In a recording studio, audio engineers take sound waves and transform them in extreme ways — clipping off negative parts, mirroring halves, or inverting amplitudes. The mathematics behind these effects is exactly what you will master in this lesson: absolute-value transformations, reciprocal graphs, and the synthesis of every transformation technique you have learned so far.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Consider the basic parabola $y = x^2 - 4$, which has $x$-intercepts at $x = -2$ and $x = 2$, and a minimum at $(0, -4)$. How would the graph of $y = |x^2 - 4|$ differ? And how would the graph of $y = \frac{1}{x^2 - 4}$ differ? Try to describe at least two key changes for each.
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: (a + b)² = a² + b².
Right: (a + b)² = a² + 2ab + b²; the middle term 2ab is essential and commonly forgotten.
📚 Core Content
This transformation takes the absolute value of the output. Any part of the original graph that lies below the $x$-axis is reflected above it. Points that are already on or above the $x$-axis stay exactly where they are.
This transformation replaces every $x$ with $|x|$. Since $|x|$ is an even function ($|-x| = |x|$), the output $f(|x|)$ is also even. The graph for $x \geq 0$ is identical to $y = f(x)$, and the graph for $x < 0$ is a mirror image of the right side across the $y$-axis.
The reciprocal transformation is one of the most dramatic. Instead of changing the sign or position of points, it inverts the $y$-coordinates. This produces several predictable effects:
| Feature of $y = f(x)$ | Becomes on $y = \frac{1}{f(x)}$ |
|---|---|
| $x$-intercept ($f(x) = 0$) | Vertical asymptote |
| Vertical asymptote ($f(x) \to \pm\infty$) | $x$-intercept ($y \to 0$) |
| Maximum turning point | Minimum turning point (and vice versa) |
| $f(x) = 1$ | Fixed point ($y = 1$) |
| $f(x) = -1$ | Fixed point ($y = -1$) |
| $f(x) > 0$ | $\frac{1}{f(x)} > 0$ (same sign) |
| $f(x) < 0$ | $\frac{1}{f(x)} < 0$ (same sign) |
Sketching strategy for reciprocals:
🧮 Worked Examples
🧪 Activities
1 Parent: $f(x) = x - 2$. Sketch $y = |f(x)|$.
Type your answer:
Answer in your workbook.
2 Parent: $f(x) = (x - 1)^2$. Sketch $y = f(|x|)$.
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Answer in your workbook.
3 Parent: $f(x) = x^2 - 1$. Sketch $y = \\frac{1}{f(x)}$.
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Answer in your workbook.
4 Parent: $f(x) = \\sqrt{x}$. Sketch $y = f(|x|)$.
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Answer in your workbook.
1 $y = -2f(3x - 6) + 1$
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Answer in your workbook.
2 $y = |f(x - 2)| + 3$
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Answer in your workbook.
3 $y = \\frac{1}{f(-x)}$
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Answer in your workbook.
Earlier you were asked: How would $y = |x^2 - 4|$ and $y = \frac{1}{x^2 - 4}$ differ from $y = x^2 - 4$?
$y = |x^2 - 4|$: The portion of the parabola between $x = -2$ and $x = 2$ that dips below the $x$-axis is reflected above it. The minimum at $(0, -4)$ becomes a local maximum at $(0, 4)$. The $x$-intercepts at $x = \pm 2$ become sharp cusps where the graph touches but does not cross the axis.
$y = \frac{1}{x^2 - 4}$: The zeros at $x = \pm 2$ become vertical asymptotes — the graph can never touch those lines. The minimum at $(0, -4)$ becomes a maximum at $(0, -\frac{1}{4})$. As $x \to \pm\infty$, the curve approaches $y = 0$. Between the asymptotes the graph is negative; outside them it is positive.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. The graph of $y = f(x)$ has $x$-intercepts at $x = 1$ and $x = 5$, a maximum at $(3, 4)$, and a $y$-intercept at $(0, -5)$. Sketch the graph of $y = |f(x)|$, labelling the images of these key features. 3 MARKS
Type your answer below (describe the sketch):
Answer in your workbook.
9. Sketch the graph of $y = f(|x|)$ where $f(x) = (x - 2)^2 - 1$. Show all $x$-intercepts, the $y$-intercept, and any turning points. Explain why the graph is symmetric about the $y$-axis. 4 MARKS
Type your answer below:
Answer in your workbook.
10. A function $g(x)$ is defined as $g(x) = \\frac{1}{x^2 - 9}$. (a) State the equations of the vertical asymptotes. (b) Find the coordinates of any turning points. (c) Explain why $g(x)$ has no $x$-intercepts. (d) State the range of $g(x)$. 4 MARKS
Type your answer below:
Answer in your workbook.
1. $y = |x - 2|$: V-shape with vertex (cusp) at $(2, 0)$. $y$-intercept at $(0, 2)$. Slope $-1$ for $x < 2$, slope $1$ for $x > 2$.
2. $y = (|x| - 1)^2$: W-shape with minima at $(-1, 0)$ and $(1, 0)$, maximum at $(0, 1)$. $y$-intercept at $(0, 1)$.
3. $y = \frac{1}{x^2 - 1}$: VA at $x = -1$ and $x = 1$. HA at $y = 0$. Maximum at $(0, -1)$. Invariant points at $x = \pm\sqrt{2}$ ($y = 1$). Positive outside asymptotes, negative between them.
4. $y = \sqrt{|x|}$: Even function. For $x \geq 0$, same as $\sqrt{x}$. For $x < 0$, mirror of right side. Domain: all real $x$. Range: $y \geq 0$.
1. Factor: $3x - 6 = 3(x - 2)$. Transformations: horizontal dilation by factor $\frac{1}{3}$, right 2, vertical dilation by 2, reflection in $x$-axis, up 1.
2. Right 2, then absolute value of output (reflect below-axis parts up), then up 3.
3. Reflection in $y$-axis, then reciprocal transformation.
1. A — Reflect negative parts above $x$-axis.
2. A — Discard left side, mirror right side across $y$-axis.
3. A — A reciprocal is never zero; it has VAs at zeros of $f(x)$.
4. B — Minimum of $|f(x)|$ is 0 at any $x$-intercept.
5. A — Vertical asymptotes where denominator $f(x) = 0$.
Q8 (3 marks): $x$-intercepts at $(1, 0)$ and $(5, 0)$ remain but become cusps [1]. Maximum $(3, 4)$ stays at $(3, 4)$ since it is above axis [0.5]. $y$-intercept $(0, -5)$ reflects to $(0, 5)$ [0.5]. Any negative portion between $x = 1$ and $x = 5$ (if any) reflects above axis; correct sketch description [1].
Q9 (4 marks): For $x \geq 0$, $y = (x - 2)^2 - 1$ has $x$-intercepts at $x = 1$ and $x = 3$, vertex at $(2, -1)$ [1]. For $x < 0$, mirror the right side: additional $x$-intercept at $x = -1$, vertex at $(-2, -1)$ [1]. $y$-intercept at $(0, 3)$ [0.5]. The graph is symmetric about the $y$-axis because $f(|x|)$ depends only on $|x|$, and $|-x| = |x|$ [1.5].
Q10 (4 marks): (a) $x = -3$ and $x = 3$ [0.5]. (b) Parent $y = x^2 - 9$ has minimum at $(0, -9)$, so reciprocal has maximum at $(0, -\frac{1}{9})$ [1]. (c) A fraction equals zero only when its numerator is zero; here the numerator is 1, so $g(x)$ is never zero [1]. (d) The reciprocal of a parabola opening upward with minimum $-9$ gives positive values outside the asymptotes and negative values between them, with a maximum of $-\frac{1}{9}$. So range is $(-\infty, -\frac{1}{9}] \cup (0, \infty)$ [1.5].
Further Transformations & Synthesis
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