Graphical Relationships: $y = 1/f(x)$
Sketch the reciprocal of a function from its graph, identifying asymptotes and key features. A zero becomes an asymptote, an asymptote becomes a zero — and the sign of the function is always preserved. Master these rules and you can transform any graph instantly.
If $f(2) = 0$, what do you expect to happen to $y = 1/f(x)$ at $x = 2$? What if $f(x) \to \infty$ as $x \to 3$ — what would $1/f(x)$ do near $x = 3$? Write your prediction before reading on.
There are only two rules driving every reciprocal graph sketch. Lock them in now:
Rule 1: where $f(x) = 0$, the reciprocal $1/f(x)$ has a vertical asymptote. Rule 2: where $f(x) \to \pm\infty$, the reciprocal $1/f(x) \to 0$. Sign is always preserved — positive stays positive, negative stays negative.
Key facts
- How key features of $f(x)$ transform to features of $y = 1/f(x)$
- Zeros of $f$ become vertical asymptotes of $1/f$
- Invariant points occur where $f(x) = \pm 1$
Concepts
- The relationship between the sign of $f(x)$ and the sign of $1/f(x)$
- Why maxima of $f$ become minima of $1/f$ and vice versa
- How the size of $f$ determines the size of $1/f$ (large becomes small)
Skills
- Sketch $y = 1/f(x)$ given the graph of $y = f(x)$
- Identify all asymptotes, invariant points and sign regions
- Describe the relationship between max/min of $f$ and $1/f$
Every feature of $y = f(x)$ has a corresponding feature in $y = 1/f(x)$. Use this table to convert features systematically.
| Feature of $y = f(x)$ | Feature of $y = 1/f(x)$ |
|---|---|
| x-intercept ($f = 0$) | Vertical asymptote |
| Vertical asymptote ($f \to \pm\infty$) | x-intercept (approaches $y = 0$) |
| $f(x) = 1$ | Invariant point: $1/f(x) = 1$ |
| $f(x) = -1$ | Invariant point: $1/f(x) = -1$ |
| $f(x) > 0$ | $1/f(x) > 0$ (same sign) |
| $f(x) < 0$ | $1/f(x) < 0$ (same sign) |
| Maximum at $f = a$ (where $a > 0$) | Minimum at $1/f = 1/a$ |
| Minimum at $f = a$ (where $a > 0$) | Maximum at $1/f = 1/a$ |
Zeros of f(x) become vertical asymptotes of 1/f(x); Vertical asymptotes of f(x) become zeros (x-axis crossings) of 1/f(x)
Pause — copy the feature-exchange rules into your book: zeros of $f$ become vertical asymptotes of $1/f$; asymptotes of $f$ become zeros of $1/f$; turning points of $f$ flip to turning points of $1/f$; sign is preserved.
Quick check: The graph of $y = f(x)$ crosses the x-axis at $x = 3$. What does $y = 1/f(x)$ have at $x = 3$?
We just saw the feature-exchange table: zeros become asymptotes, asymptotes become zeros, and sign is preserved. That raises a question: in a sketch, where do you start placing the graph of $1/f(x)$ first, and which points are the most reliable anchors? This card answers it → start with asymptotes (from zeros) and invariant points where $f = pm 1$; then sketch branches that go large when $f$ is near zero.
The diagram below shows the key principles: when $f(x)$ is close to zero (small), $1/f(x)$ is large; when $f(x)$ is large, $1/f(x)$ is close to zero.
Feature transformations from f(x) to 1/f(x), and sketching the reciprocal of a linear function.
Step-by-step method to sketch $y = 1/f(x)$:
- Find all zeros of $f(x)$ — these become vertical asymptotes.
- Find any vertical asymptotes of $f(x)$ — these become zeros (x-intercepts) of $1/f$.
- Mark invariant points where $f(x) = 1$ or $f(x) = -1$.
- Note sign regions: $1/f$ is positive where $f$ is positive, negative where $f$ is negative.
- Convert turning points: maxima of $f$ become minima of $1/f$ (at the reciprocal $y$-value), and vice versa.
- Sketch the branches, remembering $1/f \to 0$ where $f \to \pm\infty$.
Step-by-step method: (1) zeros → asymptotes; (2) asymptotes → zeros; (3) mark f = 1 as invariant; (4) preserve sign; (5) flip turning points; Where f is large, 1/f is small (near x-axis); where f is small (near zero), 1/f is large
Pause — copy the five-step reciprocal sketching method into your book: (1) zeros → asymptotes; (2) asymptotes → zeros; (3) invariant points where $f = \pm 1$; (4) preserve sign; (5) turning points of $f$ map to turning points of $1/f$.
Did you get this? True or false: the graph of $y = 1/f(x)$ has the same x-intercepts as $y = f(x)$.
Worked examples · 3 in a row, reveal as you go
Given $f(x) = x - 2$, sketch $y = 1/f(x)$ showing all key features.
The graph of $y = f(x)$ crosses the x-axis at $x = 1$ and $x = 4$, and has a maximum at $(2.5,\, 3)$. Describe the key features of $y = 1/f(x)$.
If $f(x)$ has a minimum value of 2 at $x = 0$, what is the maximum value of $1/f(x)$? Explain your reasoning.
Fill the gap: If $f(x)$ has a maximum at $y = 5$, then $y = 1/f(x)$ has a minimum at $y = $ at the same x-value. (Enter as a decimal.)
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if $f(x)$ is always positive and has no zeros, then $y = 1/f(x)$ has no vertical asymptotes.
Activities · practice with the ideas
Given $f(x) = x + 1$, state the vertical asymptote and horizontal asymptote of $y = 1/f(x)$.
The graph of $y = f(x)$ crosses the x-axis at $x = -2$ and $x = 5$. State the vertical asymptotes of $y = 1/f(x)$.
If $f(3) = 4$, find the value of $1/f(x)$ at $x = 3$ and state whether this is a maximum or minimum if $f$ has a maximum at $x = 3$.
Find the invariant points on $y = 1/f(x)$ given that $f(x) = x - 2$.
Explain in your own words why a maximum of $f(x)$ becomes a minimum of $1/f(x)$.
Odd one out: Three of these statements about $y = 1/f(x)$ are always true. Which one is NOT always true?
Earlier you were asked: if $f(2) = 0$, what happens to $1/f(x)$ at $x = 2$?
The answer: division by zero is undefined, so $1/f(x)$ has a vertical asymptote at $x = 2$ — not a zero or a regular point. And where $f(x) \to \infty$ as $x \to 3$, the reciprocal $1/f(x) \to 0$, meaning the graph approaches the x-axis there.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Given $f(x) = x + 1$, sketch $y = 1/f(x)$ showing all key features including asymptotes and invariant points. (2 marks)
Q2. If $f(x)$ has a minimum value of 2, what is the maximum value of $1/f(x)$? Explain your reasoning. (2 marks)
Q3. The graph of $y = f(x)$ crosses the x-axis at $x = 1$ and $x = 4$, and has a maximum at $(2.5, 3)$. Describe the key features of $y = 1/f(x)$, including all asymptotes, turning points and sign regions. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. Vertical asymptote $x = -1$; horizontal $y = 0$ · 2. Vertical asymptotes at $x = -2$ and $x = 5$ · 3. $1/f(3) = 1/4$; it is a minimum of $1/f$ (maximum of $f$ becomes minimum of $1/f$) · 4. Solve $x - 2 = 1 \Rightarrow x = 3$: invariant $(3,1)$; solve $x - 2 = -1 \Rightarrow x = 1$: invariant $(1,-1)$ · 5. At a maximum, $f$ is at its largest value, so $1/f$ is at its smallest (minimum). Large number → small reciprocal.
Q1 (2 marks): $y = 1/(x+1)$. Vertical asymptote at $x = -1$ [1]. Horizontal asymptote $y = 0$; invariant point at $x = 0$ where $f = 1$: $(0, 1)$; at $x = -2$ where $f = -1$: $(-2, -1)$ [1].
Q2 (2 marks): Maximum value of $1/f(x) = 1/2$ [1]. Reasoning: a minimum of $f$ at $y = 2$ corresponds to a maximum of $1/f$ at $y = 1/2$ because the smallest positive value of $f$ gives the largest positive value of $1/f$ [1].
Q3 (3 marks): Vertical asymptotes at $x = 1$ and $x = 4$ [1]. Minimum of $1/f$ at $(2.5, 1/3)$ — the maximum of $f$ at $y = 3$ becomes a minimum of $1/f$ at $y = 1/3$ [1]. Sign: for $1 < x < 4$, $f > 0$ so $1/f > 0$; for $x < 1$ or $x > 4$, $f < 0$ so $1/f < 0$, with $1/f \to 0$ as $x \to \pm\infty$ [1].
Five timed questions on reciprocal graphs. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering reciprocal graph questions. Lighter alternative to the boss.
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