Transformations

There are three types of transformations which we will be dealing with: translations, dilations and reflections.

Translations are when we shift the entire graph left, right, up or down.

Dilations stretch or compress the graph. All dilation factors are positive.

A dilation from the y-axis by a factor larger than one ( > 1) will result in the graph being stretched away from the y-axis.

A dilation from the y-axis by a factor less than one ( < 1) will result in the graph being compressed towards the y-axis.

A dilation from the x-axis by a factor larger than one ( > 1) will result in the graph being stretched away from the x-axis.

A dilation from the x-axis by a factor less than one ( < 1) will result in the graph being compressed towards the x-axis.

Reflections simply flip the graph relative to a certain line. In methods we are generally concerned with - reflections in the x-axis, y-axis and line y=x

**Note: there are no method marks awarded for any transformation questions no matter how many marks are being awarded, they are all answer marks.**

1 | Begin with (x,y) and adjust this coordinate according to the transformations provided. Do this in order of the transformations provided. View the table below this one on how to adjust the coordinates. \[ (x,y) \rightarrow (x_1,y_1) \rightarrow (x_2,y_2) \rightarrow … \rightarrow (x_n,y_n)\\ \] |

2 | Let the final coordinate obtained equal to (x’,y’). \[(x_n, y_n) = (x’, y’)\] \[\rightarrow x_n = x’ , y_n = y’\] |

3 | Make x and y the subjects of their respective equations. |

4 | Substitute the expressions for x and y obtained in step 3 into the equation provided in the question |

5 | Rearrange the equation created in step 4 to make y the subject |

6 | Remove the dashes (‘) from x and y in your equation |

\[(x,y) \rightarrow (x+a,y)\] | Translation a units right |

\[(x,y) \rightarrow (x-a,y)\] | Translation a units left |

\[(x,y) \rightarrow (x,y+a)\] | Translation a units up |

\[(x,y) \rightarrow (x,y-a)\] | Translation a units down |

\[(x,y) \rightarrow (x,ay)\] | Dilation from the x-axis by a factor of a |

\[(x,y) \rightarrow (ax,y)\] | Dilation from the y-axis by a factor of a |

\[(x,y) \rightarrow (x,-y)\] | Reflection in the x-axis |

\[(x,y) \rightarrow (-x,y)\] | Reflection in the y-axis |

\[(x,y) \rightarrow (y,x)\] | Reflection in the line y = x |

Example 3.1: Find the rule of the image of f(x) under the following sequence of transformations:

A dilation from the x-axis by a factor of 3

A reflection in the y-axis

A translation of 1 unit in the negative direction of the x-axis

A dilation from the y-axis by a factor of 5

\[f(x)=\sqrt{x}\]
\[
\text{ }\\
(x,y) \rightarrow (x,3y) \rightarrow (-x,3y) \rightarrow (-x-1, 3y) \rightarrow (5(-x-1), 3y) \rightarrow (x’,y’) \\
\begin{aligned}
(5(-x-1), 3y) &= (x’,y’) \\
(-5x-5, 3y) &= (x’,y’) \\
-5x-5 &= x’ \\
x &= \frac{x’+5}{-5} … \boxed{1}\\
3y &= y’ \\
y &= \frac{y’}{3} … \boxed{2}\\
\text{ }\\
\text{Substitute } \boxed{1} \text{ and } &\boxed{2} \text{ into } y = \sqrt(x) \\
\text{ }\\
\frac{y’}{3} &= \sqrt{\frac{x’+5}{-5}} \\
y’ &= 3 \sqrt{\frac{-x-5}{5}} \\
y &= 3 \sqrt{\frac{-x-5}{5}} \\
f(x) &= 3\sqrt{\frac{-x-5}{5}}
\end{aligned}
\]

1 | Add dashes (‘) to x and y in the image (new transformed equation). |

2 | Transpose both equations (generally only the image needs to be adjusted) so that they are in the same ‘form’ as the original function. View the table below for more information. [This is hardest step, you will get better at recognising what to do with more practice.] |

3 | Equate the left hand side of both equations and the right hand side of both equations. |

4 | Transpose and manipulate the equations obtained in step 3 to make x’ and y’ the subject. |

5 | Find a way to get x to x’ (found in step 4) with addition/ subtraction/ multiplication/ division. Do the same thing for y. |

6 | State the sequence of transformations (with aid of the table in ‘Determining the rule). To avoid issues with the order of transformations, do all transformations which affect the x-coordinate first and then those that affect the y-coordinate. |

What we want one side of the equation to look like. The other side of the equation (with y is irrelevant)

\[(ax+b)^n\] | \[\frac{1}{(ax+b)^n}\] | \[\sqrt[c]{ax+b}\] |

\[e^{ax+b}\] | \[log_{c}{(ax+b)}\] | \[\sin(ax+b)\] |

\[\cos(ax+b)\] | \[\tan(ax+b)\] |

In general, we want one side of the equation to look like: \[f(ax+b)\]

Note: whilst ‘determine the rule’ questions may say translate in the ‘negative direction of the x-axis’ (or similar expressions) there is nothing wrong with simply saying translate left (right, up or down). For dilations and reflections stick with what we have provided in the table in the ‘determine the rule’ section.

\[
\text{Example 3.2: List a sequence of transformations that maps the graph of }y=\frac{1}{3-x}\\ \text{ to the image with equation }y=\frac{4}{x}+1\\
\text{ }\\
\text{Only the second equation (that of the image) needs to be re-rearranged.}\\
\begin{aligned}
y &= \frac{1}{3-x} ... \boxed{1} \\
y' &= \frac{4}{x'} + 1 \\
y' - 1 &= \frac{4}{x'}\\
\frac{y'-1}{4} &= \frac{1}{x'} ... \boxed{2} \\
\text{ } \\
\boxed{1} &= \boxed{2} \\
\text{ } \\
y &= \frac{y' - 1}{4} \\
y'-1 &= 4y\\
y' &= 4y + 1\\
\text{ } \\
\frac{1}{3-x} &= \frac{1}{x'} \\
3-x &= x' \\
\end{aligned}\\
\text{ } \\
x \rightarrow -x \rightarrow -x + 3\\
y \rightarrow 4y \rightarrow 4y + 1\\
\text{ }\\
\text{1. Reflection in the y-axis}\\
\text{2. Translation 3 units right}\\
\text{3. Dilation from the x-axis by a factor of 4}\\
\text{4. Translation 1 unit up}
\]

\[
\text{Example 3.3: Find the sequence of transformations that map } y = \frac{4}{x+1}+3 \\ \text{ to the image with equation } y = \frac{3}{x-4}+2\\
\text{ }\\
\text{Both equations need to be rearranged} \\
\begin{aligned}
y &= \frac{4}{x+1}+3 \\
\frac{y - 3}{4} &= \frac{1}{x+1} ... \boxed{1} \\
y' &= \frac{3}{x'-4} + 2 \\
\frac{y'-2}{3} &= \frac{1}{x'-4} ... \boxed{2} \\
\text{ }\\
\boxed{1} &= \boxed{2} \\
\text{ }\\
\frac{y-3}{4} &= \frac{y'-2}{3} \\
\frac{3(y-3)}{4} &= y'-2 \\
\frac{3y-9}{4} + 2 &= y'\\
\frac{3y -1}{4} &= y'\\
\frac{3y}{4} - \frac{1}{4} &= y' \\
\text{ } \\
\frac{1}{x+1} &= \frac{1}{x'-4} \\
x + 1 &= x' - 4 \\
x + 5 &= x' \\
\end{aligned}\\
\text{ }\\
y \rightarrow \frac{3y}{4} \rightarrow \frac{3y}{4} - \frac{1}{4} \\
x \rightarrow x + 5\\
\text{ }\\
\text {1. Translate 5 units right}\\
\text{2. Dilate from the x-axis by a factor of }\frac{3}{4}\\
\text{3. Translate }\frac{1}{4}\text{ of a unit down}
\]

**Only use a matrix approach if you are given the transformation matrices. ** Otherwise refer to “Determining the rule”. Transformation questions related to matrices will almost certainly be in multiple choice, if not at least in exam 2. As such, you will have a CAS to add or multiply matrices. However, we have outlined all of the matrix operations you will need to know below anyway.

Only matrices of the same size can be added together. When adding matrices simply add the numbers in the corresponding cells together like below: \[\begin{bmatrix}a\\b\end{bmatrix} + \begin{bmatrix}c\\d\end{bmatrix} = \begin{bmatrix}a+c\\b+d\end{bmatrix}\] When a matrix is multiplied by a constant, multiply all numbers within the matrix by that constant \[k \begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}ka&kb\\kc&kd\end{bmatrix}\] Matrix multiplication is quite complex. In a matrix transformation question, you will only ever get the matrix multiplication shown below. \[ \begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}ax+by\\cx+dy\end{bmatrix} \]

When presented with a matrix transformation question, multiply and add any matrices you are presented with until you are left with only a single matrix on both sides of your equation. After that, the top entry of the matrices on either side of the equals sign will be equal and the bottom entry of the matrices on either side of the equals sign will be equal as well. Having created equations for that, complete steps 2-6 from “determining the rule”. An example is given below.

\[ \text{Example 3.4: A transformation } T : \mathbb{R}^2 \rightarrow \mathbb{R}^2 \text{ is defined by } \begin{bmatrix}x’\\y’\end{bmatrix} = \begin{bmatrix}3&0\\0&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} + \begin{bmatrix}2\\3\end{bmatrix}.\\ \text{ Find the image of the graph with rule } y = \sqrt{x} \\ \text{ }\\ \begin{aligned} \begin{bmatrix}x’\\y’\end{bmatrix} &= \begin{bmatrix}3x\\-y\end{bmatrix}+ \begin{bmatrix}2\\3\end{bmatrix} \\ \begin{bmatrix}x’\\y’\end{bmatrix} &= \begin{bmatrix}3x+2\\-y+3\end{bmatrix}\\ \text{ } 3x + 2 &= x’ \\ x &= \frac{x’-2}{3} … \boxed{1} \\ -y+3 &= y’ \\ -y &= y’-3 \\ y &= -y’+3 … \boxed{2} \\ \text{ }\\ \text{Substitute } \boxed{1} &\text{ and } \boxed{2} \text{ into } y = \sqrt{x} \\ \text{ }\\ -y’+3 &= \sqrt{\frac{x’-2}{3}} \\ -y’ &= \sqrt{\frac{x’-2}{3}}-3 \\ y’ &= -\sqrt{\frac{x’-2}{3}}+3 \\ y &= -\sqrt{\frac{x-2}{3}}+3\\ \end{aligned} \\ \]

\[ \text{ }\\ \text{Example 3.5: List the sequence of transformations that maps the graph of } y = 2\cos(\frac{x}{2})+\pi \\ \text{ to the image with equation } y = -\sin(\frac{x}{2}) \\ \text{ } \\ \begin{aligned} y &= 2\cos(\frac{x}{2}) + \pi \\ \frac{y - \pi}{2} &= \cos(\frac{x}{2}) ... \boxed{1} \\ \text{ } \\ y' &= -\sin(\frac{x'}{2}) \\ y' &= \cos(\frac{x'}{2} + \frac{\pi}{2}) ... \boxed{2} \\ \text{ } \\ \boxed{1} &= \boxed{2} \\ \text{ }\\ \frac{y-\pi}{2} &= y' \\ \frac{y}{2} - \frac{\pi}{2} &= y' \\ \text{} \\ \frac{x}{2} &= \frac{\pi}{2} + \frac{x'}{2} \\ \frac{x}{2} - \frac{\pi}{2} &= + \frac{x'}{2} \\ x - \pi &= x' \\ \end{aligned}\\ \text{ }\\ x \rightarrow x - \pi \\ y \rightarrow \frac{y}{2} \rightarrow \frac{y}{2} -\frac{\pi}{2} \\ \text{ }\\ \text{1. Translation } \pi \text{ units left} \\ \text{2. Dilation from the x-axis by a factor of } \frac{1}{2} \\ \text{3. Translation} \frac{\pi}{2} \text{ units down} \\ \] \[ \text{ }\\ \text{Example 3.6: List a sequence of transformations that maps the graph } y = e^x \text{ to } y = 2^{x+1} +4\\ \text{ }\\ \begin{aligned} y &= e^x ... \boxed{1} \\ y' &= 2^{x'+1}+4 \\ y' &= e^{\log_{e} {2^{x'+1}}}+4 \\ y' &= e^{(x'+1)\log_{e}{2}}+4 \\ y'-4 &= e^{x'\log_{e}{2} + \log_{e}{2}} ... \boxed{2}\\ \text{ }\\ \boxed{1} &= \boxed{2} \\ \text{ }\\ y &= y' - 4 \\ y+4 &= y' \\ e^x &= e^{x'\log_{e}{2} + \log_{e}{2}} \\ x &= x'\log_{e}{2} + \log_{e}{2} \\ x - \log_{e}{2} &= x'\log_{e}{2} \\ \frac{x - \log_{e}{2}}{\log_{e}{2}} &= x' \\ \end{aligned}\\ \text{ } \\ x \rightarrow x - \log_{e}{2} \rightarrow \frac{x - \log_{e}{2}}{\log_{e}{2}}\\ y \rightarrow y+4 \\ \text{ } \\ \text{1. Translation } \log_{e}{2} \text{ units left} \\ \text{2. Dilation from the y-axis by a factor of } \frac{1}{\log_{e}{2}} \\ \text{3. Translation } 4 \text{ units up} \\ \]