Vertical stretch and compression: multiply the heights

Explain it like I am five

Imagine a drawing on a rubber sheet fastened along a horizontal line. Pull the sheet vertically so every dot is twice as far from that line. Dots above it move higher, dots below it move lower, and dots on it stay put. A Vertical stretch multiplies every signed height by a number larger than 1. A Vertical compression multiplies those heights by a positive fraction smaller than 1, bringing dots closer to the x-axis. The Scale factor says how much the distance changes. Keep each dot's horizontal address. A negative multiplier also flips the drawing across the x-axis.

Multiplying by 2 doubles each height while keeping the same horizontal coordinate.

Reminder

Signed multiplication. A positive factor keeps a height's sign: $1 over 2$ × (−6) = −3. A negative factor reverses it: (−2) × 5 = −10.
Exact fractions. One half of 5 is $5 over 2$ = 2.5 exactly. Multiplying the new height by 2 recovers 5.
Order of operations. Multiply before adding: −3 × 2 + 2 = −6 + 2 = −4. Parentheses change the job: −3(2 + 2) = −12.

Why it works. If f(u) = v, then a·f(u) = a·v, so a point moves from (u, v) to (u, a·v). Its distance from the x-axis changes from |v| to |a|·|v|. This explains why multiplying a negative height by 2 sends it farther downward, rather than upward. A negative a reverses the sign as well as changing the distance. The same original input is still evaluated, so vertical scaling preserves the domain and multiplies the range values.

RuleFor g(x) = a·f(x), (u, v) becomes (u, a·v). If |a| > 1, stretch vertically; if 0 < |a| < 1, compress vertically.
If a < 0, also reflect across the x-axis. If a = 0, all defined outputs become zero; this is a collapse, not an ordinary stretch or compression.

The same idea, five ways

Say it

A vertical scale changes how far each height lies from the x-axis. A negative factor also puts it on the opposite side.

Write it

Multiply every old output by the outside factor a while keeping its input coordinate fixed.

In math

g(x) = a·f(x); (u, v) becomes (u, a·v).
|a| > 1: Vertical stretch. 0 < |a| < 1: Vertical compression.
a < 0: also a Vertical reflection. |a| = 1: no size change.
a = 0: defined heights collapse to 0, with the original Domain retained.

Pull a rubber drawing away from the horizontal axis or press it closer while holding each sideways address fixed.

See it

Halving signed heights brings both sides of y = x closer to the x-axis.

The same idea, other ways

As distance from a mirror line

A stretch by 2 doubles distance from the x-axis. Old height 4 becomes 8, old height −4 becomes −8, and zero stays zero. The sign tells which side of the line contains the point; the magnitude tells the distance.

As distance from a mirror line

g(x) = a·f(x); (u, v) becomes (u, a·v).

Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.

As a worker after the machine

The old function gives height 5. A worker multiplying by 3 changes it to 15 and leaves the input 2 untouched. A worker multiplying by $1 over 2$ changes it to $5 over 2$ . Both jobs happen after f has returned its answer.

The vertical multiplier acts on the completed answer.

As a comparison with a shift

For $x^{2}$ , a scale by 2 keeps the zero height at zero and sends height 4 to 8. Adding 3 instead sends zero to 3 and 4 to 7. Scaling changes distances proportionally; shifting adds the same amount to every height.

As a comparison with a shift

g(x) = a·f(x); (u, v) becomes (u, a·v).

Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.

Formula	Effect	A point (u, v) goes to
a·f(x), a > 1	Vertical stretch by a	(u, a·v)
a·f(x), 0 < a < 1	Vertical compression by a	(u, a·v)
f(bx), b > 1	Horizontal compression by $1 over b$	( $\frac{u}{b}$ , v)
f(bx), 0 < b < 1	Horizontal stretch by $1 over b$	( $\frac{u}{b}$ , v)

.1Vertical stretch

Think of a graph drawn on a sheet that you pull away from the x-axis. Multiplying its heights by 2 doubles their distances from that axis. You do not move the points left or right, and a zero height stays zero. In a population model, twice as large means twice the population at the same time. It does not mean that the population reaches its stages sooner.

Formula: g(x) = a·f(x), with a > 1 for a positive Vertical stretch. The point (u, v) becomes (u, a·v).
The Domain stays the same. Multiply every Range value, including negative ones, by a.
An x-intercept is a point on the x-axis, where the output is 0. The x-intercepts stay fixed for any nonzero vertical multiplier because a × 0 = 0. For $x^{2}$ , the point (0, 0) therefore stays fixed.
For a < 0, the magnitude |a| determines the stretch or compression and the sign adds a Vertical reflection.

For 2, each original height doubles and the point at height zero remains fixed.

Worked exampleDouble a quadratic's heights

Start with f(x) = $x^{2}$ . Write its vertical stretch by 2 and move the original points (1, 1) and (−2, 4).
This asks you to find the new address or height of the given graph information.

Multiplying outside by 2 doubles heights at fixed inputs.

Write g(x) = 2f(x) = 2 $x^{2}$ .The requested vertical scale multiplies the function's output after it is calculated.
(1, 1) becomes (1, 2 × 1) = (1, 2).The input stays 1 and its height doubles.
(−2, 4) becomes (−2, 2 × 4) = (−2, 8).The negative input also stays fixed; the multiplier acts on its output 4.

Answer

g(x) = 2 $x^{2}$ .
New points: (1, 2) and (−2, 8).
Domain: (−∞, ∞).
Range: [0, ∞).

Check g(−2) = 2(−2 = 2 × 4 = 8. The original height 4 and the new height 8 have ratio 2.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A vertical stretch by 2 turns into (2x.

That changes the input and gives 4. The stated outside stretch is 2. At input 1 its correct height is 2, rather than 4.

✓ Instead: g(x) = 2.
New points: (1, 2) and (−2, 8).
Domain: (−∞, ∞).
Range: [0, ∞).

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

.2Vertical compression

Imagine lowering every shelf to half its signed height from the floor line on a drawing. A shelf at height 6 goes to 3, while a mark at height −6 goes to −3. Both are closer to the line. A Vertical compression keeps the horizontal address and multiplies the entire answer by a positive fraction below 1. It changes the distance from the axis, rather than subtracting a fixed amount.

Formula: c(x) = a·f(x), with 0 < a < 1. The point (u, v) becomes (u, a·v).
Multiplying by $1 over 2$ is dividing the whole output by 2. The same operation applies to every table entry.
Compression keeps the original Domain. If the original Range is [−6, 10], multiplication by $1 over 2$ gives [−3, 5].
The identity multiplier a = 1 leaves all heights unchanged. The multiplier a = 0 collapses every defined height to zero while retaining the original domain.

Halving y = x brings both positive and negative heights closer to the x-axis.

Worked exampleCompress a negative output too

If f(4) = 6 and f(−4) = −6, find the corresponding points on c(x) = $1 over 2$ f(x).
This asks you to find the new address or height of the given graph information.

Read each column as one x input paired with its f(x) output.

Read each column as one x input paired with its c(x) output.

At input 4, c(4) = $1 over 2$ × 6 = 3.One half of six is three, and vertical scaling keeps the input fixed.
At input −4, c(−4) = $1 over 2$ × (−6) = −3.A positive times a negative is negative, and half the distance six is three.
The new points are (4, 3) and (−4, −3).Each unchanged input is paired with its newly halved output.

Answer

(4, 6) becomes (4, 3).
(−4, −6) becomes (−4, −3).

Check Doubling the new outputs 3 and −3 recovers 6 and −6. Their distances from the x-axis both fell from 6 to 3.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Compressing old height −6 by produces −12.

Compression brings the height closer to zero. Multiply −6 by one half to get −3; doubling it instead is a stretch.

✓ Instead: (4, 6) becomes (4, 3).
(−4, −6) becomes (−4, −3).

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

.3Negative vertical scale

A negative scale does two jobs. Its size stretches or compresses the drawing, and its minus sign reflects the drawing across the x-axis. Think of pulling a dot to twice its distance from a mirror line and then placing it on the other side. A factor of −2 therefore doubles distance and reverses the height's sign. A factor of − $1 over 2$ halves distance and reverses the sign.

Write a = −|a| when a < 0. The factor |a| gives the size change; the remaining minus gives the reflection.
If a = −1, the transformation is a reflection with no size change. If a = 0, it is a collapse, not a reflection or ordinary scale.
For g(x) = a·f(x) + k, scale the old height first and add k afterward. The final distance from y = k is |a| times the original distance from the x-axis.
A negative multiplier reverses ordered range endpoints: if f has range [1, 5], −2f has range [−10, −2].

The graph − has half the distance from the x-axis and the opposite sign of .

Worked exampleCompress and reflect a quadratic

Describe g(x) = − $1 over 2$ $x^{2}$ relative to f(x) = $x^{2}$ . Move the original point (2, 4) and give the domain and range.
This asks you to change the coordinate named by each mirror move and keep the other coordinate.

Half the distance and opposite signs give new point (2, −2).

The multiplier is a = − $1 over 2$ , whose magnitude is $1 over 2$ .The magnitude measures distance change without its direction.
Compress vertically by $1 over 2$ , then reflect across the x-axis.The magnitude lies between 0 and 1, and the negative sign reverses output signs.
Keep input 2. The new output is − $1 over 2$ × 4 = −2, so the point becomes (2, −2).The outside factor acts on the whole old height.
Domain: all real numbers. Range: (−∞, 0].Squaring still accepts all real inputs; every nonnegative square now becomes nonpositive, and arbitrarily large squares give arbitrarily negative heights.

Answer

Vertical compression by $1 over 2$ and reflection across the x-axis.
(2, 4) becomes (2, −2).
Domain: (−∞, ∞).
Range: (−∞, 0].

Check g(2) = − × = − × 4 = −2. Its distance from the x-axis is 2, half the original distance 4.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The minus in − makes the new point from (2, 4) equal (−2, 2).

The multiplier is outside, so input 2 stays fixed. It halves height 4 to 2 and reverses that height to −2.

✓ Instead: Vertical compression by and reflection across the x-axis.
(2, 4) becomes (2, −2).
Domain: (−∞, ∞).
Range: (−∞, 0].

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

Strategy: step by step

1. Find the multiplier a outside the function. Its magnitude |a| sets the size change, and its sign tells you whether to reflect.
2. Keep every input coordinate. Multiply each entire signed output by a, including outputs below zero.
3. If a shift is added outside afterward, multiply the original height first and then add the shift. The shifted graph is scaled relative to the line y = k in a·f(x) + k.
4. Keep the original domain. Multiply the original range values by a, reversing their order when a is negative.
5. Check a transformed point by substituting its unchanged input into the new formula.

Strategy

Read an outside factor and move heights

Is a = 0?

YesEvery permitted input has output 0 before any final shift. Keep the original Domain.

NoUse |a| to identify the size change.

↓

Is |a| > 1?

YesStretch vertically by |a|.

NoIf 0 < |a| < 1, compress vertically by |a|. If |a| = 1, there is no size change.

↓

Is a negative?

YesAlso reflect across the x-axis by reversing every output sign.

NoKeep every output sign during the positive scale.

↓

Is a final outside shift added?

YesAdd the shift after multiplying each old output.

NoThe scaled signed output is the final height.

1. Identify the outside factor a. Its magnitude gives the distance change and its sign gives the reflection.
2. Keep each known input and multiply its output by a. This finds the new height at that same address; for (2, 5) and a = 3, the new height is 3 × 5 = 15.
3. If there is a final outside shift k, add it to the scaled height. This finishes the change in the written order; for old height 5 in −2f(x) + 3, compute −10 + 3 = −7.
4. Keep the original Domain and apply the same height calculation to the Range. A negative multiplier reverses ordered endpoints; [1, 5] scaled by −2 becomes [−10, −2].
5. Substitute an unchanged input into the transformed formula to check its output. For c(x) = $1 over 2$ f(x) and f(2) = 5, c(2) = $5 over 2$ , and doubling this result recovers 5.

Worked exampleStretch and compress the known point (2, 5)

The point (2, 5) is on f. Find the corresponding points on g(x) = 3f(x) and c(x) = $1 over 2$ f(x).
This asks you to find the new address or height of the given graph information.

On g(x) = 3f(x): (2, 15).

On c(x) = f(x): (2, ), equivalently (2, 2.5) exactly.

The card records the exact result of the worked coordinate or symmetry calculation.

The given point means f(2) = 5.A point on a function graph records its input first and its output second.
For g, keep the input 2 and multiply the output: 3 × 5 = 15.The outside factor 3 changes the height only, giving a vertical stretch.
For c, keep the input 2 and multiply the output: $1 over 2$ × 5 = $5 over 2$ = 2.5 exactly.Multiplying by one half takes half the original height, giving a vertical compression. Five halves is the terminating decimal 2.5 with no rounding.

Answer

On g(x) = 3f(x): (2, 15).
On c(x) = $1 over 2$ f(x): (2, $5 over 2$ ), equivalently (2, 2.5) exactly.

Check g(2) = 3f(2) = 3 × 5 = 15. c(2) = f(2) ÷ 2 = 5 ÷ 2 = 2.5. Dividing the first new height by 3 and doubling the second new height both recover 5.

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: triple a known height

The point (2, 5) is on f. Find its corresponding point on g(x) = 3f(x).
This asks you to find the new address or height of the given graph information.

(2, 15).

The card records the exact result of the worked coordinate or symmetry calculation.

Read f(2) = 5 from the point.The second coordinate is the output for the first coordinate.
Keep input 2 and multiply height 5 by 3: 3 × 5 = 15.An outside multiplier changes the output coordinate.

Answer

(2, 15).

Check g(2) = 3f(2) = 15, and 15 ÷ 3 = 5 recovers the original output.

Rung 2Rung 2: halve the same height exactly

The point (2, 5) is on f. Find its corresponding point on c(x) = $1 over 2$ f(x).
This asks you to find the new address or height of the given graph information.

(2, ), or (2, 2.5) exactly.

The card records the exact result of the worked coordinate or symmetry calculation.

Keep the input 2.There is no change inside the function.
Multiply the output: $1 over 2$ × 5 = $5 over 2$ = 2.5 exactly.Five halves is five divided by two; 2 × 2.5 = 5, so the decimal is exact.

Answer

(2, ), or (2, 2.5) exactly.

Check c(2) = f(2) ÷ 2 = 5 ÷ 2. Doubling the new height gives 5.

Rung 3Rung 3: halve a whole table

Use the original f table shown beside this question. List the corresponding known points on c(x) = $1 over 2$ f(x).
This asks you to attach each known output to its transformed input and calculate any requested new height.

Read each column as one x input paired with its f(x) output.

Read each column as one x input paired with its c(x) output.

Read the displayed original table column by column. Each top input belongs to the output directly below it.The picture supplies the known values used in the following calculations; keep each column paired.
Copy inputs 2, 4, 6, 8 without changing them.A vertical compression multiplies outputs, not input coordinates.
Halve each output: 1 ÷ 2 = $1 over 2$ , 3 ÷ 2 = $3 over 2$ , 7 ÷ 2 = $7 over 2$ , and 11 ÷ 2 = $11 over 2$ .The same scale factor applies to every original height.
The exact decimal forms are 0.5, 1.5, 3.5, and 5.5.Each denominator is 2, so dividing produces a terminating decimal without rounding.

Answer

(2, $1 over 2$ ) = (2, 0.5).
(4, $3 over 2$ ) = (4, 1.5).
(6, $7 over 2$ ) = (6, 3.5).
(8, $11 over 2$ ) = (8, 5.5).

Check At input 6, c(6) = × 7 = . Doubling every new output recovers the original row 1, 3, 7, 11.

Rung 4Rung 4: a different table compressed by one quarter

Use the second textbook practice table shown beside this question. For this original transformation, make the table for c(x) = $1 over 4$ f(x). This asks you to keep each input and quarter its output.

Read each column as one x input paired with its f(x) output.

Read each column as one x input paired with its c(x) output.

Read f(2) = 12, f(4) = 16, f(6) = 20, and f(8) = 0 from the four displayed columns.The input and output in the same column belong together.
Divide the first three heights by 4: 12 ÷ 4 = 3, 16 ÷ 4 = 4, 20 ÷ 4 = 5.Multiplication by $1 over 4$ quarters the entire output.
The last height stays 0 because $1 over 4$ × 0 = 0; keep inputs 2, 4, 6, 8.A vertical compression changes heights only, and multiplying zero still gives zero.

Answer

c: use the displayed compressed table.
At input 8 the output is still 0.

Check c(6) = f(6) = × 20 = 5. Multiplying the complete new output row by 4 restores the original row.

Rung 5Rung 5: twice the population at the same time

A fruit-fly population graph has reference points (0, 1), (3, 3), (6, 2), and (7, 0). A second population is twice as large at every time. Give its formula in terms of f and its corresponding points.
This asks you to attach each known output to its transformed input and calculate any requested new height.

Read each column as one t input paired with its f(t) output.

Read each column as one t input paired with its P(t) output.

Read the displayed original table column by column. Each top input belongs to the output directly below it.The picture supplies the known values used in the following calculations; keep each column paired.
Write P(t) = 2f(t).Twice as large describes an output change at the same time input t.
Keep times 0, 3, 6, 7. Multiply heights 1, 3, 2, 0 by 2 to get 2, 6, 4, 0.The population doubles at each unchanged time, including zero population.
Plot (0, 2), (3, 6), (6, 4), and (7, 0), following the same time pattern with doubled heights.These are the original reference points after vertical scaling. Their horizontal coordinates remain unchanged.

Answer

P(t) = 2f(t).
New points: (0, 2), (3, 6), (6, 4), (7, 0).

Check P(3) = 2f(3) = 2 × 3 = 6, and P(7) = 2 × 0 = 0. The known reference point at time 3 keeps that time. The four supplied points alone do not establish a greatest population.

Rung 6Rung 6: recognize a cubic scale from graph points

In this original graph example, the function is known to have the form g(x) = a· $x^{3}$ and contains (1, 4) and (2, 32). Find a and describe the transformation.
This asks you to find the new address or height of the given graph information.

The supplied points agree with the stated family g(x) = 4.

For the Cubic function f(x) = $x^{3}$ , f(1) = $1 cubed$ = 1.To find a vertical scale, compare the new and old heights at the same nonzero-output input.
Use g(1) = a × 1 = 4, so a = 4.Dividing the new height 4 by the old height 1 gives the multiplier.
Write g(x) = 4 $x^{3}$ , a vertical stretch by 4.The positive multiplier 4 is greater than 1, so it increases distances from the x-axis without reflection.
Check the other supplied point: 4 × $2 cubed$ = 4 × 8 = 32.A second known point must agree with the candidate scale. The assumed form g(x) = a· $x^{3}$ is what makes one nonzero height sufficient to determine a.

Answer

a = 4.
g(x) = 4 $x^{3}$ .
Vertical stretch by 4.

Check The original cubic gives −1 at x = −1, so the new formula predicts −4 there. The signed height is multiplied by the same factor as the positive heights. The second supplied point independently confirms a = 4.

Rung 7Rung 7: stretch the identity function, then shift

Write the formula obtained by stretching the Identity function f(x) = x vertically by 3, then shifting it down by 2.
This asks you to find the new address or height of the given graph information.

The stretched and lowered line passes through (0, −2) and (1, 1).

Start with f(x) = x. A vertical stretch by 3 gives 3f(x) = 3x.The identity function returns the same number it receives, and the outside multiplier triples that output.
Subtract 2 afterward: g(x) = 3x − 2.The instruction says to shift the stretched graph down, so the subtraction follows the multiplication.
At input 0, the original height 0 becomes −2. At input 1, the original height 1 becomes 3 × 1 − 2 = 1.These two points show the shift and the tripled change in height between neighboring inputs.

Answer

g(x) = 3x − 2.
Points: (0, −2) and (1, 1).
Domain and range: (−∞, ∞).

Check At input 2, g(2) = 6 − 2 = 4. Starting with the original point (2, 2), tripling its height gives 6 and shifting down 2 gives 4. Reversing the order would give 3(x − 2) = 3x − 6, a different function.

Rung 8Rung 8: a negative scale followed by a shift

Describe g(x) = −2 $\sqrt{x}$ + 5 relative to f(x) = $\sqrt{x}$ . Find two points and the domain and range.
This asks you to find the new address or height of the given graph information.

The endpoint stays at input 0 and moves to height 5; the curve falls below it.

The factor −2 stretches vertically by 2 and reflects across the x-axis; +5 then shifts up 5.The magnitude gives the size change, the minus reverses signed outputs, and the added number acts after multiplication.
The original point (0, 0) becomes (0, −2 × 0 + 5) = (0, 5).Keep the input and apply the outside operations to its original height.
The original point (4, 2) becomes (4, −2 × 2 + 5) = (4, 1).Multiplication precedes addition: −4 + 5 = 1.
Domain: [0, ∞). Range: (−∞, 5].The root still needs x ≥ 0. Its outputs start at 0 and grow without bound, so −2 $\sqrt{x}$ + 5 starts at 5 and decreases without bound.

Answer

Stretch vertically by 2, reflect across the x-axis, then shift up 5.
Points: (0, 5) and (4, 1).
Domain: [0, ∞).
Range: (−∞, 5].

Check g(4) = −2 + 5 = −4 + 5 = 1. Its distance from the new line y = 5 is 4, twice the original distance 2 from the x-axis.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A vertical stretch by 2 moves every point up 2.

A stretch multiplies the signed height. Old height 1 becomes 2, old height 4 becomes 8, and old height 0 stays 0. These changes are not one fixed upward distance.

✓ Instead: Write g(x) = 2f(x) and multiply outputs. For f(x) = , (1, 1) becomes (1, 2) and (−2, 4) becomes (−2, 8).

✗ Not this: Compressing height −6 by makes it −12 because it moves downward.

Compression reduces distance from the x-axis. A positive factor preserves the height's sign, and half of six is three.

✓ Instead: The new signed height is × (−6) = −3. It moves closer to the x-axis.

✗ Not this: Stretch by 3, then move down 2 means g(x) = 3(f(x) − 2).

That expression subtracts 2 first and then triples the shifted height, giving 3f(x) − 6. It also scales the downward shift.

✓ Instead: The stated order gives g(x) = 3f(x) − 2. For the Identity function and input 0, the correct height is −2, while the tempting expression gives −6.

Tips and tricks

Outside multiply, heights multiply. Keep the input fixed and apply the factor to the entire signed output.
Say farther from the x-axis for a stretch and closer to the x-axis for a compression. This works for positive and negative heights.
Split a negative multiplier into its magnitude and its sign. For − $1 over 2$ , halve the distance and reverse the height's sign.
With a final +k, multiply the old height first, then add k. Check your new point in the formula after both actions.

Trap. Saying a stretch moves every point upward. A positive scale larger than 1 sends positive outputs up and negative outputs down, increasing distance from the x-axis. Also keep a·f(x) separate from f(ax): the bottom two table rows preview the horizontal-coordinate changes explained in lesson 7.