# Graphing Related Functions

> Build new graphs from old ones. Sketch 1/f(x), |f(x)|, f(|x|) and the sum or difference of two functions.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/graphical-relationships/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The sound engineer's trick
Mina mixes music at a radio station. She has the graph of one sound wave on her screen. With one button she can flip every dip of the wave upward. With another she can add a second wave on top. She never draws a new wave from scratch. She builds it from the one she already has.

Mathematicians work the same way. If you know the graph of $y = f(x)$ well, you can quickly sketch $y = \dfrac{1}{f(x)}$, $y = |f(x)|$ and $y = f(|x|)$ without plotting a single table of values.

**In short:** Related graphs are built from a known graph $y = f(x)$. For $y = \dfrac{1}{f(x)}$, the zeros of $f$ become vertical asymptotes and big heights become small. For $y = |f(x)|$, every part below the x-axis flips up, and for $y = f(|x|)$, the right half is mirrored in the y-axis.

## Starting point: what you already know

You will use these ideas from Years 9 and 10 and from Mathematics Advanced.

- **Function notation.** $f(x)$ is the height of the graph above the point $x$. So $f(3) = 5$ means the point $(3, 5)$ is on the graph.
- **The hyperbola** $y = \dfrac{1}{x}$ has a vertical asymptote at $x = 0$ and a horizontal asymptote at $y = 0$. An **asymptote** is a line the graph gets closer and closer to but never reaches.
- **Absolute value.** $|x|$ is the distance of $x$ from 0, so it is never negative. $|5| = 5$ and $|-5| = 5$.
- The graph of $y = |x|$ is a V shape with its corner at the origin.

### Key idea: One big idea
Each new graph is made from the heights of the old one. Ask: "What happens to each height?" and you can sketch it.

## The reciprocal: y = 1/f(x)

To get $y = \dfrac{1}{f(x)}$, take every height of $f$ and turn it upside down: a height of 2 becomes $\dfrac{1}{2}$, and a height of $-4$ becomes $-\dfrac{1}{4}$. Use these rules.

1. **Zeros become asymptotes.** Where $f(x) = 0$, you would divide by zero, so $\dfrac{1}{f(x)}$ has a vertical asymptote.
2. **Big becomes small.** Where $f(x)$ is huge, $\dfrac{1}{f(x)}$ is close to 0.
3. **1 and $-1$ stay put.** If $f(x) = 1$ then $\dfrac{1}{f(x)} = 1$ too. The graphs meet where the height is 1 or $-1$.
4. **Same sign.** Above the axis stays above. Below stays below.
5. **Up becomes down.** Where $f$ goes up, $\dfrac{1}{f}$ goes down. A minimum of $f$ becomes a maximum of $\dfrac{1}{f}$.

*Diagram: Graph of y equals x squared minus 1 and its reciprocal, with vertical asymptotes at x equals minus 1 and 1. Both graphs pass through 0, minus 1.*
*f(x) = x² − 1 and its reciprocal. Where f is zero, 1/f has an asymptote.*

### Worked example: Sketch y = 1/(x² − 1)
Let $f(x) = x^2 - 1$. It is a parabola with roots $\pm 1$ and lowest point $(0, -1)$.

- Roots $x = \pm 1$ become vertical asymptotes.
- As $x \to \pm\infty$, $f$ is huge, so $\dfrac{1}{f} \to 0$. The x-axis is a horizontal asymptote.
- The minimum $(0, -1)$ becomes a maximum $(0, -1)$ of the reciprocal, because $\dfrac{1}{-1} = -1$.
- Between $-1$ and 1, $f$ is negative, so the reciprocal is below the axis there.

## The absolute value: y = |f(x)|

Absolute value makes every height positive. So **keep the parts above the x-axis and flip the parts below the axis up**, like a mirror on the x-axis.

*Diagram: The graph of y equals the absolute value of x squared minus 4. The dashed part of the parabola below the x-axis between minus 2 and 2 is flipped up above the axis.*
*y = |x² − 4|: the part under the x-axis is flipped up.*

## Mirror the right side: y = f(|x|)

Now the absolute value is on the $x$, not on the height. Since $f(|-3|) = f(3)$, the left side is a copy of the right side.

**Rule:** draw $y = f(x)$ for $x \ge 0$ only, then reflect that part in the y-axis. The result is always symmetrical about the y-axis.

### Worked example: Compare the two
Let $f(x) = x - 2$.

- $y = |x - 2|$ is a V with its corner at $(2, 0)$.
- $y = |x| - 2$ is a V with its corner at $(0, -2)$, because we keep the line for $x \ge 0$ and mirror it.

### Watch out: Do not mix them up
$|f(x)|$ flips **downward parts up** (mirror in the x-axis). $f(|x|)$ copies the **right half to the left** (mirror in the y-axis).

## Adding and subtracting graphs

For $y = f(x) + g(x)$, add the two heights at each $x$. For $y = f(x) - g(x)$, subtract them.

Useful checks:

- Where $g(x) = 0$, the sum $f + g$ sits exactly on $f$.
- Where $f(x) = -g(x)$, the sum is zero.
- The sum only exists where **both** functions exist. The domain is the overlap of the two domains.

### Exam tip: A fast sketch
Pick a few easy $x$ values, such as the zeros of each function. Add the heights there, plot those points and join them smoothly.

**Mind map**

```text
Graphing related functions
  Reciprocal 1/f(x)
    Zeros become asymptotes
    Big heights become small
    Meets f at heights 1 and -1
    Max becomes min
  |f(x)|
    Flip the parts below the x-axis up
  f(|x|)
    Keep x ≥ 0
    Mirror in the y-axis
  Sum and difference
    Add or subtract heights
    Domain is the overlap
```

### Year 11: Domain and range of a sum
Let $f(x) = \sqrt{x}$ and $g(x) = \sqrt{4 - x}$. Then $f$ needs $x \ge 0$ and $g$ needs $x \le 4$, so $h(x) = \sqrt{x} + \sqrt{4 - x}$ has domain $0 \le x \le 4$.

At the ends, $h(0) = 2$ and $h(4) = 2$. By symmetry the largest value is at $x = 2$, where $h(2) = 2\sqrt{2}$. So the range is $2 \le y \le 2\sqrt{2}$.

### Year 11: How many solutions?
To find how many solutions $|x^2 - 4| = k$ has, draw $y = |x^2 - 4|$ and the line $y = k$ and count crossings. The hump between $-2$ and 2 reaches 4.

- $k < 0$: no solutions.
- $k = 0$: two solutions, $x = \pm 2$.
- $0 < k < 4$: four solutions.
- $k = 4$: three solutions.
- $k > 4$: two solutions.

### Year 11: The graph of y² = f(x)
$y^2 = f(x)$ means $y = \pm\sqrt{f(x)}$, so draw the square root of each height, above and below the axis.
- It only exists where $f(x) \geq 0$.
- It is symmetrical about the $x$-axis.
- Heights between 0 and 1 get bigger (since $\sqrt{0.25} = 0.5$) and heights above 1 get smaller.
- Where $f(x) = 0$ with $f$ crossing the axis, the graph meets the axis with a **vertical** tangent.

For example $y^2 = x(4 - x)$ is $x^2 - 4x + y^2 = 0$, that is $(x - 2)^2 + y^2 = 4$: a circle with centre $(2, 0)$ and radius 2.

### Year 11: Inequalities and absolute values with graphs
**Solving with a sketch.** To solve $|x - 1| > 2x - 4$, sketch both sides and find where they meet. They can only meet on the right arm, $x - 1 = 2x - 4$, at $x = 3$. To the left of $x = 3$ the V is above the line, so $x < 3$.

**Sums of absolute values.** $|x - a| + |x - b|$ is the total distance from $x$ to $a$ and to $b$. Between $a$ and $b$ it is constant, equal to the gap $|b - a|$. So $|x - 1| + |x + 3| \leq 6$ holds between the points where $2x + 2 = 6$ and $-2x - 2 = 6$: $-4 \leq x \leq 2$.

**Fractions of polynomials.** For $y = \dfrac{x^2 - 9}{x^2 - 1}$: vertical asymptotes where the bottom is zero ($x = \pm 1$), the horizontal asymptote $y = 1$ from the leading terms, and the $y$-intercept 9.

**Counting solutions of $f(|x|) = k$.** Count the solutions of $f(x) = k$ with $x > 0$, then double them (and add one if $x = 0$ is a solution).

### Year 12: Where this returns in Year 12
In integration, the integral counts area below the axis as negative. To find the **total area** between a curve and the x-axis, you integrate $|f(x)|$, which is the same as flipping the negative parts up, exactly as in this chapter.

## Quick summary

- $y = \dfrac{1}{f(x)}$: zeros of $f$ become vertical asymptotes, large heights become small, and the graphs meet at heights $\pm 1$.
- $y = |f(x)|$: flip every part below the x-axis up.
- $y = f(|x|)$: keep the part where $x \ge 0$ and mirror it in the y-axis.
- $y = f(x) \pm g(x)$: add or subtract heights. The domain is where both exist.

## Quiz

Try these questions from the chapter quiz.

1. If $f(x) = 0$ at $x = 3$, what does $y = \dfrac{1}{f(x)}$ have at $x = 3$?
   - A. a zero
   - B. a vertical asymptote
   - C. a turning point
   - D. a horizontal asymptote
   - Answer: B. a vertical asymptote. You cannot divide by zero, so the reciprocal has a vertical asymptote there.

2. Where do $y = f(x)$ and $y = \dfrac{1}{f(x)}$ always meet?
   - A. where $f(x) = 0$
   - B. where $f(x) = 1$ or $f(x) = -1$
   - C. at the origin only
   - D. they never meet
   - Answer: B. where $f(x) = 1$ or $f(x) = -1$. $\dfrac{1}{1} = 1$ and $\dfrac{1}{-1} = -1$, so those heights do not change.

3. $f(x)$ has a minimum point at $(2, 4)$. The reciprocal has:
   - A. a maximum at $\left(2, \tfrac{1}{4}\right)$
   - B. a minimum at $\left(2, \tfrac{1}{4}\right)$
   - C. a maximum at $(2, 4)$
   - D. an asymptote at $x = 2$
   - Answer: A. a maximum at $\left(2, \tfrac{1}{4}\right)$. Small heights become big and big become small, so a minimum turns into a maximum with height $\dfrac{1}{4}$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/graphical-relationships/#quiz