Transformations: Part B — Modulus transformations, point mapping and rotations

This is the second of two parts. Part A covers the standard vertical and horizontal transformations, the three principles that govern them, and the first-principles method for transforming the equation of a curve. Part B assumes those principles and applies them to the two modulus transformations, then develops a direct method for mapping a point to its image, and closes with a bonus section on rotating a curve through $90^\circ$.

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Modulus transformations

The three high-level principles continue to apply exactly as written when the modulus transformations $f(x)\longmapsto|f(x)|$ and $f(x)\longmapsto f(|x|)$ are introduced. They tell us that vertical and horizontal transformations are independent, that a vertical transformation must act on the whole current function, and that a horizontal transformation must replace every occurrence of $x$, and nothing else. Therefore, any order is valid provided that every transformation is applied according to these principles and the required functional form is produced.

The earlier observation that transformations in the same direction may be performed in different orders by adjusting their parameters is not one of the three principles. It is an additional feature of the eight transformations covered in Part A, and does not generally extend to modulus transformations.

Vertical modulus transformations

The transformation from $y=f(x)$ to $y=|f(x)|$ reflects any $y$-negative part of the graph in the $x$-axis:

• Any part of the graph on or above the $x$-axis remains unchanged.

• Any part below the $x$-axis is reflected into the region above it.

Equivalently, $|f(x)|=f(x)$ when $f(x)\geq0$, while $|f(x)|=-f(x)$ when $f(x)<0$.

In terms of an individual point,

$$(p,q)\longmapsto(p,|q|).$$

In terms of the whole graph,

$y=|f(x)|$ is obtained by reflecting any $y$-negative part of the graph in the $x$-axis.

Applying vertical modulus correctly

Vertical modulus is a vertical transformation, so the second principle applies without any modification:

Apply the modulus to the whole current function.

Example 1

The graph of $y=10-|5+2f(x)|$ is obtained by transforming the graph of $y=f(x)$. Give a possible sequence of transformations that produces the required graph.

There are several valid answers. We will consider one possible sequence:

The transformations are:

• a vertical stretch by scale factor $2$;
• a vertical translation by $+5$ units;
• vertical modulus;
• a reflection in the $x$-axis;
• a vertical translation by $+10$ units.

Horizontal modulus transformations

The transformation from $y=f(x)$ to $y=f(|x|)$ is less commonly examined and requires a little more care. Its effect becomes clear once we focus on the input being passed to $f$.

For $x\geq0$, we have $|x|=x$, so $f(|x|)=f(x)$. The part of the original graph for which $x\geq0$ therefore remains unchanged.

For $x<0$, we have $|x|=-x>0$, so the function is still being evaluated at a positive input. For example,

$f(|-1|)=f(1)$ and $f(|-2|)=f(2)$.

$f(|x|)$ does not produce $f(-1)$ or $f(-2)$ at all! In fact, $|x|$ can never equal $-1$, $-2$ or any other negative number, so the part of the original graph arising from negative inputs is never used. Instead, the part arising from positive inputs is used again to construct the new left-hand part of the graph.

The resulting function is even, so its graph is symmetric about the $y$-axis. It is important to understand that the original part for which $x<0$ is not itself reflected: it is completely lost, since a negative input can never occur inside $f(|x|)$. For a particularly stark example, take $f(x)=x^3-x$. Comparing $y=f(x)$ with $y=f(|x|)$, we see that the entire part of the original graph to the left of the $y$-axis disappears. The new left-hand part is instead obtained by reflecting the original right-hand part in the $y$-axis.

A further consequence is that the point correspondence is no longer one-to-one: a single original point may produce two images, one, or none at all. We can summarise as follows.

In terms of an individual point,

If $(p,q)$ lies on $y=f(x)$, then on $y=f(|x|)$:

• if $p>0$, it produces the two points $(p,q)$ and $(-p,q)$;
• if $p=0$, it produces the single point $(0,q)$;
• if $p<0$, it produces no corresponding point at all.

In terms of the whole graph,

$y=f(|x|)$ is obtained by keeping the part of $y=f(x)$ for which $x\geq0$, reflecting it in the $y$-axis, and discarding the original part for which $x<0$.

Applying horizontal modulus correctly

Horizontal modulus is a horizontal transformation, so the third principle applies without any modification:

Replace every occurrence of $x$ in the whole current function with $|x|$.

Example 2

The graph of $y=f(6-|2x+1|)$ is obtained by transforming the graph of $y=f(x)$. Give a possible sequence of transformations that produces the required graph.

There are several valid answers. We will consider one possible sequence:

The transformations are:

• a horizontal translation by $-6$ units;
• a reflection in the $y$-axis;
• horizontal modulus;
• a horizontal translation by $-1$ unit;
• a horizontal stretch by scale factor $1/2$.

A shortcut for finding corresponding points

Drawing every intermediate graph is unnecessary if all we want to know is where a particular point ends up. Suppose $(p,q)$ lies on $y=f(x)$, so that $f(p)=q$, and write the transformed function as

where $V$ describes the vertical part of the transformation and $H(x)$ is the new input being passed to $f$. For example, $y=10-2f(3x-1)$ has $V(t)=10-2t$ and $H(x)=3x-1$.

The image of $(p,q)$ can then be written down directly, using two rules:

New $y$-coordinate: apply the vertical transformation to the old $y$-coordinate.

New $x$-coordinate: set the new input to $f$ equal to the old $x$-coordinate and solve.

In other words, $y_{\text{new}}=V(q)$, while $x_{\text{new}}$ satisfies $H(x_{\text{new}})=p$.

The second rule give us an equation to be solved, and when $H$ contains a modulus that equation may have two solutions, one, or none at all.

A vertical example

Suppose $(2,-3)$ lies on $y=f(x)$. Under $y=4f(x)+5$, the $x$-coordinate is unchanged and

The corresponding point is therefore $(2,-7)$.

A horizontal example

Suppose $(7,2)$ lies on $y=f(x)$. Under $y=f(3x-2)$, the $y$-coordinate remains $2$, while the new $x$-coordinate satisfies $3x_{\text{new}}-2=7$. Hence $x_{\text{new}}=3$, and the corresponding point is $(3,2)$.

A combined example

Suppose $(8,1)$ lies on $y=f(x)$. Find its image under

Here $V(t)=10-2t$ and $H(x)=3x-1$. For the new $y$-coordinate,

For the new $x$-coordinate,

so $x_{\text{new}}=3$. The corresponding point is therefore $(3,8)$.

A vertical modulus example

Suppose $(5,-3)$ lies on $y=f(x)$. Find its image under $y=7-|f(2x+1)|$.

The vertical part is $V(t)=7-|t|$, so $y_{\text{new}}=7-|-3|=4$. The new $x$-coordinate satisfies $2x_{\text{new}}+1=5$, giving $x_{\text{new}}=2$. The image is therefore $(2,4)$.

A vertical modulus is applied to a single number, $q$, so it can only ever produce one image.

Horizontal modulus example with two image points

A horizontal modulus behaves quite differently, because it sits inside $H$ and therefore appears inside an equation that has to be solved. For $y=f(|x|)$, the new $x$-coordinate satisfies $|x_{\text{new}}|=p$, which has two solutions when $p>0$, one when $p=0$, and none when $p<0$.

Take the same $f(x)=x^3-x$ as before. The point $(1,0)$ lies on $y=f(x)$, since $f(1)=1-1=0$. Find its image under

There is no vertical transformation, so the new $y$-coordinate remains $0$. The new $x$-coordinates satisfy

Hence $|3-2x_{\text{new}}|=9$, so $3-2x_{\text{new}}=9$ or $3-2x_{\text{new}}=-9$. This gives $x_{\text{new}}=-3$ or $x_{\text{new}}=6$, and the single original point therefore has two images: $(-3,0)$ and $(6,0)$.

Bonus: Rotating a curve through $90^\circ$ about $(a,b)$

A rotation through $90^\circ$ is not normally presented as a function transformation because the rotated curve may no longer be the graph of a function of $x$. Nevertheless, both the image of a point and the equation of the rotated curve can be found directly from basic coordinate geometry.

Take a point $(x,y)$ and consider its position relative to the centre of rotation $(a,b)$. It lies $x-a$ units horizontally and $y-b$ units vertically from the centre.

Anticlockwise rotation

Under a $90^\circ$ anticlockwise rotation:

• A movement to the right becomes the same movement upwards.
• A movement upwards becomes the same movement to the left.

Thus the relative displacement $(x-a,y-b)$ becomes $(-(y-b),x-a)$. If the image point is $(X,Y)$, then

$X=a-(y-b)=a+b-y$,

$Y=b+(x-a)=b+x-a$.

To find the equation of the rotated curve, work backwards from a new point $(X,Y)$. The original coordinates were

$x=a+Y-b,\qquad y=a+b-X$.

Since the original curve satisfies $y=f(x)$, the rotated curve satisfies

$a+b-X=f(a+Y-b)$.

Relabelling the new coordinates as $x$ and $y$, its equation is

$a+b-x=f(a+y-b)$.

It gives a perfectly acceptable $x$ as a function of $y$!

Clockwise rotation

Under a $90^\circ$ clockwise rotation, a movement to the right becomes a movement downwards, while a movement upwards becomes a movement to the right. Thus $(x-a,y-b)$ becomes $(y-b,-(x-a))$, giving

$X=a+y-b$,

$Y=a+b-x$.

Working backwards gives the equation of the rotated curve as

$x-a+b=f(a+b-y)$.

Example 3

Find the equation of the curve obtained by rotating $y=x^3-x$ through $90^\circ$ anticlockwise about $(1,2)$.

Here $a=1$ and $b=2$, so the anticlockwise point map becomes

Substituting these, together with $f(t)=t^3-t$, into the anticlockwise equation $a+b-x=f(a+y-b)$ gives

Therefore, the rotated curve has equation $x=3+(y-1)-(y-1)^3$.

Alternative method: rotate a defining point

Some curves are pinned down completely by a single distinguished point together with a shape. A parabola is determined by its vertex and the coefficient of its squared term, and a circle by its centre and radius. For such a curve there is no need to substitute into the general equation at all: rotate the defining point, work out which way the shape now faces, and write the new equation down.

What makes this work is that a rotation preserves lengths and angles. The curve is not stretched or reshaped, only carried to a new position and turned, so its shape parameter is unchanged and only the direction of its axis has to be reconsidered.

Suppose we rotate $y=(x-3)^2+7$ through $90^\circ$ anticlockwise about $(1,2)$.

The vertex is $(3,7)$. With $a=1$ and $b=2$, the anticlockwise point map $X=a+b-y$, $Y=b+x-a$ sends it to

so the new vertex is $(-4,4)$.

The original parabola opens upwards, and a $90^\circ$ anticlockwise rotation carries the upward direction to the left, so the rotated parabola opens to the left. Its coefficient still has magnitude $1$, so the rotated curve has equation $x=-4-(y-4)^2$.

This shortcut is usually the faster of the two, and it works whenever the rotated curve can be found from its defining point. That is a wider class of curves: it includes the cubic of Example 3, whose point of inflection $(0,0)$ can serve as the defining point. The anticlockwise point map sends $(0,0)$ to $(3,1)$, which is indeed the point of inflection of the rotated curve $x=3+(y-1)-(y-1)^3$.

Summary

If you take away one thing from these two articles, let it be the three high-level principles:

1. Vertical and horizontal transformations are independent of each other.

2. A vertical transformation must act on the whole current function.

3. A horizontal transformation must replace every occurrence of $x$, and nothing else.

Almost everything in both parts follows from them. The six basic transformations are these principles applied one at a time, the reflections in $y=b$ and $x=a$ are each built from two of the six, and the two modulus transformations also obey these principles. You should also remember to use the first principle point-tracking method for implicitly defined functions.