JC2 H2 Maths Bukit Timah | f(|x|) vs |f(x)|

Did you know? The bars in f(|x|) and |f(x)| are only one symbol apart, but the two graph transformations act on different things. For f(x)=2x−4, the first becomes 2|x|−4, with its turning point at (0,−4). The second becomes |2x−4|, with its turning point at (2,0). One copies the right half of a graph to the left; the other reflects parts below the x-axis upward. Confusing them moves the turning point and changes the intercepts.

This JC2 H2 Mathematics Tutor Bukit Timah guide explains exactly how to sketch y=f(|x|) and y=|f(x)| from a known graph. The short comparison and two complete examples come first, followed by changed-case graphs, a piecewise interpretation and 36 original linear-function clinics with precise coordinates, domains, ranges and intercepts. The lesson is deliberately narrower than a general chapter on all function transformations.

The SEAB Singapore–Cambridge H2 Mathematics 9758 syllabus for the 2027 examination explicitly includes relating the graphs of y=f(x), y=|f(x)| and y=f(|x|) in Section 1.2. We stay with that requirement and do not assert that every graph has to contain a cusp or that both transformations are always different. What matters is the original function’s shape and exactly where the modulus symbol acts.

Quick answer: bars inside change x; bars outside change y

Graph	What stays	What changes	Reliable sketch instruction
y=f(\|x\|)	The original right-hand part, x≥0	The left-hand part is replaced	Keep the right half, reflect it in the y-axis onto the left
y=\|f(x)\|	Every non-negative output	All negative outputs become positive	Keep above-axis parts, reflect below-axis parts in the x-axis

The first transformation modifies the input: wherever x is negative, the function receives |x|=−x, so the graph at negative x copies the graph at the corresponding positive x. The result is always symmetric about the y-axis when the required inputs exist. The second transformation modifies the output: points with f(x)≥0 stay unchanged, while points with f(x)<0 move vertically above the axis. Its result is always nonnegative, but is not necessarily symmetric about the y-axis.

Do not move everything: f(|x|) leaves the right half unchanged; |f(x)| leaves all non-negative heights unchanged.
Start with the original graph: mark intercepts, turning points and the allowed domain.
For input bars: mirror only the right-hand graph across x=0.
For output bars: reflect only the original negative-y portions across y=0.
Check sign: |f(x)| can never be negative; f(|x|) may be negative.
Check symmetry: f(|x|) is even when the requisite domain is symmetric; |f(x)| need not be even.
Check if the shapes actually coincide: some special functions make the two transformations identical, so do not claim they must differ.

Worked example one: the linear function f(x)=2x−4

The original straight line has slope 2, y-intercept −4 and x-intercept 2. For the input-modulus graph, f(|x|)=2|x|−4. When x≥0, it follows the original line y=2x−4. When x<0, it follows y=−2x−4. These two rays meet at (0,−4). The graph opens upward and crosses the x-axis at x=−2 and x=2. Its range is y≥−4.

For the output-modulus graph, |f(x)|=|2x−4|. The original line is negative when x<2, so that portion reflects above the x-axis. The unchanged portion starts from x=2 onward. The two rays now meet at (2,0), with y-intercept 4 and x-intercept only x=2. Its range is y≥0. The turning point has shifted because one transformation acted on x while the other acted on f(x).

A one-minute verification uses three x-values. At x=−2, f(|x|)=2(2)−4=0, whereas |f(x)|=|−8|=8. At x=0, f(|x|)=−4 and |f(x)|=4. At x=2, both are zero. These values prevent a mistaken drawing of the same V at the same vertex for the two expressions.

Worked example two: f(x)=−x+3

Now the original line decreases and crosses the x-axis at 3. The input-modulus expression is f(|x|)=3−|x|. It has a maximum at (0,3), opens downward like an inverted V, and reaches zero at x=−3 and x=3. Its range is y≤3 and includes negative values when |x|>3. This directly disproves the common belief that any function containing modulus bars must have a nonnegative output.

The output-modulus expression is |f(x)|=|3−x|. It is nonnegative, has its minimum at (3,0) and opens upward. At x=0 it has height 3. The two transformations are still different: the input bars created an even reflection with a highest point on the y-axis, whereas the output bars folded negative outputs above the x-axis.

A piecewise description makes the transformation exact

For any function f that accepts the required inputs, f(|x|)=f(x) when x≥0 and f(|x|)=f(−x) when x<0. This is a rule about inputs. For the other transformation, |f(x)|=f(x) when f(x)≥0 and |f(x)|=−f(x) when f(x)<0. This is a rule about outputs. The distinction between conditions on x and conditions on f(x) is the most useful formula to remember.

For f(x)=2x−4, the f(|x|) formula splits at x=0, while |f(x)| splits where 2x−4=0, namely x=2. The graph’s relevant fold location is determined by its operation, not simply by seeing the symbol | | in the question. A student who identifies the fold location before plotting normally avoids the most visible wrong-vertex error.

A quadratic can make input modulus change nothing

Consider f(x)=x²−1. Because f is even, f(−x)=f(x). The input-modulus transformation is f(|x|)=(|x|)²−1=x²−1, identical to the original parabola. The graph already has y-axis symmetry, so copying its right half onto the left changes nothing. The turning point remains (0,−1), below the x-axis.

But the output-modulus graph is |f(x)|=|x²−1|. Between x=−1 and x=1, the original parabola is negative, so that central portion reflects upward into y=1−x². Outside that interval, it remains x²−1. The resulting graph is nonnegative and has corners at (−1,0) and (1,0). This is a striking example of the two transformations acting differently on an already symmetric function.

An odd function can make both transformed graphs identical

For f(x)=x³, we have f(|x|)=|x|³. We also have |f(x)|=|x³|=|x|³. Thus both transformations produce exactly the same graph in this special case. More generally, for functions with the appropriate sign behaviour and symmetry, coincidence can happen. The instruction is not to insist every input/output modulus graph differs; it is to apply the correct definition and see what follows.

This example also shows why graph recognition beats a memorised slogan. If two methods independently produce the same function, that is a genuine mathematical identity, not automatically a mistake. It is the same reasoning used when two different calculus substitutions give the same exact answer.

A negative even function separates the transformations again

If f(x)=−x², the input-modulus graph is f(|x|)=−(|x|)²=−x², still the original downward parabola. But the output-modulus graph becomes |−x²|=x², an upward parabola. Both are symmetric about the y-axis and both pass through the origin, but their ranges differ: y≤0 versus y≥0. Symmetry alone is not enough to decide whether the transformations agree.

At x=2, the input-modulus graph gives −4 while the output-modulus graph gives +4. Using one positive test x-value is enough to expose a mistaken claim that two even-looking graphs must be identical.

What happens to the domain?

For y=|f(x)|, the domain is the same as that of f: taking the modulus of a defined output does not create new inputs. For y=f(|x|), the domain consists of those x for which |x| belongs to the original domain of f. If f is defined only on x≥1, for example, f(|x|) is defined for x≤−1 or x≥1. A graph transformation must respect the original function definition, not just its algebraic expression.

The basic straight-line clinics below use functions defined on all real numbers, so both transformations also have all-real domains. In an examination question with a restricted f-domain, check it explicitly before claiming a full reflected branch exists. The original 2027 H2 functions syllabus includes domain and range, and transformation questions can combine those ideas.

What happens to zeros and intercepts?

For |f(x)|, the x-intercepts are exactly the original x-intercepts of f because |f(x)|=0 precisely when f(x)=0. The y-intercept becomes |f(0)| when x=0 is permitted. For f(|x|), the equation f(|x|)=0 uses only nonnegative inputs of the original function; every positive original root r generates transformed roots ±r, while an original root at zero remains zero.

This explains why f(|x|)=2|x|−4 has two roots ±2 even though f(x)=2x−4 has only the positive root 2. It also explains why a negative root of the original line may disappear: f(|x|) never feeds a negative argument into f. The two transformations have different rules for producing x-intercepts.

How to handle a graph that was only given visually

If the question shows a drawn graph of y=f(x) rather than a formula, begin by placing a vertical line at x=0 and observing the right half. To construct f(|x|), retain that right half and reflect it to the left. To construct |f(x)|, instead place a horizontal line at y=0 and fold any portion below it upward. Never use the whole original left half as the input for f(|x|), because the left half is discarded and replaced.

Graphing software can confirm your result, but two near-identical plots can be hard to distinguish at a small screen scale. Label the turning point, x-intercepts and y-intercept. Those are exact features that make the graph choice unambiguous and align with the syllabus focus on symmetries, intercepts and transformations.

A three-point arithmetic diagnostic

Use f(x)=2x−4 and ask for each transformed function’s value at −2, 0 and 2. The input-modulus values are 0,−4,0. The output-modulus values are 8,4,0. That is enough to identify their different symmetry, turning points and intercept behaviour. Next change the slope to −2 while keeping the intercept −4 and predict whether f(|x|) opens upward or downward. It opens downward because the coefficient of |x| is negative.

A learner who reports that both graphs are nonnegative has mistaken bars on x for bars on y. One who places the turning point of f(|x|) at the original line’s x-intercept has folded at the wrong axis. One who reflects the original left half for f(|x|) has reversed which half is preserved. Each error has a different repair, and checking three values is usually faster than re-explaining every transformation from scratch.

Thirty-six original graph-reading clinics

Every clinic uses a fictional linear function f(x)=mx+c, with nonzero m. We compare the exact graphs y=f(|x|) and y=|f(x)|, identify their vertices, ranges, intercepts and sample values. Both are defined for every real x in these exercises. Work several at a time, then change only the sign of m or c to test whether the rules transfer beyond one memorised V shape.

1. -4x − 9: compare the two folds

Try the original line. Let f(x)=-4x − 9, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-4|x| − 9, with vertex (0,-9). Its range is y≤-9 because m=-4. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -17, -9, -17, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-4x − 9|, with vertex (-9/4,0), x-intercept x=-9/4 and range y≥0. Its y-intercept is 9. The values at x=−2, 0 and 2 are 1, 9, 17. The input-modulus graph always uses the original line’s right-hand half as its source. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

2. 4x: compare the two folds

Try the original line. Let f(x)=4x, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=4|x|, with vertex (0,0). Its range is y≥0 because m=4. The x-intercepts are x=0. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 8, 0, 8, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|4x|, with vertex (0,0), x-intercept x=0 and range y≥0. Its y-intercept is 0. The values at x=−2, 0 and 2 are 8, 0, 8. The output-modulus graph never has negative y-values, while f(|x|) may have them. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

3. x + 8: compare the two folds

Try the original line. Let f(x)=x + 8, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=|x| + 8, with vertex (0,8). Its range is y≥8 because m=1. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 10, 8, 10, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|x + 8|, with vertex (-8,0), x-intercept x=-8 and range y≥0. Its y-intercept is 8. The values at x=−2, 0 and 2 are 6, 8, 10. For m>0, f(|x|)=m|x|+c has a minimum at (0,c). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

4. -3x − 1: compare the two folds

Try the original line. Let f(x)=-3x − 1, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-3|x| − 1, with vertex (0,-1). Its range is y≤-1 because m=-3. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -7, -1, -7, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-3x − 1|, with vertex (-1/3,0), x-intercept x=-1/3 and range y≥0. Its y-intercept is 1. The values at x=−2, 0 and 2 are 5, 1, 7. For m<0, f(|x|)=m|x|+c has a maximum at (0,c). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

5. 5x + 6: compare the two folds

Try the original line. Let f(x)=5x + 6, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=5|x| + 6, with vertex (0,6). Its range is y≥6 because m=5. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 16, 6, 16, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|5x + 6|, with vertex (-6/5,0), x-intercept x=-6/5 and range y≥0. Its y-intercept is 6. The values at x=−2, 0 and 2 are 4, 6, 16. The cusp of |mx+c| is located at the zero of the original line, not necessarily at x=0. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

6. 2x − 3: compare the two folds

Try the original line. Let f(x)=2x − 3, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=2|x| − 3, with vertex (0,-3). Its range is y≥-3 because m=2. The x-intercepts are x=−3/2 and x=3/2. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 1, -3, 1, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|2x − 3|, with vertex (3/2,0), x-intercept x=3/2 and range y≥0. Its y-intercept is 3. The values at x=−2, 0 and 2 are 7, 3, 1. If c=0 and m>0, these two transformed linear graphs are identical. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

7. -2x + 4: compare the two folds

Try the original line. Let f(x)=-2x + 4, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-2|x| + 4, with vertex (0,4). Its range is y≤4 because m=-2. The x-intercepts are x=−2 and x=2. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 0, 4, 0, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-2x + 4|, with vertex (2,0), x-intercept x=2 and range y≥0. Its y-intercept is 4. The values at x=−2, 0 and 2 are 8, 4, 0. If c=0 and m<0, they are reflections of each other in the x-axis. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

8. -5x − 5: compare the two folds

Try the original line. Let f(x)=-5x − 5, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-5|x| − 5, with vertex (0,-5). Its range is y≤-5 because m=-5. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -15, -5, -15, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-5x − 5|, with vertex (-1,0), x-intercept x=-1 and range y≥0. Its y-intercept is 5. The values at x=−2, 0 and 2 are 5, 5, 15. The input-modulus transformation can create two symmetric x-intercepts from a positive original root. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

9. 3x + 2: compare the two folds

Try the original line. Let f(x)=3x + 2, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=3|x| + 2, with vertex (0,2). Its range is y≥2 because m=3. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 8, 2, 8, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|3x + 2|, with vertex (-2/3,0), x-intercept x=-2/3 and range y≥0. Its y-intercept is 2. The values at x=−2, 0 and 2 are 4, 2, 8. A negative root of the original linear function cannot be reached as a positive argument |x| in f(|x|). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

10. −x − 5: compare the two folds

Try the original line. Let f(x)=−x − 5, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=−|x| − 5, with vertex (0,-5). Its range is y≤-5 because m=-1. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -7, -5, -7, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|−x − 5|, with vertex (-5,0), x-intercept x=-5 and range y≥0. Its y-intercept is 5. The values at x=−2, 0 and 2 are 3, 5, 7. A sample x-value left of the origin is particularly effective at distinguishing the transformations. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

11. -4x + 2: compare the two folds

Try the original line. Let f(x)=-4x + 2, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-4|x| + 2, with vertex (0,2). Its range is y≤2 because m=-4. The x-intercepts are x=−1/2 and x=1/2. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -6, 2, -6, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-4x + 2|, with vertex (1/2,0), x-intercept x=1/2 and range y≥0. Its y-intercept is 2. The values at x=−2, 0 and 2 are 10, 2, 6. The range of |mx+c| with m≠0 is always [0,∞). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

12. 4x − 7: compare the two folds

Try the original line. Let f(x)=4x − 7, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=4|x| − 7, with vertex (0,-7). Its range is y≥-7 because m=4. The x-intercepts are x=−7/4 and x=7/4. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 1, -7, 1, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|4x − 7|, with vertex (7/4,0), x-intercept x=7/4 and range y≥0. Its y-intercept is 7. The values at x=−2, 0 and 2 are 15, 7, 1. The domain is all real x for every linear function in this practice set. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

13. x + 1: compare the two folds

Try the original line. Let f(x)=x + 1, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=|x| + 1, with vertex (0,1). Its range is y≥1 because m=1. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 3, 1, 3, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|x + 1|, with vertex (-1,0), x-intercept x=-1 and range y≥0. Its y-intercept is 1. The values at x=−2, 0 and 2 are 1, 1, 3. The input-modulus graph always uses the original line’s right-hand half as its source. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

14. -3x − 9: compare the two folds

Try the original line. Let f(x)=-3x − 9, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-3|x| − 9, with vertex (0,-9). Its range is y≤-9 because m=-3. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -15, -9, -15, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-3x − 9|, with vertex (-3,0), x-intercept x=-3 and range y≥0. Its y-intercept is 9. The values at x=−2, 0 and 2 are 3, 9, 15. The output-modulus graph never has negative y-values, while f(|x|) may have them. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

15. 5x: compare the two folds

Try the original line. Let f(x)=5x, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=5|x|, with vertex (0,0). Its range is y≥0 because m=5. The x-intercepts are x=0. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 10, 0, 10, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|5x|, with vertex (0,0), x-intercept x=0 and range y≥0. Its y-intercept is 0. The values at x=−2, 0 and 2 are 10, 0, 10. For m>0, f(|x|)=m|x|+c has a minimum at (0,c). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

16. 2x + 8: compare the two folds

Try the original line. Let f(x)=2x + 8, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=2|x| + 8, with vertex (0,8). Its range is y≥8 because m=2. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 12, 8, 12, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|2x + 8|, with vertex (-4,0), x-intercept x=-4 and range y≥0. Its y-intercept is 8. The values at x=−2, 0 and 2 are 4, 8, 12. For m<0, f(|x|)=m|x|+c has a maximum at (0,c). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

17. -2x − 1: compare the two folds

Try the original line. Let f(x)=-2x − 1, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-2|x| − 1, with vertex (0,-1). Its range is y≤-1 because m=-2. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -5, -1, -5, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-2x − 1|, with vertex (-1/2,0), x-intercept x=-1/2 and range y≥0. Its y-intercept is 1. The values at x=−2, 0 and 2 are 3, 1, 5. The cusp of |mx+c| is located at the zero of the original line, not necessarily at x=0. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

18. -5x + 6: compare the two folds

Try the original line. Let f(x)=-5x + 6, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-5|x| + 6, with vertex (0,6). Its range is y≤6 because m=-5. The x-intercepts are x=−6/5 and x=6/5. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -4, 6, -4, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-5x + 6|, with vertex (6/5,0), x-intercept x=6/5 and range y≥0. Its y-intercept is 6. The values at x=−2, 0 and 2 are 16, 6, 4. If c=0 and m>0, these two transformed linear graphs are identical. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

19. 3x − 1: compare the two folds

Try the original line. Let f(x)=3x − 1, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=3|x| − 1, with vertex (0,-1). Its range is y≥-1 because m=3. The x-intercepts are x=−1/3 and x=1/3. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 5, -1, 5, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|3x − 1|, with vertex (1/3,0), x-intercept x=1/3 and range y≥0. Its y-intercept is 1. The values at x=−2, 0 and 2 are 7, 1, 5. If c=0 and m<0, they are reflections of each other in the x-axis. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

20. −x + 6: compare the two folds

Try the original line. Let f(x)=−x + 6, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=−|x| + 6, with vertex (0,6). Its range is y≤6 because m=-1. The x-intercepts are x=−6 and x=6. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 4, 6, 4, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|−x + 6|, with vertex (6,0), x-intercept x=6 and range y≥0. Its y-intercept is 6. The values at x=−2, 0 and 2 are 8, 6, 4. The input-modulus transformation can create two symmetric x-intercepts from a positive original root. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

21. -4x − 3: compare the two folds

Try the original line. Let f(x)=-4x − 3, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-4|x| − 3, with vertex (0,-3). Its range is y≤-3 because m=-4. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -11, -3, -11, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-4x − 3|, with vertex (-3/4,0), x-intercept x=-3/4 and range y≥0. Its y-intercept is 3. The values at x=−2, 0 and 2 are 5, 3, 11. A negative root of the original linear function cannot be reached as a positive argument |x| in f(|x|). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

22. 4x + 4: compare the two folds

Try the original line. Let f(x)=4x + 4, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=4|x| + 4, with vertex (0,4). Its range is y≥4 because m=4. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 12, 4, 12, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|4x + 4|, with vertex (-1,0), x-intercept x=-1 and range y≥0. Its y-intercept is 4. The values at x=−2, 0 and 2 are 4, 4, 12. A sample x-value left of the origin is particularly effective at distinguishing the transformations. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

23. x − 5: compare the two folds

Try the original line. Let f(x)=x − 5, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=|x| − 5, with vertex (0,-5). Its range is y≥-5 because m=1. The x-intercepts are x=−5 and x=5. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -3, -5, -3, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|x − 5|, with vertex (5,0), x-intercept x=5 and range y≥0. Its y-intercept is 5. The values at x=−2, 0 and 2 are 7, 5, 3. The range of |mx+c| with m≠0 is always [0,∞). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

24. -3x + 2: compare the two folds

Try the original line. Let f(x)=-3x + 2, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-3|x| + 2, with vertex (0,2). Its range is y≤2 because m=-3. The x-intercepts are x=−2/3 and x=2/3. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -4, 2, -4, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-3x + 2|, with vertex (2/3,0), x-intercept x=2/3 and range y≥0. Its y-intercept is 2. The values at x=−2, 0 and 2 are 8, 2, 4. The domain is all real x for every linear function in this practice set. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

25. 5x − 7: compare the two folds

Try the original line. Let f(x)=5x − 7, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=5|x| − 7, with vertex (0,-7). Its range is y≥-7 because m=5. The x-intercepts are x=−7/5 and x=7/5. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 3, -7, 3, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|5x − 7|, with vertex (7/5,0), x-intercept x=7/5 and range y≥0. Its y-intercept is 7. The values at x=−2, 0 and 2 are 17, 7, 3. The input-modulus graph always uses the original line’s right-hand half as its source. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

26. 2x + 1: compare the two folds

Try the original line. Let f(x)=2x + 1, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=2|x| + 1, with vertex (0,1). Its range is y≥1 because m=2. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 5, 1, 5, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|2x + 1|, with vertex (-1/2,0), x-intercept x=-1/2 and range y≥0. Its y-intercept is 1. The values at x=−2, 0 and 2 are 3, 1, 5. The output-modulus graph never has negative y-values, while f(|x|) may have them. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

27. -2x − 9: compare the two folds

Try the original line. Let f(x)=-2x − 9, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-2|x| − 9, with vertex (0,-9). Its range is y≤-9 because m=-2. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -13, -9, -13, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-2x − 9|, with vertex (-9/2,0), x-intercept x=-9/2 and range y≥0. Its y-intercept is 9. The values at x=−2, 0 and 2 are 5, 9, 13. For m>0, f(|x|)=m|x|+c has a minimum at (0,c). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

28. -5x + 1: compare the two folds

Try the original line. Let f(x)=-5x + 1, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-5|x| + 1, with vertex (0,1). Its range is y≤1 because m=-5. The x-intercepts are x=−1/5 and x=1/5. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -9, 1, -9, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-5x + 1|, with vertex (1/5,0), x-intercept x=1/5 and range y≥0. Its y-intercept is 1. The values at x=−2, 0 and 2 are 11, 1, 9. For m<0, f(|x|)=m|x|+c has a maximum at (0,c). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

29. 3x − 9: compare the two folds

Try the original line. Let f(x)=3x − 9, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=3|x| − 9, with vertex (0,-9). Its range is y≥-9 because m=3. The x-intercepts are x=−3 and x=3. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -3, -9, -3, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|3x − 9|, with vertex (3,0), x-intercept x=3 and range y≥0. Its y-intercept is 9. The values at x=−2, 0 and 2 are 15, 9, 3. The cusp of |mx+c| is located at the zero of the original line, not necessarily at x=0. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

30. −x: compare the two folds

Try the original line. Let f(x)=−x, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=−|x|, with vertex (0,0). Its range is y≤0 because m=-1. The x-intercepts are x=0. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -2, 0, -2, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|−x|, with vertex (0,0), x-intercept x=0 and range y≥0. Its y-intercept is 0. The values at x=−2, 0 and 2 are 2, 0, 2. If c=0 and m>0, these two transformed linear graphs are identical. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

31. -4x + 8: compare the two folds

Try the original line. Let f(x)=-4x + 8, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-4|x| + 8, with vertex (0,8). Its range is y≤8 because m=-4. The x-intercepts are x=−2 and x=2. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 0, 8, 0, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-4x + 8|, with vertex (2,0), x-intercept x=2 and range y≥0. Its y-intercept is 8. The values at x=−2, 0 and 2 are 16, 8, 0. If c=0 and m<0, they are reflections of each other in the x-axis. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

32. 4x − 1: compare the two folds

Try the original line. Let f(x)=4x − 1, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=4|x| − 1, with vertex (0,-1). Its range is y≥-1 because m=4. The x-intercepts are x=−1/4 and x=1/4. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 7, -1, 7, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|4x − 1|, with vertex (1/4,0), x-intercept x=1/4 and range y≥0. Its y-intercept is 1. The values at x=−2, 0 and 2 are 9, 1, 7. The input-modulus transformation can create two symmetric x-intercepts from a positive original root. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

33. x + 6: compare the two folds

Try the original line. Let f(x)=x + 6, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=|x| + 6, with vertex (0,6). Its range is y≥6 because m=1. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 8, 6, 8, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|x + 6|, with vertex (-6,0), x-intercept x=-6 and range y≥0. Its y-intercept is 6. The values at x=−2, 0 and 2 are 4, 6, 8. A negative root of the original linear function cannot be reached as a positive argument |x| in f(|x|). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

34. -3x − 3: compare the two folds

Try the original line. Let f(x)=-3x − 3, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=-3|x| − 3, with vertex (0,-3). Its range is y≤-3 because m=-3. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -9, -3, -9, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|-3x − 3|, with vertex (-1,0), x-intercept x=-1 and range y≥0. Its y-intercept is 3. The values at x=−2, 0 and 2 are 3, 3, 9. A sample x-value left of the origin is particularly effective at distinguishing the transformations. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

35. 5x + 4: compare the two folds

Try the original line. Let f(x)=5x + 4, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=5|x| + 4, with vertex (0,4). Its range is y≥4 because m=5. The x-intercepts are no real x-intercepts. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively 14, 4, 14, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|5x + 4|, with vertex (-4/5,0), x-intercept x=-4/5 and range y≥0. Its y-intercept is 4. The values at x=−2, 0 and 2 are 6, 4, 14. The range of |mx+c| with m≠0 is always [0,∞). Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

36. 2x − 5: compare the two folds

Try the original line. Let f(x)=2x − 5, for all real x. Sketch y=f(|x|) and y=|f(x)|, label their vertices, and find the complete set of x-intercepts of each. Decide whether the input-modulus graph opens upward or downward and whether the two transformed graphs have the same y-intercept. Begin by locating the unchanged half of the original line.

Input-modulus answer. The first graph is y=2|x| − 5, with vertex (0,-5). Its range is y≥-5 because m=2. The x-intercepts are x=−5/2 and x=5/2. For x≥0, it copies the original right-hand ray; for x<0, it copies its reflection. Its values at x=−2, 0 and 2 are respectively -1, -5, -1, confirming symmetry about the y-axis.

Output-modulus answer and transfer. The second graph is y=|2x − 5|, with vertex (5/2,0), x-intercept x=5/2 and range y≥0. Its y-intercept is 5. The values at x=−2, 0 and 2 are 9, 5, 1. The domain is all real x for every linear function in this practice set. Change the sign of c while keeping m fixed, then explain which vertices move and why neither operation should be sketched by merely copying the previous picture without recalculating the fold location.

Ten short self-checks

For f(x)=2x−4, f(|x|) has vertex (0,−4), roots x=±2 and range y≥−4.
For the same f, |f(x)| has vertex (2,0), root x=2 and range y≥0.
At x=−2 for that f, f(|x|)=0 but |f(x)|=8.
For f(x)=−x+3, f(|x|)=3−|x| has maximum (0,3).
For the same f, |f(x)|=|3−x| has minimum (3,0).
For f(x)=x²−1, f(|x|)=f(x) because f is even.
But |x²−1| reflects the negative central arc upward.
For f(x)=x³, f(|x|)=|f(x)|=|x|³.
For f(x)=−x², f(|x|)=−x² while |f(x)|=x².
Taking |f(x)| preserves f’s domain; taking f(|x|) requires |x| to lie in the domain of f.

Five-day graph transformation practice

Day one: plot five straight lines and label their original intercepts. Day two: build f(|x|) by keeping only the right-hand half and reflecting it in the y-axis, checking symmetry and range. Day three: build |f(x)| by keeping positive output regions and reflecting negative regions above the x-axis, finding the new vertex at an original zero. Day four: add even and odd polynomial functions to discover cases where one transformation changes nothing or where both transformations agree. Day five: mix formulas with restricted domains, then sketch from a supplied original graph instead of from an equation.

In a three-learner tutorial, one student can sketch the original f, a second can construct f(|x|), and the third can construct |f(x)| and audit all three at chosen x-values. Rotate the roles. The learner should be able to explain which axis is used for reflection and whether the operation changes an input or an output, instead of using “modulus makes a V” as an unreliable rule.

Frequently asked questions

Does f(|x|) always have nonnegative values?

No. It modifies the input only. For f(x)=2x−4, the transformed output at x=0 is −4.

Does |f(x)| always produce an even graph?

No. It is nonnegative but need not be symmetric about the y-axis. For f(x)=2x−4, |f(x)| is centred at x=2.

Is f(|x|) always even?

Where the input domain is symmetric as required, yes: replacing x by −x does not change |x|, so the output is the same.

Can the two transformed graphs be identical?

Yes. For f(x)=x³, both become |x|³. Compare definitions rather than assuming they must differ.

Where is the fold for |f(x)|?

It occurs where the original graph crosses the x-axis or otherwise changes from positive to negative output; portions below the axis reflect upward.

Which part is copied for f(|x|)?

The original graph for nonnegative inputs x≥0 is kept and reflected to the negative-input side.

How do the domains compare?

|f(x)| keeps the domain of f; f(|x|) accepts only x whose absolute value belongs to f’s domain.

Is this in the 2027 H2 9758 syllabus?

Yes. Section 1.2 explicitly includes relating y=f(x), y=|f(x)| and y=f(|x|).

JC2 H2 Mathematics Tutor Bukit Timah | Graphing f(|x|) Versus |f(x)|