Did you know? When a graph equation changes from y = f(x) to y = f(2x), students often expect the graph to become twice as wide. After all, the number two looks like a stretch. But the graph actually becomes half as wide horizontally. The function now receives the same original input at half the original x-coordinate.
For students searching for a JC1 H2 Mathematics Tutor Bukit Timah, this “backwards” effect is one of the quickest ways to diagnose whether graph transformations are understood or merely memorised. The 2027 Singapore–Cambridge H2 Mathematics syllabus 9758 includes changes such as y = f(ax), y = f(x + a), y = af(x), and combinations. The subject rewards accurate mapping between points, not arrows drawn from a slogan.
This guide explains why f(2x) compresses a graph, how to map significant points and domains, why f(−2x) is different, and how combining a compression with a shift can cause mistakes. There are worked examples, independent questions, a concise tutorial sequence, answers for parents, and references to the official syllabus. For the closely related *shift* problem, read why x + a moves a graph left.
Quick answer: divide the original x-coordinates by two
A point (a, b) belongs to the original graph y = f(x) when f(a) = b. On the transformed graph y = f(2x), we want the function to receive the original input a. Therefore solve 2x = a, which gives x = a/2.
So (a, b) maps to (a/2, b).
Every original point moves to half its horizontal distance from the y-axis. The y-coordinate stays unchanged. That is a horizontal compression by factor one-half.
- For y = f(2x), map every original (a, b) to (a/2, b).
- For y = f(x/2), map every original (a, b) to (2a, b).
- For y = f(−2x), map (a, b) to (−a/2, b): a reflection in the y-axis combined with compression.
- For y = 2f(x), map (a, b) to (a, 2b): a vertical stretch, not a horizontal one.
- If the original domain is D, the new domain for f(2x) consists of x satisfying 2x ∈ D.
Remember the reason rather than a rule sheet: the equation inside f must produce the old input, so solve that equation for the new x.
Imagine the function as an input machine
A function f receives a number and returns a height. The statement f(4) = 7 says the original graph includes (4, 7). In the new function f(2x), the same height seven will be produced whenever 2x = 4.
That happens at x = 2. So the transformed graph includes (2, 7). The height has not doubled or halved; only the horizontal location changed.
If f(−6) = 3, then the original point (−6, 3) maps to (−3, 3). The negative x-coordinate also moves closer to zero. Compression happens on both sides of the y-axis.
This explanation makes the transformation predictable. It does not matter whether the original curve is a parabola, rational function, exponential graph or an unfamiliar sketch supplied in an examination.
Worked example 1: a quadratic with a clear vertex
Let f(x) = (x − 2)² + 1. Its graph is an upward-opening parabola with vertex (2, 1). It has y-intercept f(0) = 5.
Now substitute 2x into f:
f(2x) = (2x − 2)² + 1 = 4(x − 1)² + 1.
The new vertex is (1, 1). Its x-coordinate has halved from two to one, while its y-coordinate remains one. That agrees with the point mapping.
Consider the original point (3, 2), because f(3) = (3 − 2)² + 1 = 2. Under the transformation, it becomes (3/2, 2). Check directly: f(2 × 3/2) = f(3) = 2.
The original point (1, 2) becomes (1/2, 2). The left and right points around the original vertex are now closer together horizontally, explaining why the parabola looks narrower.
Why does the formula contain a factor of four?
The transformed expression is 4(x − 1)² + 1. It resembles a vertically stretched parabola, and algebraically that description can also be correct for this particular quadratic when referred to an appropriately shifted reference graph.
But the requested transformation from the specific original f(x) to f(2x) is determined by input mapping. Every original x-coordinate is halved. That is the universal geometric description.
If students only look at the coefficient four, they may miss what happened to the vertex. Starting from the original curve and mapping its points is the method that works for general functions.
Worked example 2: a straight line
Let f(x) = 3x + 2. Its graph passes through (0, 2), (1, 5), and (−1, −1).
The transformed function is f(2x) = 3(2x) + 2 = 6x + 2.
The original point (1, 5) maps to (1/2, 5). The original point (−1, −1) maps to (−1/2, −1). The y-intercept remains (0, 2).
The transformed line has gradient six rather than three. In this example a horizontal compression of the line appears as a steeper slope, but the underlying point mapping is still the same.
Try checking the point (1/2, 5) directly: 6(1/2) + 2 = 5. The calculation verifies the transformation independently of the sketch.
Worked example 3: a domain restriction changes too
Now let f(x) = √(x − 1), defined for x ≥ 1.
The transformed expression is f(2x) = √(2x − 1). For the square root to be real, we must have 2x − 1 ≥ 0. Thus the new domain is x ≥ 1/2.
The original curve begins at (1, 0). After horizontal compression it begins at (1/2, 0). That is exactly what the point rule predicts.
The point (2, 1) on the original graph maps to (1, 1), and f(2) = √1 = 1 appears at new input one. Every defined point maps correctly.
This is why a transformation is not merely changing a sketch. The domain must be transformed alongside the points. Drawing a new square-root curve to the left of x = 1/2 would invent real outputs where the formula is not defined.
Worked example 4: a vertical asymptote moves
Take f(x) = 1/(x − 4), defined for x ≠ 4. Its graph has vertical asymptote x = 4.
After replacing x with 2x, the function becomes f(2x) = 1/(2x − 4). The denominator is zero when 2x = 4, giving x = 2.
Thus the vertical asymptote moves from x = 4 to x = 2. Again its horizontal coordinate is halved.
The original point (5, 1) maps to (5/2, 1), and substituting x = 5/2 gives 1/(5 − 4) = 1.
The horizontal asymptote y = 0 remains y = 0. The transformation changes input positions; it does not multiply output heights.
For an extended explanation of how denominator zeros and other exclusions affect rational graphs, see reciprocal graph zeros and asymptotes.
Why the transformation works for negative x-values
Students sometimes compress only the right side of the original graph. But the mapping (a, b) → (a/2, b) is valid for every original input a, positive or negative.
For instance, suppose the original graph has three marked points (−4, 3), (0, −1), and (6, 2). The transformed graph must contain (−2, 3), (0, −1), and (3, 2).
The point at the origin’s vertical line stays on the same vertical line because zero divided by two remains zero. The points on either side move closer to that line.
This creates a compressed appearance without reversing the order of the points. A negative input stays negative when divided by positive two.
That last fact distinguishes f(2x) from f(−2x), where the extra negative sign also reverses the left and right sides.
What changes when the multiplier is negative?
Let g(x) = f(−2x). To recover an old input a, solve −2x = a. The new coordinate is x = −a/2.
Thus (a, b) → (−a/2, b).
If the original graph contains (4, 7), the new graph contains (−2, 7). A point previously on the right appears on the left, at half the horizontal distance from the y-axis.
A convenient description is reflect across the y-axis and compress horizontally by factor one-half. These two operations commute when expressed as scaling the x-coordinate by −1/2.
The sign and magnitude of the inner multiplier play different roles. The magnitude determines the horizontal scaling, while the negative sign reverses the horizontal orientation.
Compare f(2x) with 2f(x)
These expressions are not equivalent for most functions.
For f(x) = x², we have f(2x) = (2x)² = 4x², whereas 2f(x) = 2x². At x = 1, their outputs are four and two.
For f(x) = x + 3, we have f(2x) = 2x + 3, whereas 2f(x) = 2x + 6. The y-intercepts differ as well.
The notation tells us which operation happens first. In f(2x), multiply the input by two before applying f. In 2f(x), apply f to the original input and then multiply its output by two.
A useful diagnostic is to ask: “Does the two sit inside the function’s brackets or outside?” Inside means the x-coordinate mapping changes; outside means the height mapping changes.
Combining a shift with compression: f(2x + 4)
Consider g(x) = f(2x + 4). This is a classic place for students to choose the wrong shift.
Suppose an original point is (a, b), so f(a) = b. On g, the same output occurs when 2x + 4 = a. Solve:
x = (a − 4)/2 = a/2 − 2.
Therefore the original point maps to (a/2 − 2, b).
You can describe the operation as compressing the original x-coordinates by one-half and then shifting them left two units. It is not a shift left four followed by an unspecified compression.
A vertex check makes the order visible
Let f(x) = x², whose vertex is (0, 0). Then g(x) = f(2x + 4) = (2x + 4)² = 4(x + 2)².
The new vertex is (−2, 0). The point mapping predicts the same result: a = 0 gives x = 0/2 − 2 = −2.
If a learner shifts the vertex to −4, the result contradicts the equation. This is why a known landmark is a good check.
More generally, if g(x) = f(kx + c) with k ≠ 0, then an original x-coordinate a maps to (a − c)/k. Solving the inner-input equation removes ambiguity about operation order.
The meaning of the number one-half in a horizontal compression
Some students object that the formula says f(2x), so why should the scaling factor be one-half? The apparent contradiction disappears if we distinguish input speed from graph position.
As x increases by one unit in f(2x), the number fed to f increases by two units. The machine travels through the original input values twice as fast. So the new graph needs only half as much horizontal space to display the same input progression.
For example, the old graph reaches input six at x = 6. The transformed graph reaches that same input when x = 3. That is why it is narrower.
A horizontal compression is about where an old input value now appears, not about the numeric multiplier’s size in isolation.
An input–output table parents can use
Let the original function be f(x) = x² − 1. Its values at x = −2, −1, 0, 1 and 2 are 3, 0, −1, 0 and 3.
The transformed function f(2x) takes the same outputs at new x-values −1, −1/2, 0, 1/2 and 1.
The heights remain 3, 0, −1, 0 and 3. All the changes are in the horizontal positions.
If a child creates the transformed table by doubling the original x-coordinates instead, the mistake will be visible immediately: the new x-values should be closer to zero, not farther away.
This simple table is enough to teach the mapping before introducing unfamiliar curves and asymptotes.
What about the range?
A horizontal compression by a non-zero multiplier preserves the range of f when the transformed domain is defined as all x for which the inner input lies in f’s original domain.
Why? Every original allowed input a is still reached: choose the new input x = a/2. The output f(a) therefore remains available somewhere on the transformed graph.
The set of output heights has not changed, even though the x-coordinates where they appear have moved.
For the quadratic f(x) = (x − 2)² + 1, the range is y ≥ 1. The transformed f(2x) = 4(x − 1)² + 1 also has range y ≥ 1.
For f(x) = √(x − 1), the range is y ≥ 0, and f(2x) has the same range despite its new domain starting at one-half.
This is another useful answer check. If a student reports that the graph’s y-values were doubled by f(2x), they are confusing horizontal and vertical transformations.
Five common wrong steps and the correct repair
- Mistake: (a, b) maps to (2a, b). Repair: solve 2x = a, giving x = a/2.
- Mistake: f(2x) means multiply every height by two. Repair: compare with 2f(x); only the latter scales outputs.
- Mistake: the graph is compressed only on the right. Repair: negative x-values are halved too.
- Mistake: the original domain x ≥ 1 stays unchanged for f(2x). Repair: require 2x ≥ 1, so x ≥ 1/2.
- Mistake: f(2x + 4) shifts left four. Repair: solve 2x + 4 = a; the coordinate changes by a/2 − 2.
Each error points to a different gap. Some learners do not understand input mapping, while others understand it but forget to apply it to domains and asymptotes. A good tutorial repairs the first incorrect decision rather than repeating the entire chapter.
Ten short questions with worked answers
1. If (6, 4) is on f, where is it on f(2x)? Answer: (3, 4).
2. If (−8, 1) is on f, where is it on f(2x)? Answer: (−4, 1).
3. If f(x) = x + 5, find f(2x). Answer: 2x + 5.
4. If f(x) = x² + 2, find f(2x). Answer: 4x² + 2.
5. If f has domain x ≥ 4, what is the domain of f(2x)? Answer: x ≥ 2.
6. If f has a vertical asymptote x = −6, where is that asymptote on f(2x)? Answer: x = −3, assuming the asymptote arises from the input approaching −6 within the domain.
7. If (4, 9) is on f, where is it on f(−2x)? Answer: (−2, 9).
8. If (0, 3) is on f, where is it on f(2x + 4)? Answer: (−2, 3).
9. Does f(2x) generally preserve the output range of f? Answer: Yes, under the naturally transformed full domain with non-zero horizontal scaling.
10. Is f(2x) always equal to 2f(x)? Answer: No. For f(x) = x² they give 4x² and 2x² respectively.
These questions are designed to separate mapping, domain, graph features and operation order. A student who can explain the answers without referring to a memorised arrow chart is ready for unfamiliar functions.
An effective three-stage JC1 teaching sequence
Stage one: map points before graphing
Choose labelled points on a simple original curve. Ask the learner to write 2x = a for each point and map its x-coordinate. Do not begin with a complicated graphing calculator screen.
Stage two: apply the mapping to structure
Use a shifted quadratic, a square-root domain and a rational asymptote. The learner should correctly place the vertex, endpoint and excluded input. Then compare with algebraic substitution.
Stage three: combine transformations and check independently
Introduce f(2x + 4) and f(−2x). Require a rule mapping original points to new coordinates, followed by a sketch and a calculator or substitution check. End with one unseen graph.
The objective is transferable interpretation. If the learner can map any original point and explain why the domain changes in the same way, the topic is no longer a collection of isolated tricks.
Frequently asked questions
Why does f(2x) compress a graph instead of stretch it?
To get the old input a, the new x must satisfy 2x = a, so x = a/2. Every original horizontal coordinate is halved.
Does f(2x) change the y-intercept?
The y-intercept has height f(0) if zero belongs to the domain, because f(2·0) = f(0). It therefore stays the same.
What happens to a turning point?
An original turning point (a, b) maps to (a/2, b). Its height is unchanged while its horizontal position moves towards the y-axis.
What happens when the factor is less than one?
For f(x/2), the new x-coordinate is twice the original, giving a horizontal stretch by factor two.
Is this part of the 2027 H2 Mathematics syllabus?
Yes. The official 2027 H2 Mathematics 9758 syllabus includes transformations of the form y = f(ax) and combinations with shifts and other transformations.
Is this always taught in JC1?
Teaching order varies by junior college. The article uses JC1 as its learner audience while staying aligned with the two-year H2 syllabus.
Related reading and practical next steps
Continue with why x + a shifts the graph left, how to find the range of rational functions, and why reciprocal graphs turn zeros into asymptotes. For the full programme context, see the Bukit Timah H2 Mathematics overview and the 2027 SEAB H2 Mathematics syllabus.
The unchanged eduKateSG Secondary 1 Mathematics small-group reference illustrates a diagnosis–worked-example–independent-check teaching style. It does not set the class size, fees or schedules of a JC1 programme. Current JC1 enquiries belong on the Bukit Timah JC1 tuition page.
Remember the calculation that unlocks the graph: set the transformed inner input equal to the original input, then solve for the new x-coordinate.