Graphs of Inverse Functions
Every function has a mirror twin. Reflecting in the line $y = x$ swaps inputs and outputs — turning x-intercepts into y-intercepts, horizontal asymptotes into vertical ones, and producing a graph that is the exact reflection of the original. In this lesson you'll learn to sketch inverse graphs, track how features transform, and identify the special class of self-inverse functions.
If $(2, 5)$ is on the graph of $y = f(x)$, what point must be on the graph of $y = f^{-1}(x)$? Without looking ahead — explain why in your own words.
Everything in this lesson flows from two ideas: reflecting in $y = x$ swaps coordinates, and some functions are their own inverse.
Every inverse graph question reduces to one of two tasks: swap each coordinate to find a point on the inverse, or test $f(f(x)) = x$ to show self-inverse. Master these two moves and any inverse graph problem becomes straightforward.
Key facts
- The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in $y = x$
- Every point $(a, b)$ on $f$ maps to $(b, a)$ on $f^{-1}$
- Horizontal and vertical asymptotes swap under the reflection
Concepts
- How key features (intercepts, asymptotes) transform under the reflection
- Why points on $y = x$ are invariant under the reflection
- What it means geometrically for a function to be self-inverse
Skills
- Sketch the graph of $f^{-1}$ from the graph of $f$
- Identify and transform key features under the reflection
- Show a function is self-inverse using $f(f(x)) = x$
The graph of $y = f^{-1}(x)$ is obtained by reflecting the graph of $y = f(x)$ in the line $y = x$. This transformation swaps the coordinates of every point:
Key feature transformations:
- The x-intercept of $f$ (a point $(a,0)$) becomes the y-intercept $(0,a)$ of $f^{-1}$.
- The y-intercept of $f$ (a point $(0,b)$) becomes the x-intercept $(b,0)$ of $f^{-1}$.
- A horizontal asymptote $y = k$ on $f$ becomes the vertical asymptote $x = k$ on $f^{-1}$.
- A vertical asymptote $x = c$ on $f$ becomes the horizontal asymptote $y = c$ on $f^{-1}$.
Invariant points lie on $y = x$, so $(a, a)$ satisfies $f(a) = a$ and stays fixed under the reflection.
Reflecting $y = f(x)$ in $y = x$ gives $y = f^{-1}(x)$. Asymptote types swap; invariant points sit on $y = x$.
(a,b) on f (b,a) on f^{-1} — always swap both coordinates; x-intercept of f becomes y-intercept of f^{-1}, and vice versa
Pause — copy the reflection rule into your book: the graph of $f^{-1}$ is the reflection of $f$ in the line $y = x$; every point $(a,b)$ on $f$ maps to $(b,a)$ on $f^{-1}$, so $x$- and $y$-intercepts swap.
Quick check: The point $(3, -1)$ lies on $y = f(x)$. Which point lies on $y = f^{-1}(x)$?
We just saw that the graph of $f^{-1}$ is the reflection of $f$ in $y = x$, swapping every point $(a,b)$ to $(b,a)$. That raises a question: what happens to a function whose graph already lies on $y = x$ — can a function be its own inverse? This card answers it → a self-inverse function satisfies $f(f(x)) = x$; its graph is symmetric about $y = x$ and $f^{-1} = f$.
A function is self-inverse if $f^{-1}(x) = f(x)$, meaning its graph is symmetric about $y = x$. Equivalently:
Common self-inverse functions:
- $f(x) = x$ (the identity function — trivially its own inverse)
- $f(x) = -x$ (negation)
- $f(x) = \dfrac{1}{x}$ (reciprocal)
- $f(x) = a - x$ for any constant $a$ (reflection plus shift)
- $f(x) = \dfrac{k}{x}$ for any constant $k \ne 0$
To test whether $f$ is self-inverse, compute $f(f(x))$ and check whether it simplifies to $x$.
Self-inverse: f^{-1}(x) = f(x), equivalently f(f(x)) = x for all x in the domain; Test method: compute f(f(x)); if it equals x, the function is self-inverse
Pause — copy the self-inverse definition into your book: $f$ is self-inverse if $f^{-1}(x) = f(x)$, equivalently $f(f(x)) = x$; the graph is symmetric about $y = x$; test by computing $f(f(x))$ and checking it simplifies to $x$.
Did you get this? True or false: $f(x) = \dfrac{1}{x}$ is self-inverse because $f(f(x)) = x$.
Worked examples · 2 in a row, reveal as you go
Sketch $y = f^{-1}(x)$ given that $f$ passes through $(0, 2)$ and $(1, 3)$, has horizontal asymptote $y = 1$, and vertical asymptote $x = -2$.
Show that $f(x) = \dfrac{3}{x}$ is self-inverse.
Fill the gap: If $f(x)$ has a horizontal asymptote $y = 5$, then $f^{-1}(x)$ has a vertical asymptote $x =$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $f(x) = 4 - x$ is self-inverse because $f(f(x)) = 4 - (4 - x) = x$.
Activities · practice with the ideas
The point $(5, -2)$ lies on $y = f(x)$. Write down the corresponding point on $y = f^{-1}(x)$.
$f(x)$ has a vertical asymptote $x = 2$ and a horizontal asymptote $y = -1$. State the asymptotes of $y = f^{-1}(x)$.
Show that $f(x) = 4 - x$ is self-inverse. Use Method 2 ($f(f(x)) = x$).
$f(x)$ passes through $(-1, 4)$, $(0, 2)$, and $(3, 0)$. Sketch $y = f^{-1}(x)$ by plotting the transformed points and joining with a smooth curve.
Explain why a function that is NOT one-to-one does not have an inverse function, and why this matters when sketching the inverse graph.
Odd one out: Three of these functions are self-inverse. Which one is NOT?
Earlier you predicted which point on $f^{-1}$ corresponds to $(2, 5)$ on $f$.
The answer is $(5, 2)$ — the coordinates are simply swapped. This is because the reflection in $y = x$ swaps the roles of $x$ and $y$. Every input becomes an output and every output becomes an input. The line $y = x$ is the mirror, and any point on that line (like $(3, 3)$) stays fixed.
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. The point $(3, -1)$ lies on $y = f(x)$. Write down the corresponding point on $y = f^{-1}(x)$. (1 mark)
Q2. $f(x)$ has a vertical asymptote $x = 2$ and horizontal asymptote $y = -1$. State the asymptotes of $y = f^{-1}(x)$. (2 marks)
Q3. Show that $f(x) = 4 - x$ is self-inverse. (2 marks)
Comprehensive answers (click to reveal)
Activities: 1. $(-2, 5)$ [swap coordinates of $(5,-2)$] · 2. Vertical asymptote $x = -1$, horizontal asymptote $y = 2$ [swap asymptote types and values] · 3. $f(f(x)) = f(4-x) = 4-(4-x) = x$ ✓ · 4. Transformed points: $(4,-1)$, $(2,0)$, $(0,3)$ · 5. A non-one-to-one function fails the horizontal line test, so its "inverse" relation is not a function — it would have multiple $y$-values for a single $x$.
Q1 (1 mark): $(-1, 3)$ — swap the coordinates of $(3, -1)$. [1]
Q2 (2 marks): Vertical asymptote of $f$, $x = 2$, becomes horizontal asymptote $y = 2$ of $f^{-1}$ [1]. Horizontal asymptote of $f$, $y = -1$, becomes vertical asymptote $x = -1$ of $f^{-1}$ [1].
Q3 (2 marks): $f(f(x)) = f(4 - x) = 4 - (4 - x) = 4 - 4 + x = x$ [1]. Since $f(f(x)) = x$, $f$ is self-inverse [1].
Five timed questions on inverse graphs and self-inverse functions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering inverse function questions. A lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.