The Shape of Ordered Pairs: Last Step
Harold Mick and Ben Bazak
This article is a continuation of our previous article The Shape of Ordered Pairs (2017). At the end of that article, we invited readers to take the "last step." The context was the problem: Consider equation Y = 4(X - 3)2 - 5. Sketch its graph. We had constructed a sequence of motions (transformations) that moved the given equation's parabola to the parent parabola before we stopped developing our approach at the end of the article. As a last step, we invited the readers to use our constructed sequence of motions to construct an inverse sequence of motions that would move the parent parabola to the given equation's parabola. In this article, we develop the "last step" from our perspective, expand on the associated mathematics, and take a test consisting of three questions.
Replacement (substitution) rules, formulas, graphing calculators, and graphing software are all practical ways to deal with equations and graphs, and they dominate the transformation approach to equations and graphs found in present secondary mathematics curricula. Supporting these practical ways of dealing with equations and graphs is powerful and beautiful mathematics that is not usually dealt with in the present secondary mathematics curricula. Much of this supporting mathematics can be found in transformation geometry (1982), which brings together geometry and algebra. Our far-reaching goal is to use the power and beauty of this mathematics and adapt it to the level of high school mathematics. We tap into this power and beauty by using motions to move points in the coordinate plane. We label and describe these motions, we write ordered lists of labels for sequences of motions and use other notation to make our approach accessible to high school students. More specifically, this powerful and beautiful mathematics serves to unify the secondary mathematics curriculum by emphasizing the important "undoing" concept that leads to inverse functions, inverse motions/transformations, and symmetry.
We (the authors) are writing to secondary mathematics teachers, mathematics educators, mathematics supervisors, secondary mathematics curriculum specialists, and curious readers, but as with our previous article, we direct our attention primarily to high school mathematics teachers. We wish to provide teachers with a deeper level of understanding of motions and their application to equations and graphs in the coordinate plane. We are looking toward the future rather than suggesting template lesson plans.
Problem.
We begin developing the "last step" within the context of Problem 2 that we present at the end of our previous article The Shape of Ordered Pairs (2017). Consequently, readers will not need to refer to our previous article for the problem. However, we do use strategies and vocabulary that we developed in that article (2017).
Problem 2: Consider equation Y = 4(X - 3)2 - 5. Sketch its graph.
We observe the "square" structure of the equation's syntax and realize its graph is a parabola. We call the equation's parabola the mystery parabola since it is unknown to us and is a mystery to solve. Our goal is to sketch this mystery parabola using our theory of motions. First, we sweep all the Ystuff to the left of the equation and get (1/4)(Y + 5) = (X - 3)2. This puts the original equation in parent form Ystuff = (Xstuff)2. Take points with ordered pair (u, v) on (1/4)(Y + 5) = (X - 3)2 so that (1/4)(v + 5) = (u - 3)2. This statement shows how first coordinate, u, and second coordinate, v, are related as points with ordered pairs (u, v) move along the curvature of the mystery parabola. We call this statement the shape of ordered pair (u, v) on the mystery parabola and emphasize this by highlighting the first coordinate, u, and second coordinate, v, in red:
(1/4)(v + 5) = (u - 3)2.
Next, we change perspectives from the mystery parabola to the parent parabola and re-interpret this same statement, but from the perspective of the parent parabola Y = X2. We look at statement (1/4)(v + 5) = (u - 3)2 differently -- eyes concentrating on Y = X2 -- and see the shape of ordered pair (u - 3, 1/4(v + 5)) on the parent parabola (coordinates highlighted in red),
1/4(v + 5) = (u - 3)2,
where the second coordinate 1/4(v + 5) is the square of the first coordinate u - 3. (Again, we highlight individual coordinates in red to point out the change of perspectives. It may surprise readers that algebraic expressions such as u - 3 and 1/4(v + 5) represent individual coordinates.)
We pause for a moment and review what we have done. We took points with ordered pair (u, v) on the mystery parabola, wrote a statement, then re-interpreted the statement from the perspective of the parent parabola; this gave us points with ordered pair (u - 3, 1/4(v + 5)) on the parent parabola. In short, we have identified points (u, v) on the mystery parabola and points (u - 3, 1/4(v + 5)) on the parent parabola.
Now we can construct a sequence of motions that moves the mystery parabola to the parent parabola by moving points (u, v) on the mystery parabola to points (u - 3, 1/4(v + 5)) on the parent parabola. We move u 
u - 3 by adding -3 which corresponds to Xshift -3 (a horizontal shift of -3 units), and we move v 1/4(v + 5) by adding 5, then multiplying by 1/4, which corresponds to Xshift 5|Yscale 1/4 (a vertical shift of 5 units followed by a vertical scale by factor 1/4). We construct a sequence of motions made up of these three individual motions in their order of application that moves the mystery parabola to the parent parabola under
Xshift -3|Yshift 5|Yscale 1/4.
We have successfully constructed a sequence of motions that moves the mystery parabola to the parent parabola. This is the stage of our approach where we stopped development in our article The Shape of Ordered Pairs (2017) and posed the question, "What is the last step?" By the last step, we mean sketch the graph of the mystery parabola Y = 4(X - 3)2 - 5. So, we gave readers a clue -- we invited the reader to construct an inverse sequence of motions that could be applied to the parent parabola to sketch the mystery parabola as the image. Now, we accept our own challenge: Our task is to construct such an inverse sequence of motions that moves the parent parabola to the mystery parabola.
Last Step.
We begin by studying the sequence of motions Xshift -3|Yshift 5|Yscale 1/4 that moves the mystery parabola to the parent parabola. This sequence of three individual motions connects the two parabolas by forming a "bridge" -- a one-way bridge -- from the mystery parabola to the parent parabola. The cargo is points, but there is no cargo to move in the sense that the mystery parabola is unknown to us. How can we visually move points we don't see? On the other hand, we know the parent parabola very well; it is visually clear to us. Could we somehow "reverse" our constructed sequence of motions and use this reversed version to move the parent parabola to the mystery parabola? The image of this reversed sequence of motions would be the mystery parabola and as an image of a sequence of motions, we can sketch it. So, the question before us is how do we "reverse" the sequence of motions Xshift -3|Yshift 5|Yscale 1/4 to construct a new sequence that moves the parent parabola to the mystery parabola? Using our bridge metaphor, we want to construct a new lane but going one-way in the opposite direction. This will require switching the roles of preimage and image along with making other changes to our existing sequence of motions.
Rather than extracting the details now, let's fast forward and look at how Problem 2 ends. A solution to the "last step" lies with the concept "undoing." For instance, we undo Xshift -3 by applying the motion Xshift 3 and we undo Yshift 5|Yscale 1/4 by applying the sequence of motions Yscale 4|Yshift -5. By constructing a sequence of motions with these three individual motions, we undo
Xshift -3|Yshift 5|Yscale 1/4
by applying the sequence of motions
Yscale 4|Yshift -5|Xshift 3
to the parent parabola.
Now we can finish Problem 2. We sketch the graph of 1/4(Y + 5) = (X - 3)2 by applying the undoing sequence of motions Yscale 4|Yshift -5|Xshift 3 to the parent parabola (see Figure 1).
Figure 1.
We invite you (the readers) to follow the images in Figure 1, beginning with the red parent parabola: Yscale 4 moves the red parent parabola to the blue parabola; Yshift -5 moves the blue parabola to the green parabola; and finally, Xshift 3 moves the green parabola to the purple parabola, which is the graph of the equation 1/4(Y + 5) = (X - 3)2 or as originally stated, Y = 4(X - 3)2 - 5.
Now check the newly constructed undoing sequence of motions by applying the original sequence of motions Xshift -3|Yshift 5|Yscale 1/4 to the mystery parabola, which you can now visualize (see purple parabola in Figure 1): Xshift -3 moves the purple parabola to the green parabola; Yshift 5 moves the green parabola to the blue parabola; Yscale 1/4 moves the blue parabola to the red parent parabola. We think you will agree the doing/undoing movement of points works perfectly.
In our article, The Shape of Ordered Pairs (2017), we showed how to uncover moving directions in a given equation that move its mystery parabola to the parent. In this article, we have added undoing motions to our tool kit which permit us to sketch the mystery parabola and complete our approach for problems of the type given an equation, find its graph. In the next two sections, we explore undoing motions in more depth in recognition of their importance, not only in carrying out our approach, but in mathematics more generally, and we give details on constructing an undoing sequence of motions.
Inverse Motions.
To undo a motion means to undo a motion's movement of points by following it with another motion so that points return to their original location. In our previous article (2017), we showed the close association between a motion's movement of points and the change of expressions representing the coordinates. For instance, Xshift 2 moves points (u, v) (u + 2, v). To undo this change of expression, u u + 2 means to change u + 2 u. But changing expression u + 2 to expression u is easy; we add -2:
(u + 2) - 2 = u + ( 2 - 2) = u + 0 = u.
The motion that adds -2 to first coordinates is Xshift -2. So, we follow Xshift 2 with Xshift -2 and explore the sequence of motions, Xshift 2|Xshift -2. Does this sequence return points to their original locations? Let's check by following the change of ordered pairs:
(u, v) (u + 2, v) ((u + 2) - 2, v) = (u + (2 - 2), v) = (u + 0, v) = (u, v).
Indeed, the sequence works; it returns points to their original location.
We introduce a new motion we call identity, whose movement of points is defined by (u, v) (u, v). In other words, identity leaves points' locations unchanged -- not a very interesting motion, for sure -- but an important motion, nevertheless. For instance, in a communication sense, the identity motion permits us to write
Xshift 2|Xshift -2 = identity.
If we change the order of application of these two horizontal shifts, we still get the identity:
Xshift -2|Xshift 2 = identity.
We verify this by using our change of ordered pairs notation applied to (u, v):
(u, v) (u - 2, v) ((u - 2) + 2, v) = (u + (-2 + 2), v) = (u + 0, v) = (u, v).
Now we become more formal and say Xshift 2 and Xshift -2 are inverse motions since they undo each other in either order of application:
Xshift 2|Xshift -2 = identity = Xshift -2|Xshift 2.
The other basic motions behave similarly. For instance, Yscale 4 and Yscale 1/4 are inverse motions since
Yscale 1/4|Yscale 4 = identity = Yscale 4|Yscale 1/4,
and the flip programs are self-inverses in the sense that
Xflip|Xflip = identity = Yflip|Yflip.
(We note that the slide, flip, and scale motions are one-to-one and onto functions defined to the coordinate plane. Consequently, motions have inverse motions, and sequences of motions have inverse sequences of motions.)
Inverse Sequences of Motions.
Undoing a sequence of two or more motions adds complexity. For instance, what is the inverse of Xshift 2|Xscale 3? To investigate, we look at "unraveling" the expression that results from applying the sequence of motions Xshift 2|Xscale 3 to point (u, v):
(u, v) (u + 2, v) (3(u + 2), v).
We have arrived at points (3(u + 2, v). The challenge before us is to construct a sequence of motions that moves points (3(u + 2), v) (u, v). The expression 3(u + 2) shows the sequence of arithmetic operations, first add 2 then multiply by 3. How do we undo this expression in the sense of changing 3(u + 2) u? To undo expression 3(u + 2), we must first undo "multiplying by 3," right? But this is easy. We multiply by the multiplicative inverse of 3, or 1/3, giving us
1/3(3(u + 2)) = (1/3
3)(u + 2) = 1(u + 2) = u + 2.
Then we undo u + 2 by adding the additive inverse of 2, or -2:
(u + 2) - 2 = u + (2 - 2) = u + 0 = u.
So, it appears that the inverse sequence of motions that undoes Xshift 2|Xscale 3 is Xscale 1/3|Xshift -2. To confirm this conjecture, we apply the sequence of motions
(Xshift 2|Xscale 3)|Xscale 1/3|Xshift -2)
to points (u, v) and see if we end up with points (u, v):
(u, v) (u + 2, v)
(3(u + 2), v)
(1/3(3(u + 2)), v)
= (u + 2, v)
((u + 2) - 2, v)
= (u, v)
The sequence of motions works. Now let's see if our inverse candidate Xscale 1/3|Xshift -2 works in reverse order. We apply
(Xscale 1/3|Xshift -2)|(Xshift 2|Xscale 3)
to points (u, v):
(u, v) (1/3u, v)
(1/3u - 2, v)
((1/3u - 2) + 2, v)
= (1/3u, v)
(3(1/3u), v)
= (u, v)
This reverse order works too! We have shown that the sequence Xscale 1/3|Xshift -2 is the inverse sequence of motions for Xshift 2|Xscale 3.
Next, we present an argument using just labels to illustrate this same inverse relationship. If we consider following one motion with another motion as an operation, the commutative principle does not hold, but the associative principle does hold. Observe that we apply the associative principle a couple of times in the argument below:
(Xshift 2|Xscale 3)|(Xscale 1/3|Xshift -2) = Xshift 2|(Xscale 3|Xscale 1/3)|Xshift -2
= Xshift 2|identity|Xshift -2
= (Xshift 2|identity)|Xshift -2
= Xshift 2|Xshift -2
= identity
The reverse order, (Xscale 1/3|Xshift -2)|(Xshift 2|Xscale 3) = identity, works too! We have shown that the sequence Xscale 1/3|Xshift -2 is the inverse sequence of motions for Xshift 2|Xscale 3.
A non-mathematical example of an inverse sequence of motions is first putting on your socks and then putting on your shoes. If you want to reverse this whole process, you first remove your shoes (even though you put them on last) and then remove your socks. Communicating this real-life experience with algebraic notation, we have
(socks | shoes)-1 = shoes-1 | socks-1
where the exponent -1, as expressed in socks-1, is a symbol for the inverse motion.
This completes the development of our theory of motions approach to problems dealing with equations and graphs. It is time for a test.
Three Assessment Items.
We test the efficacy of our approach by responding to three assessment items. We refer to them as Question 1, Question 2, and Question 3.
Question 1. What sequence of motions moves the parent parabola Y = X2 to the parabola of equation Y = -2(X + 4)2 - 7?
We begin by sweeping the Ystuff to the left side of equation Y = -2(X + 4)2 - 7 to get equation -1/2(Y + 7) = (X + 4)2, which is now in parent form Ystuff = (Xstuff)2. We take points (u, v) on -1/2(Y + 7) = (X + 4)2 so that
-1/2(v + 7) = (u + 4)2.
Next, we change perspective from the mystery parabola to the parent parabola and re-interpret this same statement from the perspective of the parent parabola Y = X2,
-1/2(v + 7) = (u + 4)2,
which shows us the shape of ordered pair (u + 4, -1/2(v + 7)) on the parent parabola. To move the mystery parabola to the parent parabola, we move points (u, v) to points (u + 4, -1/2(v + 7)) by applying the sequence of motions
Xshift 4|Yshift 7|Yscale 1/2|Yflip.
To move the parent parabola to the mystery parabola, we apply the inverse sequence of motions
Yflip|Yscale 2|Yshift -7|Xshift -4.
Question 2. Find an equation for the image graph of Y = |X| under the sequence of motions Yflip|Xshift 3|Yshift 4|Xscale 2.
To find an equation for the image graph, we seek the shape of ordered pairs for points on the image graph. To that end, we take points with ordered pair (u, v) on the image graph and ask the question, "What is the shape of (u, v) on the image graph?" We know the shape of ordered pairs for points on the parent absolute value graph, so can we somehow move points (u, v) that lie on the image graph to points on the parent absolute value graph? Fortunately, the inverse sequence of motions
Xscale 1/2|Yshift -4|Xshift -3|Yflip
will do the trick:
(u, v) (1/2u - 3, -(v - 4)).
The shape of ordered pairs (1/2u -3, -(v - 4)) on the parent absolute value graph is
-(v - 4) = |1/2u - 3|.
Next, we change perspective back to the image graph and re-interpret this statement as the shape of ordered pair (u, v) on the image graph:
-(v - 4) = |1/2u - 3|.
We describe this relationship between first coordinates, u, and second coordinates, v, on the image graph with equation
-(Y - 4) = |1/2X - 3|.
Question 3. Consider the graph of Y = f(X). Apply a horizontal shift of h units and a vertical shift of k units to this "parent" graph. Find an equation of the image graph.
We take points (u, v) on the image graph. We want to find the shape of ordered pair (u, v) on this image graph. Since we know the shape of ordered pairs on the parent graph, we seek how to move points (u, v) from the image graph to the parent graph. But this is easy to do. Since Xshift h|Yshift k moves the parent graph to the image graph, the inverse sequence of motions, Yshift -k|Xshift -h, will move the image graph back to the parent graph. We do it; we apply a vertical shift of -k units followed by a horizontal shift of -h units to the image graph, which moves points (u, v) (u - h, v - k). We observe that image points (u - h, v - k) lie on the parent graph so that the shape of ordered pair (u - h, v - k) on the parent graph is v - k = f(u - h). This statement also shows the shape of (u, v) on the image graph, v - k = f(u - h). We describe this relationship between first coordinate, u, and second coordinate, v, on the image graph with equation
Y - k = f(X - h).
We have completed our responses to the three assessment questions. Did we pass the test? To help you (the readers) sort this out, Question 1 and Question 2 are assessment items and Question 3 is an extension and connection item -- all three items take from Virginia Department of Education Standards of Learning (SOL) Web post: Mathematics Enhanced Scope and Sequence, Algebra II, Transformational Graphing, Assessment Questions, Extensions and Connections (2011). The questions as listed on the SOL Web post read:
Question 1. What transformations will map y = x2 onto y = -2(x + 4)2 - 7?
Question 2. I am a function. My parent is y = |x|. My parent function is mapped onto me by a reflection over the line y = 0, then a horizontal shift 3 units to the right, a vertical shift 4 units up, and finally a horizontal stretch with a factor of 2. Who am I?
Question 3. Demonstrate the transformations from y = f(x) onto y = f(x + h) + k.
We re-phrased SOL Question 1 and SOL Question 2 to fit the language of our theory of motions. We took the liberty of interpreting the meaning of "demonstrate" in SOL Question 3 to make the statement: When you shift a graph horizontally h units and shift vertically k units, you change it equation by replacing X with X - h and by replacing Y with Y - k. In other words, we justified the shift replacement rule. The other replacement rules can be justified similarly with our theory of motions. The key is inverse motions.
Wrapping Up.
This concludes our series of two articles in which we have developed our theory of motions approach for solving the fundamental problems given a graph, find its equation and given an equation, find its graph. Our approach extends to other problems dealing with equations and graphs as our responses to the three SOL questions above have shown. We have restricted our development largely to a discussion of parabolas, but our strategies can be broadened to include other families of elementary curves that have parent equations. For instance, the family of lines provide an excellent family of equations and graphs to explore with its parent Y = X.
Our emphasis throughout this discussion has been on a conceptual development of the undoing concept that ties together different areas of mathematics throughout the school mathematics curricula. For instance, the undoing concept shows up in the elementary mathematics curriculum with subtraction undoing addition and division undoing multiplication. Later, as sets of numbers expand, subtraction becomes the additive inverse on the set of integers and division becomes the multiplicative inverse on the set of rational numbers. In high school, the undoing concept can be applied to solving equation in Algebra I. Later in the curriculum, inverse functions, inverse motions/transformations, and symmetry are studied. Motions bring together geometry and algebra, woven together into a beautiful tapestry of mathematics. We just need to open our eyes to see it.