# Graph Transformations

> Moving, flipping and stretching graphs. Translations, reflections and dilations of any function.

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

### Story time: The phone photo editor
Aisha takes a photo of her cat and opens the editor. She drags it to the right. She flips it like a mirror. She pinches to stretch it taller. The cat is still the same cat. Only its position and size changed.

Graphs work the same way. Once you know the shape of $y = x^2$, you can drag it, flip it and stretch it to get thousands of other graphs. You do not need to plot a table every time. You just ask: what was done to the basic shape?

**In short:** A graph transformation moves, flips or stretches a graph. $f(x) + c$ moves it up by $c$, $f(x - b)$ moves it right by $b$, $-f(x)$ flips it in the x-axis, and $kf(x)$ stretches it vertically. Changes inside the bracket do the opposite of what they look like.

## Starting point: Years 9 and 10

Start with $y = x^2$, vertex at the origin.

- $y = x^2 + 3$ moves it **up** 3.
- $y = x^2 - 3$ moves it **down** 3.
- $y = (x - 2)^2$ moves it **right** 2.
- $y = (x + 2)^2$ moves it **left** 2.
- $y = -x^2$ **flips** it upside down.
- $y = 3x^2$ makes it **narrower** (steeper). $y = \dfrac{1}{3}x^2$ makes it wider.

*Diagram: The parabola y equals x squared and a copy moved 2 right and 1 up*
*Right 2 and up 1: the vertex moves from (0, 0) to (2, 1).*

### Key idea: Inside goes the opposite way
A change **outside** the function moves the graph the way you expect: $+3$ is up. A change **inside** with $x$ does the opposite: $(x - 2)$ moves it right, not left. Think of it as: "what $x$ makes the bracket zero?" Here $x = 2$.

### Worked example: Vertex from the equation
Where is the vertex of $y = -(x + 1)^2 + 4$?

Inside: $x + 1 = 0$ at $x = -1$, so move left 1. Outside: up 4. The minus sign flips it. Vertex $(-1, 4)$, opening downward, so it is a maximum.

### Watch out: The sign inside the bracket
$y = (x + 5)^2$ has its vertex at $x = -5$, not $x = 5$.

**Mind map**

```text
Transformations
  Translations
    f(x) + c: up c
    f(x - b): right b
  Reflections
    -f(x): in the x-axis
    f(-x): in the y-axis
  Dilations
    k f(x): vertical stretch by k
    f(ax): horizontal by 1/a
  Combined
    y = k f(a(x + b)) + c
    Inside first, then outside
```

### Year 11: Translations of any function
These rules work for every function, not just parabolas.

- $y = f(x) + c$: translate **up** $c$ units (down if $c < 0$).
- $y = f(x - b)$: translate **right** $b$ units (left if $b < 0$).

**Example.** $y = \dfrac{1}{x}$ moved right 3 and down 2 is $y = \dfrac{1}{x - 3} - 2$. The asymptotes move with it: $x = 3$ and $y = -2$.

$y = \sqrt{x}$ moved left 4 is $y = \sqrt{x + 4}$, which starts at $(-4, 0)$.

### Year 11: Reflections and dilations
**Reflections**
- $y = -f(x)$ reflects in the x-axis (every $y$ value changes sign).
- $y = f(-x)$ reflects in the y-axis (every $x$ value changes sign).

**Dilations**
- $y = k f(x)$ stretches **vertically** by a factor of $k$. Every y-value is multiplied by $k$. Point $(2, 4)$ becomes $(2, 4k)$.
- $y = f(ax)$ stretches **horizontally** by a factor of $\dfrac{1}{a}$. Every x-value is divided by $a$. Point $(2, 4)$ on $y = f(x)$ becomes $\left(\dfrac{2}{a}, 4\right)$.

So $y = \sin 2x$ is $y = \sin x$ squashed to half its width.

### Year 12: Combining transformations
The general form is

$$y = k f\big(a(x + b)\big) + c$$

Work from the inside out:
1. Horizontal dilation by $\dfrac{1}{a}$ (and reflection in the y-axis if $a < 0$).
2. Horizontal translation by $-b$.
3. Vertical dilation by $k$ (and reflection in the x-axis if $k < 0$).
4. Vertical translation by $c$.

**Mapping a point.** Under $y = 2f(3(x - 1)) + 5$, the point $(6, 4)$ on $y = f(x)$ goes to $x = \dfrac{6}{3} + 1 = 3$ and $y = 2(4) + 5 = 13$. New point: $(3, 13)$.

**Watch the brackets.** $f(2x - 6) = f(2(x - 3))$ is a shift right **3**, not 6. Always factor out $a$ first.

Transformations also change the domain and range. If $f$ has range $[0, 2]$, then $3f(x) - 1$ has range $[-1, 5]$.

**Working backwards.** To find the original point from an image, undo the steps in reverse order. If $(5, 2)$ is on $y = g(3(x - 1)) - 4$, then the original $y$ is $2 + 4 = 6$ and the original $x$ is $3(5 - 1) = 12$, so the point was $(12, 6)$.

**Reflections with a shift.** $f(2 - x) = f\big(-(x - 2)\big)$: reflect in the $y$-axis, then move right 2.

**From an expanded equation.** Complete the square to see the transformations: $y = 2x^2 + 12x + 13 = 2(x + 3)^2 - 5$ is $y = x^2$ stretched vertically by 2, moved left 3 and down 5.

**Logs and exponentials can be rewritten.** $\ln(2x - 2) = \ln 2 + \ln(x - 1)$, so the same graph is $y = \ln x$ moved right 1 and up $\ln 2$. And $e^{x + \ln 3} = 3e^x$, so moving $y = e^x$ left $\ln 3$ is the same as stretching it vertically by 3.

**Tangents move too.** A translation does not change gradients. If the tangent to $y = f(x)$ at $x = 1$ is $y = 2x + 1$, then $y = f(x - 3) + 4$ has a tangent of gradient 2 at $x = 4$, through the moved point $(4, 7)$: $y = 2x - 1$.

## Quick summary

- $f(x) + c$ moves up $c$. $f(x - b)$ moves right $b$.
- $-f(x)$ flips in the x-axis. $f(-x)$ flips in the y-axis.
- $k f(x)$ stretches vertically by $k$. $f(ax)$ stretches horizontally by $\dfrac{1}{a}$.
- Changes inside the bracket act on $x$ and do the opposite of what they look like.
- For $y = k f(a(x + b)) + c$, factor out $a$ and work from the inside out.

## Quiz

Try these questions from the chapter quiz.

1. How is $y = x^2 + 5$ related to $y = x^2$?
   - A. moved up 5
   - B. moved down 5
   - C. moved right 5
   - D. moved left 5
   - Answer: A. moved up 5. Adding outside the function moves it up.

2. Where is the vertex of $y = (x - 3)^2$?
   - A. $(-3, 0)$
   - B. $(3, 0)$
   - C. $(0, 3)$
   - D. $(0, -3)$
   - Answer: B. $(3, 0)$. The bracket is zero at $x = 3$, so it moves right 3.

3. Where is the vertex of $y = (x + 2)^2 - 7$?
   - A. $(2, -7)$
   - B. $(2, 7)$
   - C. $(-2, -7)$
   - D. $(-2, 7)$
   - Answer: C. $(-2, -7)$. Left 2 and down 7.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/graph-transformations/#quiz