Put everything together: factor first, then move points

Explain it like I am five

Think of moving a printed map after changing its size on a copier. First you resize it around the original axes. Then you slide it to its new location. Reversing those jobs can resize the slide too, so the order matters. A Sequence of transformations is a set of changes performed in a stated order. The expression a·f(b(x − h)) + k keeps the jobs visible: resize or reflect the horizontal coordinates, slide them, resize or reflect the heights, then lift them. If the inside arrives as 2x − 6, use Factoring to rewrite it as 2(x − 3). That shows the final sideways slide is three, not six.

The final graph has its highest point at (3, 4) because the square's heights were reflected before the upward shift.

Reminder

Factoring every inside term. Divide both terms by the same coefficient: 2x − 6 = 2(x − 3). Distributing 2 back into the parentheses recovers 2x − 6.
Dividing by a signed fraction. For b = $1 over 2$ , u ÷ b = 2u. For b = −2, 4 ÷ b = −2. The signed division handles width and reflection together.
Outside arithmetic order. For old height 5 in −2f(x + 1) + 3, calculate −2 × 5 + 3 = −10 + 3 = −7. The final addition follows multiplication.
The square root restriction. A real square root requires a nonnegative inside. For −2(x + 4) ≥ 0, divide by −2 to find allowed inputs, reverse the comparison, and subtract 4: x ≤ −4. At −4, the inside is 0.
Extracting a square factor. Keep exact roots: $square root of 8$ = $square root of 4 × 2$ = 2 $square root of 2$ , because (2 $square root of 2$ $) squared$ = 8 and 2 $square root of 2$ is nonnegative.

Why it works. Take an old point (u, v), so f(u) = v. To reproduce its old input inside g, solve b(x − h) = u. Divide by b and add h: x = + h. Then the outside gives a·v + k. Both coordinates follow from the formula. Without factoring, bx + c = u gives x = , so the final shift is −. Vertical order matters too: 2(v + 3) = 2v + 6 differs from 2v + 3. Horizontal and vertical changes remain independent because they act on different coordinates.

RuleFor g(x) = a·f(b(x − h)) + k with b ≠ 0, the Point rule is (u, v) to ( + h, a·v + k). Factor first; scale or reflect each coordinate before shifting that coordinate.

The same idea, five ways

Say it

Resize or reflect each coordinate first, then move it to its final address. Factoring makes the horizontal slide visible.

Write it

In g(x) = a·f(b(x − h)) + k, an old point (u, v) moves to horizontal coordinate u ÷ b + h and height a·v + k, with b ≠ 0.

In math

b(X − h) = u gives X = $\frac{u}{b}$ + h. Y = a·v + k.
(u, v) becomes ( $\frac{u}{b}$ + h, a·v + k).
bx + c = b(x + $\frac{c}{b}$ ), so h = − $\frac{c}{b}$ , with b ≠ 0.
Graph words: scale or reflect horizontally, shift horizontally; scale or reflect vertically, shift vertically.

Resize a printed map around its original axes, then slide the resized map into its final position.

See it

The moved square points show right 3, reflected double heights, and a final upward shift of 4.

The same idea, other ways

As a picture

The old square graph first flips downward and doubles its heights. Then its old lowest point at (0, 0) moves to (3, 4). The new graph opens down because the coefficient of its heights is negative.

The old zero height becomes four, while an old height one becomes two after −2·1 + 4.

As two address jobs

Horizontal address: divide, then add. Height: multiply, then add. For old point (2, 5) and −2f(x + 1) + 3, these jobs give 2 − 1 = 1 and −2·5 + 3 = −7.

As two address jobs

b(X − h) = u gives X = + h. Y = a·v + k.

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

Why it must be true

Set the inside expression equal to old input u. The equation b(x − h) = u gives x = u ÷ b + h. Once f receives u, it produces old output v, and the remaining outside work produces av + k.

Why it must be true

b(X − h) = u gives X = + h. Y = a·v + k.

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

With an order comparison

Start with old height 1. Double and then add 3: 2·1 + 3 = 5. Add 3 and then double: 2(1 + 3) = 8. The slide is doubled in the second calculation, which is why the order cannot be swapped on the same coordinate.

With an order comparison

b(X − h) = u gives X = + h. Y = a·v + k.

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

A memory device

Factor, divide, multiply, then shift. Factoring exposes h. Divide the horizontal coordinate by b; multiply the height by a. Shift each coordinate after its scale. Keep the one-page table handy while learning and rebuild the point rule from its equation.

A memory device

b(X − h) = u gives X = + h. Y = a·v + k.

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

Put on the cheat sheet	Coordinate effect or check
f(x) + k	(u, v) to (u, v + k)
f(x − h)	(u, v) to (u + h, v)
−f(x)	(u, v) to (u, −v), reflection across x-axis
f(−x)	(u, v) to (−u, v), reflection across y-axis
a·f(x), a ≠ 0	Height factor \|a\|; reflect vertically when a < 0
f(bx), b ≠ 0	Width factor $1 over \|b\|$ ; reflect horizontally when b < 0
a·f(b(x − h)) + k	(u, v) to ( $\frac{u}{b}$ + h, a·v + k)
bx + c = b(x + $\frac{c}{b}$ )	Final horizontal shift h = − $\frac{c}{b}$
Even function	Symmetric domain and f(−x) = f(x)
Odd function	Symmetric domain and f(−x) = −f(x)
$square root of b(x − h)$	Require b(x − h) ≥ 0
$1 over b(x − h)$	Require b(x − h) ≠ 0

.1Factor the inside

Pull the multiplier on x outside the entire parentheses. Like sorting equal bundles, divide every inside term by the same number so the expression keeps its value.

2x − 6 = 2(x − 3), because 2·3 = 6.
The final shift is right 3 when the inside is 2(x − 3).
In bx + c, b ≠ 0, the final horizontal shift is − $\frac{c}{b}$ .
If b = 0, the inside no longer varies with x; the usual point rule does not apply.

2x − 6 = 2(x − 3)

b = 2, h = 3

new x = old x ÷ 2 + 3

The multiplier two belongs to both terms, so the shift is three after factoring.

Worked exampleFind the hidden shift

Describe f(2x − 6).
This asks which horizontal resizing and shift are encoded in the given input expression. Factor it to reveal both changes.

Horizontal compression by , then right 3.

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

Rewrite 2x − 6 as 2(x − 3).Dividing both terms by 2 leaves x − 3 inside.
Read b = 2 and h = 3.The factored form matches b(x − h).
Compress the horizontal coordinates by $1 over 2$ , then shift right 3.New x is old x ÷ 2 + 3.

Answer

Horizontal compression by , then right 3.

Check Distribute 2(x − 3) to recover 2x − 6. An old input zero appears at x = 3 because 2·3 − 6 = 0.

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

✗ Not this: f(2x − 6) compresses first, then moves right 6.

Factoring gives 2(x − 3). The final shift after division by 2 is 3; old input 0 is reached at new input 3, not 6.

✓ Instead: Horizontal compression by , then right 3.

Tips and tricks

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

.2Move the horizontal coordinate

Recover the old input before thinking about the height. A negative divisor reverses left and right; then the shift carries the already resized coordinate to its final address.

New horizontal coordinate: X = $\frac{u}{b}$ + h.
When b < 0, divide with its sign and then add h.
The alternative unfactored equation bx + c = u gives the same X.
Do not add h before dividing unless you have correctly adjusted the formula.

Matching old input four fixes the new horizontal coordinate without guessing the direction.

Worked exampleMove the draft's known input

Old point (4, 1) is on f. Move it to f(2x − 6).
This asks you to find the new address or height of the given graph information.

(5, 1).

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

Solve 2X − 6 = 4.The old output 1 belongs to old input 4.
Add 6: 2X = 10. Divide by 2: X = 5.Undo subtraction and multiplication in order.
Keep output 1.There is no outside change.

Answer

(5, 1).

Check Using the factored rule instead gives 4 ÷ 2 + 3 = 5, the same address.

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

✗ Not this: Old point (4, 1) becomes (7, 1) on f(2x − 6).

At new input 7 the inside is 8, not the supplied old input 4. Solving 2X − 6 = 4 gives X = 5.

✓ Instead: (5, 1).

Tips and tricks

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

.3Move the vertical coordinate

After the old function gives its height, the outside operations finish the job. Multiply that height, including any sign, and then add the final lift.

New vertical coordinate: Y = a·v + k.
If a < 0, reflect vertically and scale by |a| before shifting.
If a = 0, every permitted input has output k; this is a collapse, not an ordinary stretch.
The domain of the expression still requires its inner function to be defined.

The old height is multiplied first, then shifted upward by three.

Worked exampleFinish a negative height scale

For old height 5, find the new height in −2f(x + 1) + 3.
This asks you to find the new address or height of the given graph information.

New height: −7.

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

Multiply 5 by −2 to get −10.The outside coefficient reflects and doubles the height.
Add 3 to get −7.The upward shift follows the multiplication.

Answer

New height: −7.

Check Undo the outside changes: (−7 − 3) ÷ (−2) = 5, the old height.

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

✗ Not this: For old height 5 in −2f(x + 1) + 3, add 3 first and then multiply: −2(5 + 3) = −16.

The written outside operations multiply before adding. Compute −2 × 5 + 3 = −10 + 3 = −7.

✓ Instead: New height: −7.

Tips and tricks

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

.4Build and sketch the whole graph

Choose landmarks from a familiar shape, move them using both coordinate jobs, and keep the original connections. A shifted endpoint or corner gives the graph an anchor.

The vertex of a·(x − h $) squared$ + k is (h, k) when a ≠ 0.
Positive a opens the quadratic up; negative a opens it down.
For a < 0, the largest quadratic height is k, so its range is y ≤ k.
Horizontal and vertical work can be interleaved, as long as each coordinate keeps its own scale-before-shift order.

The three moved landmarks anchor a downward-opening quadratic with highest height four.

Worked exampleUse three points to fix the graph

Move the old square points (−1, 1), (0, 0), (1, 1) to −2(x − 3 $) squared$ + 4.
This asks you to find the new address or height of the given graph information.

The moved landmarks are (2, 2), (3, 4), and (4, 2).

Add 3 to each old horizontal coordinate: 2, 3, 4.The inside x − 3 shifts the graph right 3.
Transform heights 1, 0, 1 using −2v + 4 to get 2, 4, 2.The outside reflection and stretch occur before the upward shift.
Draw a downward-opening quadratic through (2, 2), (3, 4), (4, 2).The negative height coefficient reverses the original opening.

Answer

Points: (2, 2), (3, 4), (4, 2).
Vertex: (3, 4).

Check At x = 2 and x = 4, the formula gives −2·1 + 4 = 2, matching both side points.

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

✗ Not this: The graph −2(x − 3 + 4 has its lowest point at (3, 4) and opens upward.

The negative outside coefficient turns all nonzero square heights downward. Its vertex is the highest point, and the side heights at inputs 2 and 4 are 2, below 4.

✓ Instead: Points: (2, 2), (3, 4), (4, 2).
Vertex: (3, 4).

Tips and tricks

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

Strategy: step by step

1. Recognize f and its domain, range, and landmarks.
2. Factor the entire input into b(x − h). Distribute it back out to verify the factoring.
3. Label a, b, h, and k, keeping every sign.
4. For each old point (u, v), compute new x = u ÷ b + h.
5. Compute new y = a·v + k. Multiply before adding because that is the outside formula's order.
6. Move at least three points and any asymptotes or endpoints; draw the transformed shape.
7. Find the new domain from b(x − h) belonging to the old domain, and the new range from a·v + k over old range values.
8. Substitute a moved point into the original unfactored formula as a second check.

Strategy

Factor and transform a whole graph or table

Is the inside already written as b(x − h)?

YesRead b and h with their signs.

NoFactor the nonzero coefficient on x out of every term, then distribute to check it.

↓

Is b nonzero?

YesFor each known old input, divide by b and then add h.

NoThe usual point rule is unavailable. Determine whether the constant inside is accepted by f before evaluating the formula.

↓

Is an outside multiplier present?

YesMultiply the old signed height by a, then add k.

NoUse a = 1 and add only the outside shift k.

↓

Does the original function have an endpoint or a forbidden input?

YesTransform that landmark and solve the inner restriction to find the new Domain.

NoUse the original allowed-input information and the inner equation to determine the new Domain.

↓

Is a complete original graph supplied?

YesMove its landmarks and keep its original connections.

NoReport the transformed known entries and leave unavailable heights undetermined.

1. Name the original Function and its available points. Record restrictions before moving the graph so a forbidden input cannot become a supplied value.
2. Factor the inside into b(x − h), with b ≠ 0. This reveals the shift after the horizontal scale; 2x − 6 = 2(x − 3) identifies b = 2 and h = 3. Distribute to check the rewrite.
3. Solve b(X − h) = u for each old input u. This finds where its old output appears: divide by b and add h to obtain X = $\frac{u}{b}$ + h.
4. Compute Y = a·v + k from its matching old output v. This applies the outside scale before the final shift.
5. Move landmarks and several other points, or build the transformed table column by column. Keep every old input paired with its own old output.
6. Rebuild the Domain from the inner restrictions and the Range from the transformed heights. For −2(x − 3 $) squared$ + 4, all inputs remain allowed and the nonnegative square shows that heights cannot exceed 4; the vertex reaches 4.
7. Check using the original expression. For old point (4, 1) and g(x) = f(2x − 6), new input 5 gives g(5) = f(4) = 1, confirming the moved point.

Worked exampleBuild the draft's four-change quadratic

Start with $x^{2}$ . Reflect across the x-axis, stretch vertically by 2, move right 3, and move up 4. Write the formula and sketch the graph.
This asks you to write the new rule from the specified moves and check its graph landmarks.

The new graph opens downward with highest point (3, 4).

Reflect and stretch with coefficient a = −2: −2 $x^{2}$ .The negative sign reverses heights and magnitude 2 doubles their distances from the original x-axis.
Replace x with x − 3: −2(x − 3 $) squared$ .A rightward shift needs new inputs three larger to reproduce the old inputs.
Add 4 outside: g(x) = −2(x − 3 $) squared$ + 4.The final upward shift raises the already reflected and stretched heights.
Move (0, 0) to (3, 4), (1, 1) to (4, 2), and (−1, 1) to (2, 2).New x = u + 3 and new y = −2v + 4.
Domain is all real numbers and range is y ≤ 4.Squaring accepts all real inputs, and −2(x − 3 $) squared$ ≤ 0 gives a maximum height of 4.

Answer

g(x) = −2(x − 3 $) squared$ + 4.
Vertex: (3, 4).
Opening: downward.
g(4) = 2.
Domain: (−∞, ∞).
Range: (−∞, 4].

Check g(4) = −2(4 − 3 + 4 = −2 + 4 = 2. Also every square is nonnegative, so no point can rise above 4; g(3) reaches 4.

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: move one point with inside and outside changes

The point (2, 5) is on f. Where does it go on g(x) = −2f(x + 1) + 3?
This asks you to find the new address or height of the given graph information.

(1, −7).

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

Solve X + 1 = 2, giving X = 1.The input inside f must recover old input 2.
Compute Y = −2·5 + 3 = −10 + 3 = −7.Multiply the old height before adding the outside shift.

Answer

(1, −7).

Check g(1) = −2f(2) + 3 = −2·5 + 3 = −7.

Rung 2Rung 2: factor an input before reading its shift

Describe g(x) = f(2x − 6) and move old point (4, 1).
This asks you to find the new address or height of the given graph information.

Compress horizontally by , then shift right 3.

(4, 1) becomes (5, 1).

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

Factor 2x − 6 = 2(x − 3).The coefficient on x multiplies the whole parentheses.
Read b = 2 and h = 3.The factored expression is in the standard input form.
Compute new X = 4 ÷ 2 + 3 = 5 and keep Y = 1.Divide the horizontal coordinate first, then shift it; there is no outside change.

Answer

Compress horizontally by $1 over 2$ , then shift right 3.
(4, 1) becomes (5, 1).

Check g(5) = f(2·5 − 6) = f(4) = 1.

Rung 3Rung 3: transform a complete table

Create the table for g(x) = 2f(3x) + 1. Use the original f table displayed beside this question.
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 f(3x) output.

Read each column as one x input paired with its 2f(3x) 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.
Divide inputs by 3: 2, 4, 6, 8.3X must equal each old input.
Double outputs: 20, 28, 30, 34.The outside coefficient 2 stretches the old heights.
Add 1 to the doubled outputs: 21, 29, 31, 35.The final vertical shift follows the stretch.

Answer

Use the displayed transformed table
each column records one new input and its corresponding output.

Check g(6) = 2f(18) + 1 = 2·15 + 1 = 31, checking a full inside-to-outside calculation.

Rung 4Rung 4: stretch, slide left, and lower a graph

The graph of f contains (0, 2) and (2, 0). Move these points to g(x) = f( $1 over 2$ x + 1) − 3.
This asks you to find the new address or height of the given graph information.

Horizontal stretch by 2, then left 2, then down 3.

(0, 2) becomes (−2, −1).

(2, 0) becomes (2, −3).

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

Factor $1 over 2$ x + 1 = $1 over 2$ (x + 2).Multiplying $1 over 2$ by 2 returns the original constant 1.
Use b = $1 over 2$ , h = −2, a = 1, k = −3.x + 2 is x − (−2).
Stretch horizontal coordinates by 2: (0, 2) stays (0, 2), while (2, 0) becomes (4, 0).Divide old inputs by $1 over 2$ .
Shift left 2: the points become (−2, 2) and (2, 0).Add h = −2 after the horizontal scaling.
Shift down 3: the points become (−2, −1) and (2, −3).Subtract 3 from the heights after the horizontal work.

Answer

Horizontal stretch by 2, then left 2, then down 3.
(0, 2) becomes (−2, −1).
(2, 0) becomes (2, −3).

Check g(−2) = f(0) − 3 = 2 − 3 = −1 and g(2) = f(2) − 3 = −3. No missing graph shape is assumed from these two points.

Rung 5Rung 5: build a reflected and shifted quadratic

Reflect $x^{2}$ across the x-axis, stretch it vertically by 2, then move it right 3 and up 4.
This asks you to write the new rule from the specified moves and check its graph landmarks.

Reflect and stretch before raising the vertex to (3, 4).

Use −2 to reflect and stretch the original heights.The sign gives the flip and the magnitude gives the scale.
Use x − 3 inside squaring and add 4 after multiplying: g(x) = −2(x − 3 $) squared$ + 4.The horizontal shift is undone inside; the final vertical shift changes the completed height.
Find g(3) = 4 and g(4) = 2.The vertex uses zero inside and a neighboring point uses one inside.

Answer

g(x) = −2(x − 3 $) squared$ + 4.
Vertex: (3, 4).
Opening: downward.
g(4) = 2.

Check The old point (1, 1) moves to (4, −2·1 + 4) = (4, 2), agreeing with direct substitution.

Rung 6Rung 6: negative inside, negative outside, and exact fractions

Starting from f(x) = $\sqrt{x}$ , graph g(x) = −3 $square root of −2(x + 4)$ + 5. Move old points (0, 0), (1, 1), (4, 2), and (8, 2 $square root of 2$ ). State its domain and range.
This asks you to find the new address or height of the given graph information.

The endpoint is (−4, 5), and the curve extends leftward and downward.

Read a = −3, b = −2, h = −4, and k = 5.x + 4 equals x − (−4), so the factored input already exposes the shift.
Use X = u ÷ (−2) − 4 and Y = −3v + 5.The point rule divides and shifts the horizontal coordinate, then multiplies and shifts the height.
Move (0, 0) to (−4, 5). Move (1, 1) to (− $9 over 2$ , 2).0 ÷ (−2) − 4 = −4; 1 ÷ (−2) − 4 = − $1 over 2$ − $8 over 2$ = − $9 over 2$ . Heights are 5 and 2.
Move (4, 2) to (−6, −1) and (8, 2 $square root of 2$ ) to (−8, 5 − 6 $square root of 2$ ).4 ÷ (−2) − 4 = −6 and 8 ÷ (−2) − 4 = −8. Apply −3v + 5 to each old height.
Solve −2(x + 4) ≥ 0. Dividing by −2 reverses the inequality: x + 4 ≤ 0, so x ≤ −4.The radicand must be nonnegative, and multiplying an inequality by a negative reverses its order.
The range is y ≤ 5.The nonnegative root is multiplied by −3, producing nonpositive values, then raised by 5; unbounded roots give arbitrarily low heights.

Answer

Endpoint: (−4, 5).
Other points: (− $9 over 2$ , 2), (−6, −1), (−8, 5 − 6 $square root of 2$ ).
Domain: (−∞, −4].
Range: (−∞, 5].

Check At X = −6 the inside is −2(−6 + 4) = 4, so g(−6) = −3·2 + 5 = −1. At X = −8 the inside is 8, whose exact root is 2, giving the stated exact height.

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

✗ Not this: The graph of f(2x − 6) is compressed horizontally and then shifted right 6.

The inside constant is still part of the multiplier's expression. Factoring gives 2(x − 3), so the final shift after dividing coordinates by 2 is 3.

✓ Instead: Compress by , then move right 3. Old point (4, 1) becomes (5, 1), and 2 × 5 − 6 = 4 checks its new input.

✗ Not this: Move the old input right 3 and then halve it for f(2(x − 3)).

This would compute (u + 3) ÷ 2. Solving the actual input equation 2(X − 3) = u gives X = + 3, so the coordinate division comes first.

✓ Instead: For old input 4, halve 4 to get 2 and then add 3 to get 5. Shifting first would give , whose inside 2( − 3) = 1 fails to recover old input 4.

✗ Not this: The outside expression 2f(x) + 3 permits adding 3 to the old height before doubling.

Doubling a shifted height also doubles the shift. For old height 1, 2(1 + 3) = 8, while the given expression gives 2 × 1 + 3 = 5.

✓ Instead: Multiply each old height by 2, then add 3. Order matters for a scale and a shift on the same coordinate.

Tips and tricks

Write a, b, h, and k beside the factored formula before moving any point. Include each negative sign.
Use two lines for the point work: new x = old x ÷ b + h, and new y = a·old y + k.
Factor, divide, multiply, then shift is a memory cue. Keep each coordinate's scaling before its shift; horizontal and vertical work can be done independently.
Check a moved point in the original unfactored input. This can catch both a factoring error and a reversed transformation order.
Track a corner, endpoint, or asymptote and more than one other point. A few supplied dots determine those dots' new positions, while a complete sketch also requires the original graph's shape.

Trap. Using 6 as the final shift in f(2x − 6), or shifting a height before scaling it. Factor the input and write the two coordinate equations. For a·f(b(x − h)) + k, divide then add h horizontally, and multiply then add k vertically.