Tag: Transformations

Let’s do a couple of examples using the knowledge from my last 3 posts and show how the order of transformations make a difference.

The image of the equation that follows will be generated by the following transformations:

a. Dilate along y-axis by factor 1/2
b. Reflect along y-axis (across x-axis)
c. Dilate along x-axis by factor 3
d. Translate along x-axis +4 units
e. Translate along y-axis -2 units

\[y=\sqrt{3x-4}\]

In the previous post, I showed that given an equation y = f(x), its image under a general transformation is given by

\[f(x)\Rightarrow af\left[\frac{1}{n}(x-h)\right]+k\]

where
a = dilation\reflection factor along y-axis (from x-axis)
n = dilation\reflection factor along x-axis (from y-axis)
h = translation along x-axis
k = translation along y-axis

This assumes that dilations/reflections are done first. So if

\[f(x)=\sqrt{3x-4}\]

then

\[f(x)\Rightarrow af\left[\frac{1}{n}(x-h)\right]+k=-\frac{1}{2}\sqrt{3\left [ \frac{1}{3}(x-4)\right]-4}-2\]

This can be simplified to

\[y=-\frac{1}{2}\sqrt{ x-8}-2\]

In this form, the pre-image (the equation that this one came from before the transformation) is lost. The question could be asked, what are the transformations required to go from

\[y=-\frac{1}{2}\sqrt{ x-8}-2\Rightarrow y=\sqrt{3x-4}\]

One would expect that if we just do the opposite of the transformations above, we would get the original equation. Let’s see. Let’s do the following:

a. Dilate along y-axis by factor 2
b. Reflect along y-axis (across x-axis)
c. Dilate along x-axis by factor 1/3
d. Translate along x-axis -4 units
e. Translate along y-axis +2 units

These undo the previous transformations. Putting these in our model

\[f(x)\Rightarrow af\left[\frac{1}{n}(x-h)\right]+k\]

We get

\[f(x)\Rightarrow -2f\left[3(x+4)\right]+2=\sqrt{3(x+4)-8}+4+2=\sqrt{3x+4}+6\]

Not exactly what we wanted. What went wrong? Well, the model we used assumes that dilations go first. If we want to undo the previous transformations, not only do we use the values we just used, but they must be applied in reverse order as well: the translations go first then the dilations/reflections. Otherwise the dilations affect the translations before they are applied.

I’ll leave this as an exercise for the reader, but the model for transforming y = f(x) assuming that translations go first is

\[y=af\left( \frac{x}{n}-h \right)+ak\]

So undoing the original transformations in reverse order:

a. Translate along y-axis +2 units
b. Translate along x-axis -4 units
c. Dilate along x-axis by factor 1/3
d. Reflect along y-axis (across x-axis)
e. Dilate along y-axis by factor 2

gives the result

\[f(x)\Rightarrow -\frac{1}{2}f\left(3x+4\right)\Rightarrow-\frac{1}{2}\sqrt{3x+4-8}-2+2\Rightarrow\sqrt{3x-4}\]

which is the original equation.

So the order of transformations steps will change the final result.

I hope this series of posts helps you better understand transformations.

In my last post, I took a general point (x,y) and through the series of transformations:

A. (x,y)
B. Dilate/reflect by factor a along y-axis
C. Dilate/reflect by factor n along x-axis
D. Translate k units along y-axis
E. Translate h units along x-axis

the point was changed to:

A(x, y) —-> B(x, ay) —-> C(nx, ay) —-> D(nx, ay+k) —->
E(nx+h, ay+k).

Now let’s restrict the original point to be one that satisfies the equation y = x². How does this equation change so that all of its points are transformed correctly?

Think of the transformation process as changing old points to new ones. So the old points, (x, y), are changed to the new points (nx+h, ay+k) = (x‘, y‘), where the apostrophe ‘ is used to distinguish the new x‘s and the new y‘s from the old ones. So under this transformation, x‘ = nx+h and y‘ = ay+k.

Now the equation y = x² is the equation that the old points satisfy. To find the equation that the new points satisfy, we need to find what the old variables are in terms of the new ones. Solving for the old x and y in the above equations, we get

\[x’=nx+ h\Longrightarrow x = \frac{1}{n}\left(x’-h\right)\]
\[y’=ay+ k\Longrightarrow y = \frac{1}{a}\left(y’-k\right)\]

If we replace the old variables in the equation with the right side of the above equations, then we will get an equation with the new variables, which can be cleaned up with a little algebra:

\[\frac{1}{a}(y’-k)=\left[\frac{1}{n}(x’-h)\right]^2\Longrightarrow y’=a\left[\frac{1}{n}(x’-h)\right]^2+k\]

Now that we have the new equation, we don’t need the ‘ anymore. So the new equation is:

\[y=a\left[\frac{1}{n}(x-h)\right]^2+k\]

Notice that if we were given this new equation and were asked, “what are the transformations that generated this equation from y = x² ?”, then we can pick off the dilations and the translations. Be aware though, that we must assume that the dilations occurred first. Textbooks will tell you this to keep answers consistent. With this caution, if a = 2, n = 3, h =-5, and k =4, then the new equation would be

\[y=2\left[\frac{1}{3}(x+5)\right]^2+4\]

The graph below shows the effect of these transformations on the original curve:

So given a set of transformations, the steps that I did above for the particular equation y = x², can be done for any equation.

What if the problem is reversed: given the end result, what is the set of transformations that created the new equation from and original one? Using functional notation, notice that:

\[f(x)={x}^{2}\Longrightarrow af\left[\frac{1}{n}(x-h)\right]+k=2{\left[\frac{1}{3}(x+5)\right]}^{2}+4\]

If you were given this final result, you can pick off the dilations along both axes (These are assumed to be first) and the translations along both axes. If either the a and/or the n were negative, then there would be reflections as well.

In my last post, I took specific points and transformed them using dilations/reflections and translations. The goal is to transform an equation into a new equation. But before we get there, let’s again look at a single point. But this time, this will be a general point (x, y).

So starting with general point A(x, y), let’s do the same transformation as I did in my last post:

A. (x,y)
B. Dilate by factor 2 along y-axis
C. Dilate by factor 1/3 along x-axis
D. Reflect across x-axis
E. Reflect across y-axis
F. Translate 3 units up
G. Translate 3 units to the left

In this order, the new transformed points become:

A(x, y) —-> B(x, 2y) —-> C(x/3, 2y) —-> D(x/3, -2y) —->
E(-x/3, -2y) —-> F(-x/3, -2y+3) —-> G(-x/3-3, -2y+3)

From my last post, if you use the particular point A(3,1), and use these values in the result above, you get the same point G(-4,1).

This can be generalised more by using letters to represent the dilation factors and the translation amounts:

a = dilation/reflection factor along y axis
n = dilation/reflection factor along x axis
h = translation along x axis
k = translation along y axis

Note that I have combined dilations and reflections. This is because a reflection can be viewed as a negative dilation. If a is negative, then this is a reflection along the y axis (across the x axis) as well as a dilation. If n is negative, then this is a reflection along the x axis (across the y axis) as well as a dilation. This does slightly restrict the flexibility of transformations as we cannot separate the order of a dilation and a reflection, but for most problems, this is not an issue. Also, if h is negative, that is a translation to the left. If k is negative, that is a translation down.

Now I’ll apply this to the same set of transformations as above except that I will combine the dilation and reflection steps:

A. (x,y)
B. Dilate/reflect by factor a along y-axis
C. Dilate/reflect by factor n along x-axis
D. Translate k units along y-axis
E. Translate h units along x-axis

This leads to the new point:

A(x, y) —-> B(x, ay) —-> C(nx, ay) —-> D(nx, ay+k) —->
E(nx+h, ay+k)

If these transformations are reversed:

A. (x,y)
B. Translate h units along x-axis
C. Translate k units along y-axis
D. Dilate/reflect by factor n along x-axis
E. Dilate/reflect by factor a along y-axis

you get the new point:

A(x, y) —-> B(x+h, y) —-> C(x+h, y+k) —-> D[n(x+h), y+k] —->
E[n(x+h), a(y+k)]

This is different and you will generally get a different ending point for a specific point and transformation values.

Some terminology is needed here. We have an original point (x,y) and then under a set of transformations, a new point is created. In our first set of transformations, this new point was (nx+h, ay+k). Point (nx+h, ay+k) is the image of (x,y) under this transformation. Or (x,y) is the pre-image of (nx+h, ay+k).

It may help to think of this as a black box where (x,y) enters and is magically transformed to (nx+h, ay+k). Though we know it is not magic at all but just maths:

As mentioned, the goal here is to transform a set of points, usually defined by an equation, to a new set (new equation). The above mental image (pun intended) will help keep in focus what is being done.

In my next post, I will transform equations and show how that is done.

From my experience teaching mathematics, I think that the topic most students find the most difficult is circular functions (trigonometry). The next most difficult topic is transformations: transforming a function to another via dilations, reflections, and translations. This is the first of several posts to address this topic.

First, by way of looking at individual points in a Cartesian coordinate system, let’s do some transformations and define some terms.

Consider the following points A, B, C, and D:

The first transformation will be to dilate these points. This means that we will change their distance from the x or y axes by multiplying the appropriate coordinate by a number. This number is called the dilation factor. If we dilate along the x-axis, we will change the x coordinate of each point by multiplying it by the dilation factor. Be aware that this is also called “dilating from the y-axis” because this dilation changes the point’s distance from the y-axis

So let’s dilate these points along the x-axis by a factor of 2. This makes each point twice as far from the y-axis. So point A becomes A1(6,1), B becomes B1(-4,3), C becomes C1(-8,-1), and D becomes D1(2,-2):

If we had used a factor of 1/2, then the new points would be half the distance from the y-axis than the original points.

If the factor we use is negative, not only is the dilation happening, but the points are also reflected across the y-axis. So if the dilation factor is -2 along the x-axis, the new points are A1(-6,1), B1(4,3), C1(8,-1), and D1(-2,-2):

Another transformation is translation: moving a point left or right or up or down. So if you translate a point 3 units to the right (along the x-axis), then the new points are A1(6,1), B1(1,3), C1(-1,-1), and D1(4,-2):

Now all of these transformations can be done along the y-axis (from the x-axis) as well. So doing all transformations along the y-axis, I will dilate point A by a factor of 2, reflect point B with no dilation, translate point C up 3 units, and dilate point D by a factor of 1/2. The new points will then be A1(3,2), B1(-2,-3), C1(-4,2), and D1(1,-1):

The previous example just changed the x or the y coordinate of each point. Let’s look at point A and mix up these transformations. This will also show that the order of the transformations can affect the final result. Consider the following sequence of transformations of point A:

A. (3,1)
B. Dilate by factor 2 along y-axis
C. Dilate by factor 1/3 along x-axis
D. Reflect across x-axis
E. Reflect across y-axis
F. Translate 3 units up
G. Translate 3 units to the left

The following graph shows the sequence of these transformations .The new point’s letter refer to the result of the transformation’s letter above:

Now let’s do the previous steps in reverse:

A. (3,1)
B. Translate 3 units to the left
C. Translate 3 units up
D. Reflect across y-axis
E. Reflect across x-axis
F. Dilate by factor 1/3 along x-axis
G. Dilate by factor 2 along y-axis

We end up at a very different place. Also notice that if a point is on the y-axis, reflections across the y-axis and dilations along the x-axis, have no effect.

I will generalise this for a generic point (x,y) in my next post.