What is the primary difference in the range of $y = |f(x)|$ compared to $y = f(|x|)$?

The range of $y = |f(x)|$ is restricted to non-negative values ($y \ge 0$) because the modulus is applied to the final output. In contrast, $y = f(|x|)$ can have negative y-values because the modulus only ensures the input to the function is positive, not the resulting output.

When sketching $y = f(|x|)$, what happens to the original part of the graph where $x < 0$?

The original part of the graph where $x < 0$ is completely discarded. It is replaced by a mirror image of the portion of the graph where $x \ge 0$, resulting in a graph that is perfectly symmetrical about the y-axis.

How does the vertex of $y = |x+3|-2$ differ from the vertex of $y = |x-3|+2$?

The vertex of $y = |x+3|-2$ is at $(-3, -2)$, while the vertex of $y = |x-3|+2$ is at $(3, 2)$. The vertex is found by identifying the value of $x$ that makes the modulus term zero and the resulting $y$ value.

What is a common mistake when reflecting $y = f(x)$ to find $y = |f(x)|$?

A common error is reflecting the entire graph across the x-axis. Only the portions of the graph that lie below the x-axis ($y < 0$) should be reflected; the portions already above the x-axis must remain in their original positions.

Why is it incorrect to assume $y = f(|x|)$ will never go below the x-axis?

The modulus in $f(|x|)$ only affects the input $x$, ensuring the function evaluates $f$ at positive values. If the function $f(x)$ produces negative outputs for positive inputs (e.g., $f(2) = -5$), then $y = f(|x|)$ will also have negative outputs (e.g., $y = f(|-2|) = -5$).

What error occurs if a student reflects $y = f(x)$ across the y-axis to find $y = |f(x)|$?

This is a confusion of transformation axes. $y = |f(x)|$ is a vertical transformation requiring reflection across the x-axis (the line $y=0$), whereas reflection across the y-axis is related to horizontal transformations like $f(-x)$ or $f(|x|)$.

Define the modulus function $|x|$ algebraically.

The modulus function is defined piecewise: $|x| = x$ for $x \ge 0$, and $|x| = -x$ for $x < 0$. This ensures the result is always the magnitude of the number without its sign.

What is the 'vertex' of a modulus graph?

The vertex is the sharp point or 'corner' of a modulus graph where the gradient changes abruptly. For a linear modulus function $y = a|x+p|+q$, it is the point $(-p, q)$ where the expression inside the modulus equals zero.

What does the parameter 'a' represent in the function $y = a|x+p|+q$?

The parameter 'a' determines the vertical stretch and the orientation of the graph. If $a > 0$, the graph opens upwards (V-shape); if $a < 0$, it opens downwards (inverted V-shape), and its magnitude determines the steepness of the lines.

Why is the graph of $y = f(|x|)$ always an even function?

An even function satisfies $f(x) = f(-x)$. Since $|x| = |-x|$, the input to the function $f$ is identical for both $x$ and $-x$, resulting in identical outputs and perfect symmetry across the y-axis.

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Sketching Graphs of Modulus Functions

Summary

The modulus function, also known as the absolute value function, transforms any real number input into its non-negative equivalent. In graphing, this operation results in specific geometric transformations—either reflecting negative output values across the x-axis or creating symmetry across the y-axis—depending on whether the modulus is applied to the entire function or just the independent variable.

1. Definition & Core Concepts

2. Sketching $y = |f(x)|$

The transformation $y = |f(x)|$ ensures that the output $y$ is never negative. Geometrically, this means no part of the graph can exist below the x-axis.
Step 1: Sketch the original function $y = f(x)$ using a light or dashed line to serve as a guide.
Step 2: Identify all portions of the graph where $y < 0$ (the parts below the x-axis) and reflect them vertically across the x-axis into the positive y-region.
Step 3: Retain the original portions of the graph where $y \ge 0$ and erase or ignore the original negative portions.

Comparison of y=|f(x)| and y=f(|x|) transformations. The left shows a reflection of negative y-values across the x-axis. The right shows a reflection of the positive x-region across the y-axis.

3. Sketching $y = f(|x|)$

4. Linear Modulus Graphs: $y = a|x + p| + q$

5. Key Distinctions

6. Exam Strategy & Tips

Sketching Graphs of Modulus Functions

Q: When sketching $y = f(|x|)$, what happens to the original part of the graph where $x < 0$?

The original part of the graph where $x < 0$ is completely discarded. It is replaced by a mirror image of the portion of the graph where $x \ge 0$, resulting in a graph that is perfectly symmetrical about the y-axis.

Q: How does the vertex of $y = |x+3|-2$ differ from the vertex of $y = |x-3|+2$?

The vertex of $y = |x+3|-2$ is at $(-3, -2)$, while the vertex of $y = |x-3|+2$ is at $(3, 2)$. The vertex is found by identifying the value of $x$ that makes the modulus term zero and the resulting $y$ value.

Q: What is a common mistake when reflecting $y = f(x)$ to find $y = |f(x)|$?

A common error is reflecting the entire graph across the x-axis. Only the portions of the graph that lie below the x-axis ($y < 0$) should be reflected; the portions already above the x-axis must remain in their original positions.

Q: Why is it incorrect to assume $y = f(|x|)$ will never go below the x-axis?