.
  3. 3. If m < 0, reflect the heights across the x-axis as part of that multiplication.
  4. 4. Add b to every new height. Positive b moves up; negative b moves down.
  5. 5. Verify a point in the final formula. Multiplication must occur before this stated shift.
Strategy
Strategy: use transformations in formula order
1
Is |m| greater than 1?
YesStretch vertically by |m|.
NoCompress if 0 < |m| < 1; keep size if |m| = 1; flatten to 0 if m = 0.
↓
2
Is m negative?
YesReflect across the x-axis during multiplication.
NoKeep each height's direction during multiplication.
↓
3
Is b negative?
YesShift down |b| after multiplying.
NoShift up b if positive; do not shift if zero.
  1. Start with y = x.
  2. Multiply its heights by m, interpreting size and sign separately.
  3. Add b to the new heights.
  4. Substitute one input in the final formula and compare.
Worked exampleCompress, then move down

You need to graph f(x) = 12x − 3 by changing y = x.

−4−2246−6−4−22468(0, −3)(6, 0)
The solid line is the new graph; the dashed line is the earlier graph.
  1. Begin with y = x and its point (2, 2).The identity graph repeats the input as output.
  2. Multiply the height by 12: (2, 2) becomes (2, 1).The positive factor between 0 and 1 compresses vertically without reflecting.
  3. Subtract 3 from that height: (2, 1) becomes (2, −2).The outside term −3 shifts the already compressed graph down.
  4. The origin similarly becomes (0, −3). Draw the line through (0, −3) and (2, −2).Two transformed points identify the final straight line.
Answer
  • Graph: y = 12x − 3.
  • Transformations: compress by 12, then shift down 3.
Check The rise from (0, −3) to (2, −2) is 1 and the run is 2, giving slope 12. At input 2 the formula gives 1 − 3 = −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: stretch only

You need to make y = 3x from y = x.

−4−2246−6−4−22468(0, 0)(−0, 0)
The solid line is the new graph; the dashed line is the earlier graph.
  1. Multiply every old output by 3; (1, 1) becomes (1, 3).The size 3 is greater than 1, so it stretches vertically.
  2. Keep (0, 0), and draw through (0, 0) and (1, 3).Multiplying height 0 by 3 keeps it 0.
Answer
Stretch vertically by 3.
Check At input 2, the old height 2 becomes 6, and 3 × 2 = 6.
Rung 2Rung 2: compress only

You need to make y = 13x from y = x.

−4−2246−6−4−22468(0, 0)(−0, 0)
The solid line is the new graph; the dashed line is the earlier graph.
  1. Multiply each height by 13; (3, 3) becomes (3, 1).A factor between 0 and 1 pulls each height toward the x-axis.
  2. Draw through (0, 0) and (3, 1).The horizontal input positions stay unchanged.
Answer
Compress vertically by 13.
Check The rise is 1 for a run of 3, agreeing with the coefficient.
Rung 3Rung 3: compression followed by a shift

You need to make y = 12x − 3 and keep the order correct.

−4−2246−6−4−22468(0, −3)(6, 0)
The solid line is the new graph; the dashed line is the earlier graph.
  1. Compress (2, 2) to (2, 1).Multiplication comes first in the final formula.
  2. Move it down 3 to (2, −2), and move the origin to (0, −3).The same shift is added to each compressed output.
Answer
Compress by 12, then shift down 3.
Check At input 0, the formula gives −3; the final graph has that intercept.
Rung 4Rung 4: reflection, stretch and shift

You need to make y = −2x + 3 from the identity line.

−4−2246−6−4−22468(0, 3)(1.5, 0)
The solid line is the new graph; the dashed line is the earlier graph.
  1. Multiply the height 2 at input 2 by −2 to obtain −4.The negative sign reflects and the size 2 stretches.
  2. Add 3 to obtain −1, so the final point is (2, −1).The upward shift comes after multiplying.
  3. The origin becomes (0, 3); draw the line through both final points.The point at input 0 checks the intercept independently.
Answer
Reflect across the x-axis, stretch by 2, then shift up 3.
Check Between (0, 3) and (2, −1), slope = −1−32−0 = −42 = −2, and −2 × 2 + 3 = −1.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: shift y = x down 3, then compress by 12 to obtain 12x − 3.
The compression also halves the earlier shift. At input 0 it gives −32, not −3.
✓ Instead: Compress first, then subtract 3. The final output at input 0 is −3.
✗ Not this: Counterexample: y = −2x + 3 is only a stretch by 2.
A positive stretch cannot change positive heights into negative ones.
✓ Instead: Reflect across the x-axis and stretch by 2, then shift up 3.
✗ Not this: Use |−3| = −3 as a vertical scale factor.
Absolute value is a distance from 0, and a distance cannot be negative.
✓ Instead: Use size |−3| = 3 for the stretch, then use the original minus sign for the reflection.
Tips and tricks
  • Tip: Memory cue: multiply, then move. That is the order used to evaluate mx + b.
  • Tip: Use input 0 to catch an accidentally multiplied shift.
  • Understand, then rebuild it when needed: the transformation description follows the equation's arithmetic; you do not need to memorize each graph.
Trap. Doing the stated downward shift first and then halving the heights. That produces 12(x − 3) = 12x − 32, a different line.