Horizontal stretch and compression: divide the input coordinates

Explain it like I am five

Imagine playing a recording at double speed. Something that used to happen four minutes into the recording now happens at two minutes. The event has the same content, but you reach it sooner. In g(x) = f(2x), the old function runs through its input addresses twice as fast. Its graph therefore takes half the horizontal space. This is a Horizontal compression. With f( $1 over 2$ x), the old function runs at half speed. Events need twice as much horizontal space, a Horizontal stretch. You find a dot's new position by undoing the multiplication inside. Multiplying the input expression and multiplying the dot's coordinate are different jobs. Keeping track of which job you are doing prevents the most common scaling mistake.

An inside multiplier two divides the point's horizontal coordinate by two, while its height stays two.

Reminder

Dividing by a fraction. Divide by $1 over 2$ by multiplying by 2: 2 ÷ $1 over 2$ = 4. Check by multiplying back: $1 over 2$ × 4 = 2.
Solving for the new input. To reach old input 4 in f(2x), set 2x = 4 to find the matching address. Divide both sides by 2 to get x = 2, then check 2 × 2 = 4.
Flipping an inequality. A negative divisor reverses order. For $square root of −2 x$ , solve −2x ≥ 0 to find allowed inputs: x ≤ 0. Input −2 gives inside 4, which is allowed.

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

Halving each old input produces the corresponding known heights for f(2x).

Doubling each old input produces the corresponding known heights for f(x).

Why it works. Suppose f(u) = v. For g(x) = f(bx), the same height v appears where bx = u. Dividing both sides by b gives x = . Thus the new horizontal coordinate is the old one divided by b. If b = 2, the coordinate is halved. If b = , division by a half doubles it. A negative b also changes the coordinate's sign, reflecting left and right. The size factor is , the reciprocal of the coefficient's magnitude. You cannot divide by b = 0, so this point rule requires b ≠ 0.

Ruleg(x) = f(bx), b ≠ 0, sends (u, v) to (, v). |b| > 1 compresses horizontally; 0 < |b| < 1 stretches horizontally. If b < 0, also reflect across the y-axis.

The same idea, five ways

Say it

An inside multiplier changes where an old height appears. Faster input changes squeeze the graph; slower input changes spread it out.

Write it

For g(x) = f(bx), divide each old horizontal coordinate by the nonzero coefficient b and keep the old height.

In math

bx = u gives x = $\frac{u}{b}$ , with b ≠ 0.
(u, v) becomes ( $\frac{u}{b}$ , v). Width factor: $1 over |b|$ .
|b| > 1: Horizontal compression. 0 < |b| < 1: Horizontal stretch.
b < 0: also a Horizontal reflection. |b| = 1: no width change.

A recording at double speed reaches the same event in half the time.

See it

The same height 2 appears at half its old horizontal coordinate.

The same idea, other ways

As a picture

For $square root of 2 x$ , the old square root point (4, 2) appears at (2, 2). The height 2 survives, but its horizontal distance from the axis is halved.

Doubling inside reaches the old height at half its old horizontal address.

As a recording

Twice as fast means each old event takes half as long to reach. Half as fast means each event takes twice as long. The graph's width measures when events occur, so the width changes by the reciprocal speed.

As a recording

bx = u gives x = , with b ≠ 0.

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

With a fractional coefficient

To make $1 over 2$ x equal old input 6, solve $1 over 2$ x = 6. Multiply by 2 to get x = 12. A smaller inside coefficient therefore requires a larger horizontal coordinate.

With a fractional coefficient

bx = u gives x = , with b ≠ 0.

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

Why it must be true

The new function must receive exactly u inside f to reproduce f(u). Writing bx = u and dividing gives x = u ÷ b. This equation forces the reciprocal factor for every point.

Why it must be true

bx = u gives x = , with b ≠ 0.

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

A memory device

Inside multiply, coordinates divide. Check the phrase with old input 2 and b = 2: the new coordinate 1 feeds 2·1 = 2 into f. This numerical check takes only a few seconds.

A memory device

bx = u gives x = , with b ≠ 0.

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

Old input u	Old height	New input for f(2x)	New input for f( $1 over 2$ x)
2	1	1	4
4	3	2	8
6	7	3	12
8	11	4	16

.1Horizontal compression

Squeeze the left-to-right distances toward the vertical axis. Each old height arrives closer to that axis because the inside multiplier reaches old inputs faster.

Formula: f(bx) with |b| > 1.
Width factor: $1 over |b|$ .
For b = 2, every old input coordinate is divided by 2.
A compression changes distances, while a horizontal shift adds the same amount to all addresses.

The height four now occurs at horizontal distance one instead of two.

Worked exampleCompress the draft's known point

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

(1, 5).

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

Write 2x = 2.The input expression must equal old input 2 to give old height 5.
Divide by 2 to get x = 1.Division undoes the multiplication inside.
Keep height 5.No outside operation changes the output.

Answer

(1, 5).

Check g(1) = f(2·1) = f(2) = 5.

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

✗ Not this: For f(2x), the old point (2, 5) becomes (4, 5).

At new input 4 the inside would be 8. To reach old input 2, solve 2x = 2; input 1 returns the known height 5.

✓ Instead: (1, 5).

Tips and tricks

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

.2Horizontal stretch

Pull the addresses away from the vertical axis. A half-size input multiplier means you travel twice as far before the old machine receives the same number.

Formula: f(bx) with 0 < |b| < 1.
Width factor: $1 over |b|$ .
For b = $1 over 2$ , divide each old input by $1 over 2$ , which multiplies it by 2.
The old input zero remains zero when scaling without a shift.
A horizontal stretch by 4 uses f( $1 over 4$ x), because the inside coefficient is the reciprocal of the width factor 4.

The same height two appears twice as far from the vertical axis.

Worked exampleStretch the draft's known point

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

(4, 5).

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

Solve $1 over 2$ x = 2.The old function must receive its known input 2.
x = 2 ÷ $1 over 2$ = 2·2 = 4.Dividing by a fraction multiplies by its Reciprocal.
Keep height 5.Only the input has changed.

Answer

(4, 5).

Check g(4) = f(·4) = f(2) = 5.

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

✗ Not this: For f(x), the old point (2, 5) becomes (1, 5).

At input 1 the inside is one half, not 2. Divide the old input by one half to get 4; at input 4 the inside is 2.

✓ Instead: (4, 5).

Tips and tricks

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

.3Negative horizontal scale

First think about the amount of squeezing or stretching. Then switch the address to the opposite side of the vertical axis because a negative input multiplier reverses left and right.

The sign of b supplies a Horizontal reflection.
The magnitude |b| supplies a reciprocal width factor.
For g(x) = $square root of −2 x$ , the domain is x ≤ 0 and the range is y ≥ 0.
Changing inputs does not by itself make positive square root outputs negative.

The square root curve is reflected left and compressed to half its old horizontal distances.

Worked exampleReflect and compress a square root

Move old point (4, 2) of $\sqrt{x}$ to g(x) = $square root of −2 x$ .
This asks you to change the coordinate named by each mirror move and keep the other coordinate.

The old point (4, 2) becomes (−2, 2).

Set −2x = 4.Inside the root must equal old input 4.
Divide by −2 to get x = −2.The division halves the distance and reverses its sign.
Keep the old height 2.There is no minus sign outside the square root.

Answer

(−2, 2).

Check = = 2.

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

✗ Not this: For , the old square-root point (4, 2) becomes (−2, −2).

The inside change negates and halves the horizontal coordinate only. There is no outside minus; the square root still returns the positive height 2.

✓ Instead: (−2, 2).

Tips and tricks

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

Strategy: step by step

1. Find b, the multiplier of the input before f receives it.
2. Set bx = u for each known old input u, because this recreates its old height.
3. Divide the old input by b, including the sign. A fractional b calls for division by a fraction.
4. Keep the old height unless an outside change is also present.
5. Move endpoints, corners, and several other points before drawing the curve.
6. Substitute the new input into bx to check that it returns u.

Strategy

Convert an inside coefficient into a width change

Is b = 0?

YesThe usual point rule cannot be used. Check whether f(0) is defined before evaluating the formula.

NoDivide each old input coordinate by b.

↓

Is |b| > 1?

YesThe horizontal distances shrink by .

NoIf 0 < |b| < 1, the distances grow by . If |b| = 1, there is no width change.

↓

Is b negative?

YesThe signed division also reflects inputs across the y-axis.

NoThe division keeps inputs on their original side of zero.

↓

Does a requested new input feed a listed old input into f?

YesUse the output from that same table column.

NoState that the supplied table cannot determine the output; do not guess between entries.

1. Identify b and verify b ≠ 0 before using the coordinate division rule. This prevents dividing by zero.
2. Set bx = u for a known old input u. This equation finds the new address that reproduces its old height.
3. Divide both sides by b, including its sign. For u = 4 and b = −2, x = −2, and substituting back gives −2 × (−2) = 4.
4. Keep the matching old output. For old point (4, 2) of $\sqrt{x}$ , this gives new point (−2, 2) on $square root of −2 x$ .
5. Move the other landmarks using the same division, and rebuild the Domain from the allowed old inputs. In $square root of −2 x$ , require −2x ≥ 0 to find valid inputs, then divide by −2 and reverse the comparison: x ≤ 0.
6. Check a new input inside the formula and use only known table entries. For g(x) = f( $1 over 2$ x), g(12) = f(6) = 7 is known, while g(2) needs the unavailable f(1).

Worked exampleCompare inside scaling with outside scaling

Use old point (2, 4) of f(x) = $x^{2}$ to compare f(2x) with 2f(x). Explain why they are different.
This asks you to find the new address or height of the given graph information.

Inside doubling halves the horizontal address for a given old height.

Outside doubling keeps the address and doubles the old height.

For f(2x), solve 2x = 2 to get x = 1 and keep height 4.An inside multiplier changes where the old input occurs.
For 2f(x), keep x = 2 and multiply height 4 by 2 to get 8.An outside multiplier acts on the completed output.
Write f(2x) = (2x $) squared$ = 4 $x^{2}$ and 2f(x) = 2 $x^{2}$ .Squaring 2x squares both factors: (2x)(2x) = 4 $x^{2}$ .
At x = 1 the first formula gives 4 and the second gives 2.A shared test input reveals that the resulting graphs differ.

Answer

f(2x): horizontal compression by $1 over 2$ , point (1, 4).
2f(x): vertical stretch by 2, point (2, 8).
For $x^{2}$ only, (2x $) squared$ = 4 $x^{2}$ , so this squeezed U can also be read as a taller U. The 2 sits inside, so name the move a horizontal compression.

Check The first graph has height 4 at x = 1; its old height 4 was at x = 2. The second doubles the old height 1 at x = 1 to 2. This agrees with both formulas.

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: divide an input by two

If f(2) = 5, find its matching point on f(2x).
This asks you to find the new address or height of the given graph information.

(1, 5).

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

Solve 2x = 2: x = 1.The new expression must send 2 into f.
Use the old output 5.Inside scaling leaves the corresponding height unchanged.

Answer

(1, 5).

Check f(2·1) = f(2) = 5.

Rung 2Rung 2: divide an input by a half

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

(4, 5).

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

Solve $1 over 2$ x = 2.The old height needs old input 2.
Divide by $1 over 2$ : x = 2 × 2 = 4.A fraction's reciprocal undoes multiplication by it.

Answer

(4, 5).

Check ·4 = 2, returning the old input.

Rung 3Rung 3: stretch a table without inventing values

Use the original f table shown beside this question. Give known values of g(x) = f( $1 over 2$ x). Can this table determine g(2)?
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 g(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.
Double the old input coordinates: 4, 8, 12, 16.Solve $1 over 2$ x = u by multiplying u by 2.
Keep outputs 1, 3, 7, 11.Each new input supplies the corresponding known old input to f.
g(2) = f(1), which is not supplied.The table contains no output at old input 1; its gaps do not imply a formula.

Answer

Use the displayed stretched g table.
g(2) cannot be determined from the given table.

Check g(12) = f(6) = 7. The transformed third column uses a known value instead of guessing between entries.

Rung 4Rung 4: A horizontal stretch by a factor of four

For this separate original point example, suppose F(3) = 7. Write a horizontal stretch by 4 and find where that known point appears. This asks you to spread each input address to four times its old value while keeping its matched height.

The separate original illustration supplies only the point (3, 7).

A fourfold horizontal stretch reaches that same known height at input 12.

Use inside coefficient b = $1 over 4$ , giving G(x) = F( $1 over 4$ x).The horizontal distance factor is the reciprocal of the inside coefficient: 1 ÷ $1 over 4$ = 4.
Solve $1 over 4$ X = 3 by multiplying both sides by 4: X = 12.This finds the new input that sends old input 3 into F. At X = 12, the inside $1 over 4$ × 12 is 3.
Keep the known old height 7, so (3, 7) becomes (12, 7).No outside operation changes the output. The separate function name F distinguishes this illustration from the earlier supplied table.

Answer

G(x) = F( $1 over 4$ x).
Horizontal stretch by 4.
(3, 7) becomes (12, 7).

Check G(12) = F( × 12) = F(3) = 7. Dividing the new input 12 by the requested width factor 4 recovers old input 3.

Rung 5Rung 5: recognize horizontal compression from a graph

An original graph comparison is known to have g(x) = f(bx) with b > 0 and no other changes. The old right endpoint (9, 0) corresponds to the new endpoint (3, 0). Find b and the new point corresponding to old point (6, 4).
This asks you to find the new address or height of the given graph information.

b = 3.

g(x) = f(3x).

Horizontal compression by .

(6, 4) becomes (2, 4).

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

Set 3b = 9.At new input 3, the inside must reach old endpoint input 9.
Divide by 3: b = 3.The coefficient is old input divided by its corresponding new input.
Write g(x) = f(3x), a horizontal compression by $1 over 3$ .The width factor is the reciprocal of the inside coefficient.
Move old point (6, 4) to (2, 4).Divide its input 6 by 3 and keep its output 4.

Answer

b = 3.
g(x) = f(3x).
Horizontal compression by $1 over 3$ .
(6, 4) becomes (2, 4).

Check g(2) = f(6) = 4, and g(3) = f(9) = 0. The stated family and corresponding endpoint are essential; one dot alone cannot identify an arbitrary unknown graph.

Rung 6Rung 6: build a horizontal scale from words

Write the Square root function stretched horizontally by a factor of 4. Find three graph points and its domain and range.
This asks you to write the new rule from the specified moves and check its graph landmarks.

The root height 2 now occurs at input 16 instead of 4.

Choose b = $1 over 4$ .The width factor 4 is the reciprocal of the inside coefficient.
Write g(x) = $square root of x over 4$ .Replace the old input by one-fourth of the new input.
Move old points (0, 0), (1, 1), (4, 2) to (0, 0), (4, 1), (16, 2).Divide each old input by $1 over 4$ , multiplying its coordinate by 4.
Require $x over 4$ ≥ 0, so x ≥ 0. Outputs are nonnegative.The positive horizontal scale does not change the direction of the square root's allowed inputs.

Answer

g(x) = $square root of x over 4$ .
Points: (0, 0), (4, 1), (16, 2).
Domain: [0, ∞).
Range: [0, ∞).

Check g(16) = = 2. A width factor 4 requires inside, not 4.

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

✗ Not this: f(2x) doubles every horizontal coordinate.

The inside expression must return the old input. To reach old input 2, solve 2x = 2, which finds new input 1.

✓ Instead: Divide old coordinates by 2. The old point (2, 5) becomes (1, 5), and substituting 1 gives f(2 × 1) = f(2) = 5.

✗ Not this: f(x) halves the outputs in the table.

The half is inside, so it acts before f returns an output. Reaching old input 2 requires x = 2, giving new input 4; the corresponding table height stays 1, because its displayed old value is f(2) = 1.

✓ Instead: Double the horizontal coordinates and keep matching heights. Halving the outputs would instead give f(x).

✗ Not this: For g(x) = f(x), the table entry f(2) = 1 tells us g(2) = 1.

At new input 2, the inside is 1. The given table has no f(1), so it cannot supply the requested height.

✓ Instead: g(4) = f(2) = 1 is determined. g(2) = f(1) is not determined by the supplied entries.

Tips and tricks

Inside multiply, coordinates divide. Put bx = old input on the exam when a memorized direction feels uncertain.
A width factor 4 needs inside coefficient $1 over 4$ . The coefficient and the width factor multiply to 1.
Divide by the signed b. For old input 4 and b = −2, new input is −2, which handles both the compression and the reflection.
Keep the original heights attached to their matching inputs. A table's gaps remain unknown after the transformation.

Trap. Calling f(2x) a horizontal stretch because 2 is bigger than 1. The multiplier is bigger, but the coordinate must be smaller: 2x = u gives x = u ÷ 2. Also, f(2x) and 2f(x) generally give different graphs.