What is the primary difference in the reflection axis between $y = |f(x)|$ and $y = f(|x|)$?

$y = |f(x)|$ reflects the portion of the graph below the x-axis ($y 0$) onto the left side.

How does the range of $y = |f(x)|$ compare to the range of $y = f(|x|)$?

The range of $y = |f(x)|$ is restricted to $[0, \infty)$, meaning it can never be negative. However, $y = f(|x|)$ can have negative y-values if the original function $f(x)$ produces negative outputs for positive $x$.

Why is solving $|f(x)| = k$ algebraically potentially misleading without a sketch?

Algebraic methods (like squaring or taking $\pm$) can produce 'extraneous solutions' that do not actually exist on the graph. A sketch confirms the exact number of intersections and their approximate locations.

What common error occurs when sketching $y = f(|x|)$ regarding the left side of the graph?

Students often try to reflect the existing left side of the graph. In reality, the original left side ($x 0$).

What mistake is made when a student reflects $y = |f(x)|$ across the y-axis instead of the x-axis?

Reflecting across the y-axis changes the input behavior ($f(-x)$), whereas $|f(x)|$ is an output transformation. This error results in the wrong symmetry and often ignores negative y-values that should be reflected upward.

Why is it incorrect to assume a modulus graph always looks like a 'V'?

Only linear functions with a modulus (e.g., $y = |ax+b|$) create a 'V' shape. Modulus versions of quadratics or trig functions (e.g., $y = |x^2-4|$) will have curved reflected sections, forming 'W' or 'bouncing' shapes.

Define the modulus function piecewise.

The modulus function $|x|$ is defined as $x$ if $x \ge 0$, and $-x$ if $x < 0$. This ensures the result is always the non-negative magnitude of the input.

What are the coordinates of the vertex for the general linear modulus function $y = a|x + p| + q$?

The vertex is located at $(-p, q)$. This is the point where the expression inside the modulus is zero, representing the minimum (if $a>0$) or maximum (if $a<0$) point of the 'V' or '^'.

What is the condition for a modulus graph $y = a|x + p| + q$ to have exactly two x-intercepts?

For two roots to exist, the vertex must be on one side of the x-axis and the 'arms' must point toward it. For example, if $a > 0$ (V-shape), the vertex y-coordinate $q$ must be less than 0.

Explain why the graph of $y = f(|x|)$ must be symmetrical about the y-axis.

Because the modulus $|x|$ yields the same value for $x$ and $-x$, the function $f(|x|)$ calculates the same output for both. This identity $f(|x|) = f(|-x|)$ is the definition of y-axis symmetry.

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International A-Level

Sketching Graphs of Modulus Functions

Summary

The modulus function, or absolute value, is a transformation that maps every real input to its non-negative magnitude. In coordinate geometry, this creates distinct visual patterns: the graph of $y = |f(x)|$ reflects negative outputs across the x-axis, while $y = f(|x|)$ creates y-axis symmetry by reflecting the positive domain across the vertical axis. Mastering these sketches is essential for solving modulus equations and understanding non-linear piecewise behavior.

1. Definition & Core Concepts

2. Underlying Principles

Diagram showing the transformation of a linear function into a modulus function by reflecting the negative portion across the x-axis.

3. Method: Sketching y = |f(x)|

Step 1: Sketch the graph of the 'inner' function $y = f(x)$ using a light pencil or dashed line, ignoring the modulus at first.
Step 2: Identify the portions of the graph that lie strictly below the x-axis (where $y < 0$ ).
Step 3: Reflect these negative portions vertically across the x-axis so they become positive; the x-intercepts remain fixed as they are the 'pivot' points.
Step 4: Erase or ignore the original dashed lines below the x-axis to leave the final 'V' or 'W' shaped curve.

4. Method: Sketching y = f(|x|)

5. Key Distinctions

6. Exam Strategy & Tips