Did you know? Moving two small vertical bars from inside a function to outside it can completely change an H2 Mathematics graph. The expressions f(|x|) and |f(x)| may look like near twins, but they describe different transformations. One changes which input the function sees. The other changes the sign of its output.
For JC1 H2 Mathematics learners in Bukit Timah, this distinction belongs to the Singapore-Cambridge A-Level 9758 functions-and-graphs strand. Students may know how to sketch y = f(x) but hesitate when asked to obtain y = f(|x|) or y = |f(x)| from it. The repair is not a longer list of memorised arrows. It is recognising the difference between input and output.
This quick-reading guide answers that exact question, works through contrasting examples, checks domain restrictions, explains what happens to roots and turning points, and gives an independent mini-test. It builds on the separate JC1 H2 guide about horizontal graph shifts without repeating the same problem.
The 20-second answer
For y = f(|x|), the function receives a non-negative input, whether x is positive or negative. Keep the original graph for x ≥ 0. Reflect that right-hand half across the y-axis to supply the left-hand half. The result is symmetric about the y-axis wherever defined.
For y = |f(x)|, keep the original graph wherever f(x) ≥ 0. Reflect the parts below the x-axis upward. The resulting graph never has a negative y-value, but it need not be symmetric about the y-axis.
- f(|x|) changes the input: copy the right half into the left half.
- |f(x)| changes the output: flip negative heights above the x-axis.
- f(|x|) is an even function on its resulting symmetric domain, whenever it is defined.
- |f(x)| always has non-negative outputs, but it is not necessarily an even function.
- Always check the original domain and allowed endpoints before sketching either transformation.
That is enough to start. The examples below make the two rules difficult to confuse.
Think of f as a machine
Imagine a machine that takes an x-value and returns a height. If you feed it |x| instead of x, the machine is never given a negative input. Feeding x = −3 and x = 3 both sends the value three into f. Therefore f(|−3|) = f(3) and f(|3|) = f(3). That explains the mirror symmetry.
But if the machine first computes f(x), and only afterward someone takes |f(x)|, the input remains exactly what it was. A negative output, such as −5, becomes height five; a positive output such as two stays two. That explains why the reflection is across the x-axis rather than the y-axis.
A useful tutoring question is: “Where are the modulus bars—around the input, or around the completed output?” The location answers which part of the graph changes. Nothing magical is hidden in the notation.
Worked example 1: f(x) = x − 2
Start with the straight line f(x) = x − 2. It crosses the y-axis at −2 and the x-axis at two.
For y = f(|x|), substitute |x| into the rule: y = |x| − 2. On the right (x ≥ 0), keep y = x − 2. On the left (x < 0), the rule becomes y = −x − 2. These two rays meet at (0, −2). The graph is a V-shape, its lowest point is (0, −2), and its x-intercepts are (−2, 0) and (2, 0).
Now take y = |f(x)| = |x − 2|. This time the negative part of the original line is reflected upward. The V-shape has its lowest point at (2, 0). Its y-intercept is (0, 2), and its only root is x = 2.
Two modulus expressions, two different vertices. If a student sketches both as V-shapes and stops, the answer still lacks the most important detail: where the V is located.
The simple point check that exposes confusion
Choose x = −2. For f(|x|), we get f(2) = 2 − 2 = 0. For |f(x)|, we get |f(−2)| = |−4| = 4. So the point on the first graph is (−2, 0), while the point on the second graph is (−2, 4).
One input gives two very different outputs. That one-line substitution is an excellent check when a sketch is uncertain. Before spending time drawing a perfect curve, compare a negative input on both transformed expressions.
For an H2 Mathematics question based on a supplied sketch rather than a given formula, choose a labelled point on the right-hand portion and reflect it correctly. The graph is made of points, not just an outline.
Worked example 2: f(x) = x² − 4
Take f(x) = x² − 4, an upward-opening parabola with vertex (0, −4), roots −2 and 2, and symmetry about the y-axis.
For f(|x|), the transformed expression is (|x|)² − 4 = x² − 4. Nothing visible changes. The original function is already even, so replacing x by |x| gives the same values for every real x. This is a useful counterexample to the idea that every transformation must visibly move a graph.
For |f(x)|, the new expression is |x² − 4|. It equals 4 − x² for −2 ≤ x ≤ 2, because x² − 4 is non-positive there. It equals x² − 4 for x < −2 or x > 2. The middle of the original parabola is folded upward, producing a central arch above the x-axis.
The transformed graph is always non-negative. It has zeros at x = −2 and x = 2, and the central point is now (0, 4), not (0, −4). It is still symmetric about the y-axis—but only because the original example has the necessary symmetry, not because taking |f(x)| automatically forces that symmetry.
Piecewise working without guesswork
Sometimes the exam question asks you to sketch a graph with accuracy rather than write a piecewise rule. Even then, a short piecewise observation helps.
For f(x) = x − 2, we have f(|x|) = x − 2 when x ≥ 0, and f(|x|) = −x − 2 when x < 0. The graph consists of two straight rays meeting at a lowest point. For |f(x)|, the rule is 2 − x for x < 2 and x − 2 for x ≥ 2. This produces a different joining point.
The joining location depends on which expression changes sign. In f(|x|), the input |x| changes piecewise at x = 0. In |f(x)|, the sign switch occurs wherever f(x) = 0. That distinction generalises far beyond straight lines.
When a function has several roots, |f(x)| may flip more than one section. With f(|x|), however, the underlying rule still uses only the non-negative input side. Draw the sections only after deciding which rule applies.
What happens to the domain?
Transformations are never an excuse to forget what inputs are permitted. Suppose the original function is f(x) = √(x − 1), with domain x ≥ 1.
For f(|x|), the input restriction requires |x| ≥ 1. That means x ≤ −1 or x ≥ 1. The transformed graph has two branches with a gap around the origin; you cannot simply fill the missing middle with a connected curve.
For |f(x)|, the original function is non-negative wherever defined, so |f(x)| = f(x). Its domain remains x ≥ 1. There is no new left-hand branch.
This is a particularly revealing diagnostic. A student who knows the “copy right side” slogan but draws values where |x| < 1 has missed the domain condition. A correct method checks both what the formula does and where the formula exists.
What happens to roots and intercepts?
If the original graph y = f(x) has a root at a positive input a, then f(|x|) has roots at x = a and x = −a, provided the original function is defined at a. The origin needs separate attention: a root at zero stays at zero, not two distinct roots.
For |f(x)|, a value is zero exactly when f(x) is zero. The x-intercepts therefore remain at the same inputs, although parts of the curve are moved above the x-axis. A root that was a crossing may become a touch point after the reflection.
The y-intercept can also behave differently. For f(|x|), substitute x = 0 to obtain f(0), if zero belongs to the original domain. For |f(x)|, the y-intercept is |f(0)|. If f(0) is negative, one graph starts below the x-axis and the other starts above it.
Two common wrong solutions
Mistake A: A student sees f(|x|), draws the left-hand side of the original graph and mirrors it to the right. That is backwards. At x = −3, the new function receives input three, so it is the original right-hand part that determines both sides.
Mistake B: A student sees |f(x)| and reflects everything below the y-axis across the y-axis. This changes the input direction, not the output sign. To build |f(x)|, leave x-coordinates unchanged and make each y-coordinate non-negative.
The repair is to test a point with a negative input and another point whose original output is negative. The two cases expose whether the learner is attending to inputs or outputs. More identical worksheets will not fix the error if the distinction is never explained.
Five small practice questions with answers
Question 1. If f(x) = x + 3, where is the vertex of y = f(|x|)? Answer: y = |x| + 3 has vertex (0, 3).
Question 2. For the same f, where is the vertex of y = |f(x)|? Answer: y = |x + 3| has vertex (−3, 0).
Question 3. Does f(|x|) always have y ≥ 0? Answer: No. For f(x) = x² − 4, the value at zero remains −4.
Question 4. Does |f(x)| always have symmetry about the y-axis? Answer: No. The example |x − 2| has its vertex at (2, 0), not on the y-axis.
Question 5. If f(x) = √(x − 1), what is the domain of f(|x|)? Answer: All x with |x| ≥ 1, namely x ≤ −1 or x ≥ 1.
An independent learner should be able to explain each answer in one sentence without relying on a diagram memorised from a past paper.
A practical tutorial sequence in Bukit Timah
Start by drawing y = x − 2. Ask the learner to calculate transformed outputs at x = −2, 0, 2 and 4. Then sketch f(|x|) from those points. Repeat for |f(x)| and compare the vertices. Only after the graphs differ clearly do we introduce a parabola.
Next, give an original function that is already even. This tests whether the student understands that f(|x|) may leave a graph unchanged. Finally introduce the restricted square-root domain to check that the learner is not creating invalid new inputs.
This is how a strong JC1 mathematics tutorial can move from meaning, to worked examples, to exceptions, to an unseen question. It is not about drawing the largest number of curves in a sitting. It is about developing a method the student can reproduce independently.
Frequently asked questions
Are f(|x|) and |f(x)| equivalent?
Not in general. The first modifies the input to f; the second modifies the output. They can coincide for special functions, but cannot be treated as interchangeable identities.
Which graph becomes symmetric about the y-axis?
f(|x|) is symmetric about the y-axis on its resulting domain. |f(x)| is symmetric about the y-axis only if the original output magnitudes happen to have that symmetry.
Is this examined in H2 Mathematics?
Yes. The 2027 A-Level H2 Mathematics syllabus 9758 lists the relationships between y = f(x), y = |f(x)|, y = f(|x|) and y = 1/f(x) in its graphs-and-transformations strand.
Do I always have to draw piecewise graphs?
A piecewise rule helps verify a sketch, but the required level of working depends on the question. Show the domain restrictions, significant points, shape and transformations requested.
Your next useful Maths guides
For linked JC1 skills, read why a horizontal graph shift seems backwards, modulus inequalities as distance and negative common ratios in geometric series. The Bukit Timah H2 Mathematics overview connects these foundations to the broader A-Level route.
The eduKateSG Clementi small-group Mathematics tutorial reference illustrates an approach centred on diagnosing understanding and building independent practice in a Secondary 1 setting. It does not by itself specify JC1 programme arrangements. For current JC1 questions, use the Bukit Timah JC1 tuition page.
Keep one question in your head: Did the modulus act before the function or after it? Once you know that, you know which side of the graph needs changing.