Further Transformations & Synthesis
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: absolute-value transformations, reciprocal graphs, and the synthesis of every transformation technique you have learned so far.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
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.
Further Transformations
Key insight: Absolute value and reciprocal transformations are not dilations or translations — they change the fundamental shape of the graph in predictable but non-linear ways.
Key facts
- The effect of $y = |f(x)|$, $y = f(|x|)$, and $y = \frac{1}{f(x)}$ on a graph
- How zeros, turning points, and asymptotes behave under these transformations
- The complete toolkit of graph transformations
Concepts
- Why $|f(x)|$ creates "cusps" at $x$-intercepts
- Why $f(|x|)$ always produces an even graph
- Why the reciprocal has vertical asymptotes where $f(x) = 0$
Skills
- Sketch $y = |f(x)|$, $y = f(|x|)$, and $y = \frac{1}{f(x)}$ from $y = f(x)$
- Identify the complete sequence of transformations in any equation
- Solve mixed transformation problems under exam conditions
There are two distinct absolute value transformations — one on the outside of the function, and one on the inside.
$y = |f(x)|$ — Outside the Function
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.
- $x$-intercepts of $f(x)$ become points where the graph touches but does not cross the $x$-axis (often forming a sharp "cusp")
- Any minimum below the $x$-axis becomes a local maximum on the reflected portion
- The range of $|f(x)|$ is always a subset of $[0, \infty)$
$y = f(|x|)$ — Inside the Function
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.
- If $f(x)$ is already even, then $f(|x|) = f(x)$ — no change
- If $f(x)$ is not even, the left side of the original graph is discarded and replaced
- This can create unexpected turning points at $x = 0$
$y = |f(x)|$: reflect below-axis portions above $x$-axis; $x$-intercepts become cusps; range $\subseteq [0, \infty)$; $y = f(|x|)$: always even; keep right side ($x \geq 0$); mirror right side across $y$-axis
Pause — copy both absolute value rules: $|f(x)|$ reflects below-axis parts up (cusps at $x$-intercepts), and $f(|x|)$ keeps the right half and mirrors it left into your book.
True or false: The graph of $y = |f(x)|$ can have $y$-values below the $x$-axis.
Quick check: When graphing $y = f(|x|)$, what happens to the left side of the original graph $y = f(x)$?
We just saw that $y = |f(x)|$ folds the graph above zero and $y = f(|x|)$ forces even symmetry. That raises a question: what happens to a graph if instead of modifying the output we divide 1 by it? This card answers it → zeros become vertical asymptotes, large values shrink to near zero, and turning points swap between max and min.
$y = \dfrac{1}{f(x)}$
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:
- Mark the zeros of $f(x)$ — these become vertical asymptotes
- Mark where $f(x) = 1$ and $f(x) = -1$ — these are invariant points
- Mark turning points of $f(x)$ — they become opposite turning points on the reciprocal
- Sketch the reciprocal curve in each interval, making sure it approaches asymptotes and passes through invariant points with the correct general shape
Zeros of $f(x)$ → vertical asymptotes of $\frac{1}{f(x)}$; Maxima $\leftrightarrow$ minima — turning points swap type; $y$-coord inverts
Pause — copy the three reciprocal rules: zeros become vertical asymptotes, maxima and minima swap type, and the $y$-coordinate inverts into your book.
Match up: For $y = \frac{1}{f(x)}$, match each feature of $f(x)$ to its image.
| $x$-intercept of $f(x)$ | → | Vertical asymptote |
| Maximum of $f(x)$ | → | Minimum (inverted $y$-coord) |
| $f(x) = 1$ | → | Invariant point ($y = 1$) |
Review the table above, then confirm you understand these three mappings before moving on.
Worked examples · reveal as you go
Sketch the graph of $y = |x^2 - 4|$.
Sketch the graph of $y = (|x| - 1)^2$.
Given $f(x) = x^2 - 4$, sketch $y = \dfrac{1}{f(x)}$.
Common mistakes · the 4 traps that cost marks
Thinking $y = |f(x)|$ reflects the whole graph in the $x$-axis
Only the parts below the $x$-axis are reflected. Points above or on the axis stay fixed. This is a selective reflection, not a global one.
✓ Fix: Draw the original graph lightly, then "fold up" everything below the axis.
Believing $y = f(|x|)$ reflects the whole graph in the $y$-axis
You do not reflect the whole graph. You discard the left side entirely ($x < 0$) and replace it with a mirror image of the right side ($x \geq 0$).
✓ Fix: Only keep or redraw the $x \geq 0$ portion, then mirror that across the $y$-axis.
Drawing the reciprocal curve crossing its asymptotes
The reciprocal graph must approach vertical asymptotes but never touch or cross them. Some students draw curves that look like parabolas passing through the zeros — this is completely wrong.
✓ Fix: Remember: zeros of $f(x)$ become forbidden lines (asymptotes) for $y = \frac{1}{f(x)}$.
Forgetting that reciprocal maxima and minima swap
A maximum on $f(x)$ becomes a minimum on $\frac{1}{f(x)}$ (and vice versa), provided the maximum is not at zero. The $y$-coordinate inverts: if $f(x)$ has a minimum of $-4$, the reciprocal has a maximum of $-\frac{1}{4}$.
✓ Fix: When transforming turning points, take the reciprocal of the $y$-coordinate and flip the nature (max $\leftrightarrow$ min).
Activity 1 — Absolute Value & Reciprocal Transformations
For each function below, describe the transformation from the given parent and list the key features (intercepts, asymptotes, turning points).
Parent: $f(x) = x - 2$. Sketch $y = |f(x)|$.
Parent: $f(x) = (x - 1)^2$. Sketch $y = f(|x|)$.
Parent: $f(x) = x^2 - 1$. Sketch $y = \frac{1}{f(x)}$.
Parent: $f(x) = \sqrt{x}$. Sketch $y = f(|x|)$.
Fill the blanks: drag each token into the correct blank.
Under $y = |f(x)|$, $x$-intercepts become ___. Under $y = f(|x|)$, the result is always an ___ function. Under $y = \frac{1}{f(x)}$, zeros of $f(x)$ become vertical ___. The minimum value of $|f(x)|$ is ___.
Quick-fire practice · 5 reps +2 XP per reveal
What is the range of $y = |x^2 - 9|$?
If $f(x) = x - 3$ has a zero at $x = 3$, what feature does $\frac{1}{f(x)}$ have at $x = 3$?
Is $y = f(|x|)$ always an even function? Why?
If $f(x)$ has a maximum of $8$ at $x = 2$, what is the corresponding point on $y = \frac{1}{f(x)}$?
The graph of $y = f(x)$ passes through $(\sqrt{5}, 1)$ and $(-\sqrt{5}, 1)$. Are these invariant points of $y = \frac{1}{f(x)}$?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q8. 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)
Q9. 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)
Q10. A function $g(x)$ is defined as $g(x) = \dfrac{1}{x^2 - 9}$. (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)
Comprehensive answers (click to reveal)
Multiple choice — drill bank
MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.
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 $f(x) = 0$.
Activity 1 — model answers
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: $y \geq 0$.
Short answer model answers
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) Range is $(-\infty, -\frac{1}{9}] \cup (0, \infty)$ [1.5].
Five timed questions on further transformations. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
Enter the arenaClimb platforms, hit checkpoints, and answer transformation questions. Quick recall, lighter than the boss.
Mark lesson as complete
Tick when you've finished the practice and review.