Topic 2: Modulus and Piecewise Graphs

The starting point for every modulus expression is

$$\boxed{ |A|= \begin{cases} A,&A\geq0,\\ -A,&A<0. \end{cases} }$$

When a modulus appears in an equation, the first question is:

When is the expression inside the modulus non-negative, and when is it negative?

These conditions determine which parts of the equation apply in different regions.

A modulus graph can therefore be understood by constructing its pieces. Reflections and symmetries then follow from the algebra rather than needing to be memorised as separate rules.

The general approach is

$$\boxed{ \text{identify the expression} \longrightarrow \text{find its sign} \longrightarrow \text{construct the pieces} \longrightarrow \text{use symmetry}. }$$

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1. Modulus Applied to the Output

Consider $y=|f(x)|$.

The expression inside the modulus is $f(x)$.

By definition,

$$\boxed{ |f(x)|= \begin{cases} f(x),&f(x)\geq0,\\ -f(x),&f(x)<0. \end{cases} }$$

This means that the graph of $y=|f(x)|$ is constructed from the graph of $y=f(x)$ as follows:

• Keep the parts on or above the $x$-axis unchanged.
• Reflect the parts below the $x$-axis upwards.
• Keep the roots in their original positions.

Example: A Quadratic Graph

Let $f(x)=x^2-4$.

The original graph crosses the $x$-axis at $x=-2 \qquad\text{and}\qquad x=2$.

The quadratic is non-negative when $x\leq-2 \qquad\text{or}\qquad x\geq2$, and negative when $-2<x<2$.

Therefore,

$$\boxed{ |x^2-4|= \begin{cases} x^2-4,&x\leq-2,\\ 4-x^2,&-2<x<2,\\ x^2-4,&x\geq2. \end{cases} }$$

The roots divide the original graph into three pieces. Only the middle piece is reflected.

The vertex $(0,-4)$ becomes $(0,4)$, while the roots remain at $x=-2$ and $x=2$.

Key principle

$$\boxed{ y=|f(x)|: \text{ reflect only the parts of }y=f(x) \text{ below the }x\text{-axis}. }$$

The decision depends on the sign of the output $f(x)$, not on the sign of $x$.

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2. Modulus Applied to the Input

Now consider $y=f(|x|)$.

This time the modulus is applied to the input.

Since

$$|x|= \begin{cases} x,&x\geq0,\\ -x,&x<0, \end{cases}$$

we obtain

$$\boxed{ f(|x|)= \begin{cases} f(x),&x\geq0,\\ f(-x),&x<0. \end{cases} }$$

For non-negative inputs, nothing changes.

For negative inputs, the function receives the corresponding positive input instead.

Consequently, the graph for $x\geq0$ determines the entire transformed graph.

Example: Constructing the Graph from Its Right-Hand Part

Let $f(x)=x^2-2x$.

Then $f(|x|)=|x|^2-2|x|$.

For $x\geq0$, $f(|x|)=x^2-2x$.

For $x<0$, $f(|x|)=x^2+2x$.

Thus

$$\boxed{ f(|x|)= \begin{cases} x^2+2x,&x<0,\\ x^2-2x,&x\geq0. \end{cases} }$$

The transformed graph is symmetric about the $y$-axis because $f(|-x|)=f(|x|)$.

Key principle

$$\boxed{ y=f(|x|): \text{ keep the right-hand part of }y=f(x) \text{ and reflect it across the }y\text{-axis}. }$$

Comparing the Two Transformations

Although the expressions look similar, they act differently.

For $y=|f(x)|$, the input remains unchanged. Negative outputs become positive.

For $y=f(|x|)$, negative inputs are replaced by their positive counterparts.

In particular:

• $y=|f(x)|$ is always non-negative.
• $y=f(|x|)$ is always symmetric about the $y$-axis.
• $y=f(|x|)$ can still take negative values.
• $y=|f(x)|$ need not be symmetric about the $y$-axis.

Always identify whether the modulus acts on the input or the output before drawing the graph.

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3. Nested Modulus Graphs

Some expressions contain more than one modulus sign.

For these graphs, work from the inside out. Construct the result of each operation before applying the next one.

Consider $y=\left||1-x|-1\right|$.

The outer modulus acts on the whole expression $|1-x|-1$, not just on $1-x$.

Step 1: Draw the Inner Graph

Start with $y=|1-x|=|x-1|$.

This is a V-shaped graph with vertex $(1,0)$.

Step 2: Translate the Graph Down by 1

Next construct $y=|1-x|-1$.

Every point moves down by one unit.

The vertex becomes $(1,-1)$.

To find where the graph crosses the $x$-axis, solve $|1-x|-1=0$.

Thus $|1-x|=1$, giving $x=0 \qquad\text{or}\qquad x=2$.

The intermediate graph is negative precisely when $0<x<2$.

Step 3: Apply the Outer Modulus

Now construct $y=\left||1-x|-1\right|$.

The outer modulus leaves the non-negative parts of the intermediate graph unchanged.

It reflects the negative section between $x=0$ and $x=2$ upwards.

Therefore:

• The graph is unchanged for $x\leq0$ and $x\geq2$.
• The section between $x=0$ and $x=2$ is reflected in the $x$-axis.
• The roots at $x=0$ and $x=2$ remain fixed.
• The point $(1,-1)$ becomes $(1,1)$.

The final graph has two global minima: $(0,0) \qquad\text{and}\qquad (2,0)$.

It also has a local maximum at $(1,1)$.

The outer branches continue upwards, so this local maximum is not a global maximum.

Step 4: Confirm the Graph Using a Piecewise Expression

The inner modulus changes its rule at $x=1$:

$$|1-x|= \begin{cases} 1-x,&x\leq1,\\ x-1,&x\geq1. \end{cases}$$

After subtracting $1$, the intermediate graph is

$$|1-x|-1= \begin{cases} -x,&x\leq1,\\ x-2,&x\geq1. \end{cases}$$

The outer modulus introduces two further boundaries at $x=0$ and $x=2$.

Hence

$$\boxed{ \left||1-x|-1\right|= \begin{cases} -x,&x\leq0,\\ x,&0<x\leq1,\\ 2-x,&1<x\leq2,\\ x-2,&x>2. \end{cases} }$$

The four straight-line pieces meet at $(0,0),\qquad (1,1),\qquad (2,0)$, as shown in the graph.

The piecewise expression confirms the construction. It is not necessary to derive every piece before beginning the graphical transformation.

General Principle

For a nested modulus of the form $y=\left||f(x)|-a\right|, \qquad a\geq0$, construct the graph in three stages.

First draw $y=|f(x)|$.

Next translate it down by $a$: $y=|f(x)|-a$.

Finally, reflect any part of this intermediate graph below the $x$-axis upwards.

Before the final reflection, identify the zeros and determine where the intermediate expression is negative.

$$\boxed{ \text{work from the inside out} \longrightarrow \text{locate the negative sections} \longrightarrow \text{reflect them upwards}. }$$

The essential rule is to apply each operation to the current graph, not to all the modulus signs simultaneously.

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4. Putting the Modulus on (y)

Now consider $|y|=f(x)$.

This is different from $y=|f(x)|$.

Since $|y|\geq0$, solutions of $|y|=f(x)$ are possible only when $f(x)\geq0$.

If $f(x)>0$, there are two possible values: $y=f(x) \qquad\text{and}\qquad y=-f(x)$.

If $f(x)=0$, there is only one value: $y=0$.

If $f(x)<0$, there are no solutions.

Thus

$$\boxed{ |y|=f(x) \quad\Longleftrightarrow\quad \begin{cases} y=\pm f(x),&f(x)>0,\\ y=0,&f(x)=0,\\ \text{no point},&f(x)<0. \end{cases} }$$

Example: A Relation with Two Branches

Consider $|y|=x^2-1$.

The right-hand side must be non-negative: $x^2-1\geq0$.

Therefore, $x\leq-1 \qquad\text{or}\qquad x\geq1$.

There are no points satisfying the relation when $-1<x<1$.

For the allowed values of $x$, $y=\pm(x^2-1)$.

The graph is symmetric about the $x$-axis.

However, it is generally a relation rather than a function of $x$, because a single input may correspond to two different outputs.

For example, when $x=2$, $|y|=3$, so $y=3 \qquad\text{or}\qquad y=-3$.

Compare the Two Expressions

For $y=|f(x)|$, every input in the domain of $f$ produces exactly one output. The graph lies on or above the $x$-axis.

For $|y|=f(x)$, inputs where $f(x)<0$ produce no points, while inputs where $f(x)>0$ produce two points.

The distinction is: $y=|f(x)|: \text{ negative outputs are reflected upwards}.$

$$\boxed{ |y|=f(x): \text{ negative values of }f(x)\text{ are forbidden,} \text{ and positive values give two branches}. }$$

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5. Two Modulus Signs Divide the Plane into Regions

Modulus can also appear in equations involving both coordinates.

Consider $|x|+|y|=4$.

There are two expressions inside modulus signs.

Their forms depend on the signs of both $x$ and $y$, so the coordinate axes divide the plane into four regions.

Each equation must be drawn only in its corresponding region.

Step 1: Construct the First-Quadrant Piece

In the first quadrant, $x\geq0,\qquad y\geq0$.

Therefore, $|x|=x, \qquad |y|=y$, and the equation becomes $x+y=4$.

The relevant part is only the segment joining $(0,4) \qquad\text{and}\qquad (4,0)$.

The rest of the infinite line does not belong to this sign case.

Step 2: Construct the Other Three Pieces

In the second quadrant, $-x+y=4$.

This gives the segment from $(-4,0)$ to $(0,4)$.

In the fourth quadrant, $x-y=4$.

This gives the segment from $(4,0)$ to $(0,-4)$.

In the third quadrant, $-x-y=4$.

This gives the segment from $(-4,0)$ to $(0,-4)$.

Together, the four segments form a diamond.

The four pieces can be written as

$$\boxed{ |x|+|y|=4 \quad\Longleftrightarrow\quad \begin{cases} x+y=4,&x\geq0,\ y\geq0,\\ -x+y=4,&x<0,\ y\geq0,\\ x-y=4,&x\geq0,\ y<0,\\ -x-y=4,&x<0,\ y<0. \end{cases} }$$

Each boundary point belongs to the graph and is included by the relevant non-strict inequalities.

Step 3: Recognise the Symmetry

Now examine the original equation.

Since $|-x|=|x|, \qquad |-y|=|y|$, changing the sign of either coordinate leaves the equation unchanged.

Therefore the graph is symmetric about both coordinate axes.

Once the first-quadrant segment is understood, the remaining three segments can be constructed by reflection.

This gives a faster method:

$$\boxed{ \text{construct the first-quadrant piece} \longrightarrow \text{reflect it across the axes}. }$$

The sign cases explain why the graph has this structure. Symmetry allows us to complete it without separately solving every case.

From an Equation to a Region

A closely related inequality is $|x|+|y|\leq4$.

The equation gives the diamond-shaped boundary.

The inequality includes all points inside that boundary as well.

For example, in the first quadrant, $x+y\leq4, \qquad x\geq0,\quad y\geq0$, describes the triangular region between the coordinate axes and the segment $x+y=4$.

Reflecting this triangular region across both axes produces the full diamond-shaped region.

The distinction between the boundary and the entire region is important in problems involving intersections and areas.

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6. The General Relation (f(|x|,|y|)=0)

The same reasoning extends beyond expressions such as $|x|+|y|=4$.

Consider a relation $f(|x|,|y|)=0$, where $f$ is an expression involving two variables.

Because both modulus inputs are non-negative, begin with the relation $f(u,v)=0, \qquad u\geq0,\quad v\geq0$.

This identifies the part of the graph in the first quadrant, including the axes.

The Four Sign Regions

The original equation has the following forms.

These four cases construct the complete relation.

However, once the first-quadrant piece is known, symmetry gives a shorter route.

Suppose $(a,b), \qquad a>0,\quad b>0$, satisfies $f(a,b)=0$.

Then the four points $(a,b),\qquad (-a,b),\qquad (a,-b),\qquad (-a,-b)$ all satisfy $f(|x|,|y|)=0$.

This is because each point gives the same pair of modulus values: $(|x|,|y|)=(a,b)$.

Points on the axes are shared by reflected pieces and should be counted only once.

The resulting relation is symmetric about both coordinate axes because $|-x|=|x|$ and $|-y|=|y|$.

General method

$$\boxed{ \text{construct the first-quadrant part} \longrightarrow \text{reflect it across both axes}. }$$

This method applies to equations and inequalities involving $|x|$ and $|y|$, provided the original relation is defined for the required non-negative inputs.

The first-quadrant piece may be a curve, a line segment or an entire region. The same symmetry principle applies.

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Practical Approach

When faced with an unfamiliar modulus graph or relation, begin with the definition of modulus rather than a memorised reflection rule.

1. Identify the expression inside each modulus sign. Decide whether the modulus acts on an input, an output or a coordinate.

2. Find where that expression is non-negative or negative. Roots and coordinate axes often determine the boundaries.

3. Rewrite the expression in the relevant regions. Each piece must be used only where its sign conditions hold.

4. Construct the graph from those pieces. Check the roots, endpoints and points where the rule changes.

5. Use symmetry when appropriate. Once the piecewise structure is understood, reflection may eliminate unnecessary repeated calculations.

6. Check whether the result is a function or a relation. In particular, an equation such as $|y|=f(x)$ can give two outputs for one input.

For nested modulus expressions, apply one operation at a time, beginning with the innermost expression.

$$\boxed{ \text{identify signs} \longrightarrow \text{construct the pieces} \longrightarrow \text{use symmetry} \longrightarrow \text{verify}. }$$

The central principle: A modulus graph is first a piecewise graph. Understanding where expressions change sign explains the reflections, symmetries and restrictions that appear in the final graph.