Reflections: the minus sign tells you which coordinate changes

Explain it like I am five

Imagine tracing a drawing on clear plastic and turning it over along one edge. You get a mirror image, with each dot the same distance from that edge on the other side. A Reflection does that to a graph. A Vertical reflection flips heights across the horizontal x-axis. A Horizontal reflection flips left and right across the vertical y-axis. You can tell which flip to make by finding the minus sign. A minus outside the function changes its answer. A minus inside changes the input that reaches the old answer. The axis names describe the mirror line; vertical and horizontal describe the direction the points move.

Reflecting across the vertical axis moves its allowed inputs to t ≤ 0 and keeps its heights nonnegative.

Reminder

Taking the opposite. The opposite reverses the sign: −(−3) = 3 and −0 = 0. Do not subtract a fixed distance.
Replacing every input. For f(x) = (x − 6 $) squared$ , substitute (−x) everywhere x appears: f(−x) = (−x − 6 $) squared$ . The constant −6 is part of the original rule.
Reversing an inequality. Multiplying or dividing both sides by a negative reverses order: −x ≥ 0 becomes x ≤ 0 after multiplying by −1. For numbers, 4 > 2 becomes −4 < −2.
Distributing an outside minus. Negating the whole output changes every term: −(x − 4) = −x + 4 because −1 multiplies both x and −4.

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

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

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

Why it works. If f(u) = v, then −f(u) = −v. The same input now gives the opposite output, so (u, v) becomes (u, −v). For f(−x), solve −x = u: x = −u. At that new input the function receives u and returns v, so the point becomes (−u, v). Zero stays zero under either sign change. Distances from the mirror line stay equal because a number and its negative have the same size.

RuleVertical reflection: g(x) = −f(x), with (u, v) becoming (u, −v).
Horizontal reflection: h(x) = f(−x), with (u, v) becoming (−u, v).

The same idea, five ways

Say it

Say: across the x-axis, keep the input and take the opposite height. Across the y-axis, take the opposite input and keep the height.

Write it

A Reflection puts each graph point the same distance from a mirror axis on the other side of that axis.

In math

Vertical reflection: g(x) = −f(x); old (u, v), new (u, −v).
Horizontal reflection: H(x) = f(−x); old (u, v), new (−u, v).
For $square root of −x$ : Domain x ≤ 0, (−∞, 0], {x | x ≤ 0}.
Range y ≥ 0, [0, ∞), {y | y ≥ 0}.
Graph words: flip heights across the horizontal axis, or flip addresses across the vertical axis.

A mirror puts a dot on the opposite side at the same distance. The location of the mirror decides whether height or sideways address reverses.

See it

Across the y-axis, old input 4 becomes new input −4 while the height stays 2.

The same idea, other ways

As a mirror

A Vertical reflection puts each height on the opposite side of the horizontal x-axis. The point (4, 4) becomes (4, −4). A Horizontal reflection uses the vertical y-axis as its mirror and sends (4, 4) to (−4, 4). Name the mirror line separately from the coordinate that moves.

A vertical flip negates the height and keeps its distance from the x-axis equal.

As a before-or-after machine change

For −f(x), run f first and take the opposite of its answer afterward. For f(−x), take the opposite of the input first and then run f. Take f(x) = 2 $\sqrt{x}$ . At new input −4, f(−x) = 2 $square root of −x$ hands f the input 4, and f returns 4. At input 4, −f(x) = −2 $\sqrt{x}$ returns −4.

outside: −2 = −4

inside: 2 = 2 = 4

after f: change the height

before f: change the input

Moving the minus sign changes which job is performed before or after the original function.

As opposite addresses

To keep old output f(4) = 2 in f(−x), solve −x = 4. Multiplying both sides by −1 gives x = −4. The equation therefore forces the reflected horizontal address. For −f(x), no input equation changes: keep 4 and negate the old height 2.

Across the y-axis, old input 4 becomes new input −4 while the height stays 2.

Original input u	Original output f(u)	−f(u) at input u	Reflected input −u	f(−x) at x = −u
2	1	−1	−2	1
4	3	−3	−4	3
6	7	−7	−6	7
8	11	−11	−8	11

.1Vertical reflection

Picture a dot three steps below a tabletop. Its mirror image is three steps above the tabletop. A Vertical reflection works with signed heights in the same way: a negative answer becomes positive, and a positive answer becomes negative. You keep the horizontal address. This is an Outside change because it happens after the function has returned its answer.

Formula: g(x) = −f(x). Point rule: (u, v) becomes (u, −v).
The Domain stays the same. The Range contains the negatives of the original range values.
A point on the x-axis stays fixed because −0 = 0. A reflection is a flip, not a shift by a fixed number.

The line y = x − 4 reflects to y = 4 − x, and their shared point on the x-axis stays fixed.

Worked exampleA negative height becomes positive

Reflect f(x) = x − 4 vertically and follow its point at x = 1.
This asks you to find the new address or height of the given graph information.

The point (1, −3) reflects to (1, 3).

f(1) = 1 − 4 = −3, so the original point is (1, −3).The input is the horizontal coordinate, and the function value is the signed height.
Write g(x) = −(x − 4) = −x + 4.Negate the whole output. Distributing −1 changes the signs of both terms.
The reflected point is (1, 3).The input stays 1 and the opposite of −3 is 3.

Answer

g(x) = −x + 4.
(1, −3) becomes (1, 3).

Check g(1) = −1 + 4 = 3. The two points are both three units from the x-axis.

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

✗ Not this: The vertical reflection of f(x) = x − 4 is −x − 4.

The outside minus acts on the whole output: −(x − 4) = −x + 4. At input 1, the old height −3 becomes 3, not −5.

✓ Instead: g(x) = −x + 4.
(1, −3) becomes (1, 3).

Tips and tricks

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

.2Horizontal reflection

Picture reflecting a street map in a mirror: an address two blocks east now appears two blocks west, while its northward distance stays the same. A Horizontal reflection negates every horizontal coordinate. This is an Inside change because the function receives −x. Substitute that expression everywhere the old input appears, including inside a square or a root.

Formula: h(x) = f(−x). Point rule: (u, v) becomes (−u, v).
The Domain is reflected across zero; the Range stays the same.
A graph already Symmetric about the y-axis is unchanged by this reflection. For example, (−x $) squared$ = $x^{2}$ , so the Quadratic function $x^{2}$ looks the same after the flip.

The vertex of (x − 6 at (6, 0) moves to (−6, 0) when the horizontal coordinate changes sign.

Worked exampleReflect a shifted quadratic

Reflect f(x) = (x − 6 $) squared$ horizontally. Find the new vertex.
This asks you to change the coordinate named by each mirror move and keep the other coordinate.

The vertex moves from x = 6 to x = −6 when the horizontal coordinate changes sign.

Replace x with (−x): h(x) = (−x − 6 $) squared$ .A horizontal reflection changes the input before the original function is evaluated.
Factor inside the square: −x − 6 = −(x + 6), so h(x) = (−(x + 6) $) squared$ = (x + 6 $) squared$ .Squaring a negative expression multiplies two negatives and gives the same result as squaring its positive opposite.
The original vertex (6, 0) becomes (−6, 0). Domain: all real numbers. Range: [0, ∞).Negate the horizontal coordinate. Squares are still nonnegative, and every real input remains allowed.

Answer

h(x) = (x + 6 $) squared$ .
Vertex: (−6, 0).
Domain: (−∞, ∞).
Range: [0, ∞).

Check h(−6) = 0, and h(−7) = 1 = f(7). The point (7, 1) has moved to (−7, 1).

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

✗ Not this: Horizontally reflecting (x − 6 leaves its vertex at (6, 0).

Replacing every x by −x gives (−x − 6 = (x + 6, whose vertex is (−6, 0). The horizontal address reverses sign.

✓ Instead: h(x) = (x + 6.
Vertex: (−6, 0).
Domain: (−∞, ∞).
Range: [0, ∞).

Tips and tricks

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

.3Both reflections and a shift

Imagine flipping a paper drawing left to right, then top to bottom, and finally sliding it upward. Keep a record of each dot after each action. For L(n) = −f(−n) + 1, you negate its horizontal address, negate its height, then add 1 to that height. You can trace known points even when you do not know a formula for f.

Formula: L(n) = −f(−n) + 1. A point (u, v) becomes (−u, 1 − v).
You can trace a learning-model transformation from known points even when an original formula is not given. Here the original points are (0, 1) and (1, 2).
A learning model needs nonnegative practice-session inputs and an appropriate mastery scale. A transformed point at n = −1 is an algebraic point, not a physically allowed negative practice session.
Reflecting $x^{3}$ across both axes returns its graph to the same place: the new formula −(−x $) cubed$ equals $x^{3}$ . The next lesson calls this matching behavior an odd function and explains the full test.

L(n) = −f(−n) + 1

(0, 1) → (0, 1) → (0, −1) → (0, 0)

(1, 2) → (−1, 2) → (−1, −2) → (−1, −1)

Follow each point through the horizontal reflection, vertical reflection, and upward shift in that order.

Worked exampleTrace two supplied learning-model points

Given f(0) = 1 and f(1) = 2, trace their points to L(n) = −f(−n) + 1 without guessing a formula for f.
This asks you to find the new address or height of the given graph information.

(0, 1) becomes (0, 0).

(1, 2) becomes (−1, −1).

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

Horizontal reflection: (0, 1) becomes (0, 1), and (1, 2) becomes (−1, 2).The inside minus negates each input coordinate; −0 = 0.
Vertical reflection: those points become (0, −1) and (−1, −2).The outside minus negates the output coordinates.
Shift up 1: the final points are (0, 0) and (−1, −1).Adding 1 outside gives −1 + 1 = 0 and −2 + 1 = −1.

Answer

(0, 1) becomes (0, 0).
(1, 2) becomes (−1, −1).

Check L(0) = −f(0) + 1 = −1 + 1 = 0. L(−1) = −f(1) + 1 = −2 + 1 = −1. These checks use only the two supplied values.

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

✗ Not this: For −f(−n) + 1, the supplied point (1, 2) becomes (−1, 3).

The outside minus must negate 2 before the addition. The new height is −2 + 1 = −1, while the inside minus moves input 1 to −1.

✓ Instead: (0, 1) becomes (0, 0).
(1, 2) becomes (−1, −1).

Tips and tricks

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

.4A specified learning-model equation

Treat practice like filling a progress gauge. Mastery means how much of a skill you have learned. A percentage measures an amount out of 100. This original gauge starts at 0 percent, grows with more practice, and approaches 100 percent. A formula supplies its entire curve; two isolated points could not tell you all that. Practice is continuous here, meaning every real amount, including part of a session, is allowed.

This practice model uses an explicitly specified old formula and a stated contextual input restriction.
L(n) = −f(−n) + 100 with f(x) = $300 over 3 − x$ gives L(n) = 100 − $300 over n + 3$ .
With continuous nonnegative practice amounts, the contextual domain is [0, ∞) and the range is [0, 100).
If n counts completed whole sessions only, use only whole-number inputs, with the separate output values at those inputs.

The original continuous practice model starts at 0 percent and approaches 100 percent from below.

Reminder

Function substitution. Replace every old x by (−n): 3 − x becomes 3 − (−n) = 3 + n.

The same idea, five ways

Say it

Say: mastery is one hundred minus the unfilled amount, with zero or more practice.

Write it

This original learning-model equation gives a continuous percentage that starts at zero and stays below one hundred.

In math

L(n) = 100 − $300 over n + 3$
n ≥ 0; contextual domain [0, ∞).
0 ≤ L(n) < 100; contextual range [0, 100).

A progress gauge whose remaining gap keeps shrinking.

See it

The original continuous practice model starts at 0 percent and approaches 100 percent from below.

Worked exampleAn original mastery model with both reflections and an upward shift

For this original example let f(x) = $300 over 3 − x$ . Write L(n) = −f(−n) + 100 and calculate L(0) and L(3). Here n is a nonnegative amount of practice measured in sessions, with partial sessions allowed, and L is a percentage of mastery. This asks you to substitute the opposite input, negate the answer, and add the final 100.

The original continuous practice model starts at 0 percent and approaches 100 percent from below.

Replace x by −n: f(−n) = $300 over 3 − (−n)$ = $300 over 3 + n$ .The inside reflection substitutes the opposite input everywhere. Subtracting a negative adds.
Negate that output and shift up 100: L(n) = 100 − $300 over n + 3$ .The outside minus reflects heights and the outside addition follows it. This is a completely specified original formula.
L(0) = 100 − $300 over 3$ = 100 − 100 = 0.Zero practice gives the starting mastery 0 percent, and the denominator 3 is nonzero.
L(3) = 100 − $300 over 6$ = 100 − 50 = 50.At three sessions of practice the denominator is 6, so the reciprocal term is 50 and the mastery is 50 percent.
For n ≥ 0, the denominator n + 3 is at least 3. The reciprocal term is positive and at most 100. Thus 0 ≤ L(n) < 100.As practice grows, the denominator grows and that positive term shrinks toward zero. The model approaches 100 percent without reaching it in a finite amount of practice.

Answer

L(n) = 100 − $300 over n + 3$ .
L(0) = 0 percent.
L(3) = 50 percent.
Contextual domain: [0, ∞).
Contextual range: [0, 100).

Check For any proposed mastery y with 0 ≤ y < 100, the input n = − 3 is nonnegative and substitution returns L(n) = y. This is the reciprocal-equation method from the refresher. At y = 50 it gives n = 6 − 3 = 3. If only whole sessions were allowed, the graph would consist of the values at whole-number inputs and this continuous interval would not be its exact range.

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

✗ Not this: At three sessions the mastery is 100 percent because n is positive.

Substitution gives 100 − 300 ÷ 6 = 50. The reciprocal term stays positive at every finite allowed n, so the height cannot reach 100.

✓ Instead: L(3) = 50 percent. The upper boundary 100 is approached, not included.

Tips and tricks

Write the contextual input restriction before reporting a model range. Distinguish partial-session amounts from completed whole-session counts.

Strategy: step by step

1. Locate the minus sign. Outside the function means negate the output; inside with x means negate the input coordinate.
2. Choose points on the original graph. Change only the coordinate named by the reflection.
3. Check the Domain, the allowed inputs, and the Range, the resulting outputs. A horizontal reflection negates the domain values; a vertical reflection negates the range values.
4. Substitute a transformed point into the new formula. Confirm that it gives the transformed height.

Strategy

Choose the reflection, move points, and check restrictions

Is the minus sign outside f?

YesNegate the entire output. Keep the input coordinate; this is reflection across the x-axis.

NoCheck whether the input sent into f is −x.

↓

Is the old function receiving −x?

YesNegate the horizontal coordinate. Keep the old output; this is reflection across the y-axis.

NoThat formula is not the pure inside reflection considered here.

↓

Are both minus signs present?

YesApply both flips, keeping a record of the point after each. Then apply any final outside shift.

NoApply the single stated flip.

↓

Does the reflected formula contain a real square root?

YesRequire its inside ≥ 0. An inside negative can reverse which side of the axis contains allowed inputs.

NoKeep the original function's other restrictions and reflect the relevant input or output values.

1. Translate the request: identify the mirror line or locate the minus sign in the formula.
2. For an outside minus, keep each old input and negate its output. For an inside minus, negate each old input and keep its output.
3. Write the reflected formula by negating the whole original output, or by replacing every occurrence of its input with (−x).
4. Track the original restrictions. A horizontal flip negates the allowed input addresses; a vertical flip negates the possible heights.
5. If another outside shift follows, negate the old height first, then add the shift.
6. Substitute a reflected point into the formula. Check both its new height and that its input is allowed.

Worked exampleTwo reflections of s(t) = 2

Write the vertical and horizontal reflections of s(t) = 2 $\sqrt{t}$ . Give each domain and range.
This asks you to change the coordinate named by each mirror move and keep the other coordinate.

The vertical reflection reverses heights and keeps nonnegative inputs.

The horizontal reflection reverses inputs and keeps nonnegative heights.

The original function accepts t ≥ 0 and returns s(t) ≥ 0.A real square root needs a nonnegative number inside it, and doubling a nonnegative root keeps it nonnegative.
For the vertical reflection, write V(t) = −s(t) = −2 $\sqrt{t}$ .The minus sign goes outside because every original output must change sign.
V has domain [0, ∞) and range (−∞, 0]. The point (4, 4) becomes (4, −4).The inputs stay the same, while every nonnegative output becomes nonpositive: −2 $square root of 4$ = −2 · 2 = −4.
For the horizontal reflection, write H(t) = s(−t) = 2 $square root of −t$ .The minus sign goes inside because each original input coordinate changes sign.
Require −t ≥ 0, so t ≤ 0. H has domain (−∞, 0] and range [0, ∞). The point (4, 4) becomes (−4, 4).Multiplying the inequality by −1 reverses its direction. The reflected input gives H(−4) = 2 $square root of 4$ = 4.

Answer

V(t) = −2 $\sqrt{t}$
domain [0, ∞)
range (−∞, 0].
H(t) = 2 $square root of −t$
domain (−∞, 0], meaning t ≤ 0
range [0, ∞).

Check The original point (9, 6) gives V(9) = −6 and H(−9) = 6. Each changed coordinate is the opposite of its original coordinate, while the other coordinate stays fixed.

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: reflect one point

The point (3, −4) is on f. Where does it move under −f(x), and under f(−x)?
This asks you to change the coordinate named by each mirror move and keep the other coordinate.

On −f(x): (3, 4).

On f(−x): (−3, −4).

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

For −f(x), keep 3 and negate −4: (3, 4).A vertical reflection changes the output coordinate only.
For f(−x), negate 3 and keep −4: (−3, −4).A horizontal reflection changes the input coordinate only.

Answer

On −f(x): (3, 4).
On f(−x): (−3, −4).

Check −f(3) = −(−4) = 4. At input −3, f(−(−3)) = f(3) = −4.

Rung 2Rung 2: reflect a shifted square root and its allowed inputs

Reflect s(t) = $square root of t − 3$ across each axis and track the point (7, 2).
This asks you to change the coordinate named by each mirror move and keep the other coordinate.

Vertical reflection sends (7, 2) to (7, −2).

Horizontal reflection sends (7, 2) to (−7, 2).

Across the horizontal axis: V(t) = − $square root of t − 3$ , and (7, 2) becomes (7, −2).The output changes sign while the input stays 7.
Across the vertical axis: H(t) = s(−t) = $square root of −t − 3$ , and (7, 2) becomes (−7, 2).The input changes sign while the output stays 2. Every t inside s is replaced by −t, so t − 3 becomes −t − 3.
For V, require t − 3 ≥ 0, so t ≥ 3: domain [3, ∞) and range (−∞, 0]. For H, require −t − 3 ≥ 0. Add 3 to both sides: −t ≥ 3. Multiply by −1 and reverse the sign: t ≤ −3. Domain (−∞, −3] and range [0, ∞).A real square root cannot accept a negative expression, and its output is nonnegative until an outside minus negates it.

Answer

V(t) = − $square root of t − 3$ : (7, −2)
domain [3, ∞)
range (−∞, 0].
H(t) = $square root of −t − 3$ : (−7, 2)
domain (−∞, −3]
range [0, ∞).

Check V(7) = − = −2. H(−7) = = = 2. H(7) = is not real, which confirms why its domain faces left.

Rung 3Rung 3: reflect a table without inventing values

Use the original f table shown beside this question. List the corresponding known points on V(x) = −f(x) and H(x) = 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 V(x) output.

Read each column as one x input paired with its H(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.
For V, keep inputs 2, 4, 6, 8 and negate outputs 1, 3, 7, 11 to get −1, −3, −7, −11.The outside minus acts on the output row.
For H, negate inputs 2, 4, 6, 8 to get −2, −4, −6, −8, and keep outputs 1, 3, 7, 11.At a reflected input −u, the expression −x equals the original known input u.
Do not assign a value to H(2) from this table.H(2) requires f(−2), and the table supplies no value for f(−2).

Answer

V: (2, −1), (4, −3), (6, −7), (8, −11).
H: (−2, 1), (−4, 3), (−6, 7), (−8, 11).
H(2): not determined by the supplied table.

Check V(6) = −f(6) = −7. H(−6) = f(6) = 7. Both use the same supplied value f(6) = 7.

Rung 4Rung 4: reflect a table that includes zero and both signs

For this original practice question, use the intact textbook table displayed beside it. Make separate tables for V(x) = −f(x) and H(x) = f(−x). This asks you to change heights for V and sideways addresses for H.

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

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

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

Read the column under −2: f(−2) = 5. Read the remaining columns as f(0) = 10, f(2) = 15, and f(4) = 20.Each column pairs one input with its given output; no formula between columns is assumed.
For V, keep inputs −2, 0, 2, 4 and negate the outputs to get −5, −10, −15, −20.The outside minus reflects heights across the x-axis.
For H, negate the inputs to get 2, 0, −2, −4 and keep the matched outputs 5, 10, 15, 20.At the new input −u, the inside −x returns the old input u; zero remains zero.

Answer

V: use the displayed output-reflection table.
H: use the displayed input-reflection table.
The zero input remains 0 in both tables; its height is −10 for V and 10 for H.

Check V(2) = −f(2) = −15, while H(−2) = f(2) = 15. Negating either changed row again restores the original table.

Rung 5Rung 5: combine the learning-model transformations

Using only the original points (0, 1) and (1, 2), find the corresponding points on L(n) = −f(−n) + 1. Explain what the points do and do not tell you.
This asks you to find the new address or height of the given graph information.

(0, 1) becomes (0, 0).

(1, 2) becomes (−1, −1).

A complete learning formula and its contextual domain and range remain unspecified.

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

For (0, 1), solve −n = 0, so n = 0. Its new height is −1 + 1 = 0.The new input must deliver the old input, and the outside operations act on the known old height.
For (1, 2), solve −n = 1, so n = −1. Its new height is −2 + 1 = −1.The inside reflection negates the input coordinate, then the outside reflection and shift change the output.
The known transformed points are (0, 0) and (−1, −1). No full formula, domain, range, or percentage trend follows from these two points alone.Many different original functions pass through the same two points. A learning context also needs its own restriction to physically meaningful practice sessions.

Answer

(0, 1) becomes (0, 0).
(1, 2) becomes (−1, −1).
A complete learning formula and its contextual domain and range remain unspecified.

Check Substitution gives L(0) = −f(0) + 1 = 0 and L(−1) = −f(1) + 1 = −1. No value of f beyond the supplied points was needed.

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

✗ Not this: Reflection across the y-axis changes 2 to −2.

The y-axis is the vertical mirror line. Its reflection swaps left and right, so it changes the input coordinate, not the height.

✓ Instead: Use 2. The old point (4, 4) becomes (−4, 4), and 2 = 2 = 4. Across the x-axis, use −2 instead.

✗ Not this: keeps the original square root Domain [0, ∞).

The reflected curve needs −x ≥ 0. Positive x would make the inside negative.

✓ Instead: Multiply −x ≥ 0 by −1 and reverse the inequality: x ≤ 0. Check x = −4: = 2. Input 4 would require and is not allowed.

✗ Not this: A vertical reflection sends an old height −3 to −6.

That subtracts a fixed amount instead of taking the opposite. A reflection keeps the same distance from its mirror line.

✓ Instead: Negate −3 to get 3. For f(1) = −3, −f(1) = −(−3) = 3, so (1, −3) becomes (1, 3).

Tips and tricks

Write x-axis: keep x, negate y; y-axis: negate x, keep y. The mirror axis tells you which coordinate stays fixed.
A point on the mirror line stays fixed because its moving coordinate is zero and −0 = 0.
Check the Domain after an inside minus and the Range after an outside minus.
For two reflections plus a shift, record one intermediate point after each operation instead of trying to do all signs mentally.

Trap. Choosing the mirror axis from the coordinate that changes. A vertical reflection changes y but reflects across the x-axis. A horizontal reflection changes x but reflects across the y-axis. Name the mirror line and the moving coordinate separately.