Translating a Curve: Multiple Methods

(A new question of the week)

Today we’ll look at a problem that puts a little twist on the basic idea of translating a graph. The focus is on finding alternate approaches to the problem, which is an important skill in problem solving.

The problem

The question came, as many of our most interesting questions do, from Kurisada, this one in February:

If I translate the curve y = x2 – 3x + 2, and the translation goes through (1, 1) and (2, 3), what is the equation of the translation?

How many ways can I find this?

Here is what we are talking about:

The solid curve is \(y = x^2 – 3x + 2\); A and B are the two given points; and the dotted curve is a translation of the given curve (that is, it has been slid to a new position without stretching or turning) that passes through A and B. I drew this just by looking for a pair of points on the curve with the same relationship as A and B, namely moving right 1 and up 2. The points (2, 0) and (3, 2) jumped out at me, so I shifted the graph left 1 and up 1, yielding the equation \(y = (x+1)^2 – 3(x+1) + 2 + 1 = x^2 – x + 1\). A little thought convinces me that there is only one such curve, though that would not necessarily be true for any curve and any two points. And depending on a random insight, useful as it is, doesn’t seem like a good problem-solving technique!

Kurisada, who is trying to master problem-solving, is wise to ask for multiple methods, though it is impossible to answer the question if we take it literally! Doctor Rick responded, somewhat tongue-in-cheek, to that question:

I don’t think we could count the ways. We could find a minimum number of ways to solve the problem by solving it one way (so the minimum is 1), then finding another way (so the minimum is now 2), and so on. We’d have to decide, at some point, what constitutes a “different” way.

But is that what you really want to do? Let’s start with one way. To get things started, my first thought is to write the family of functions that are translations of the given function. How many parameters will be needed to select a member of that family?

If you have your own thoughts, I’ll be happy to start by discussing whatever approach you have in mind.

So we’re just looking for as many methods as we can think of, and the way to begin that is to find one way! We’ll eventually be getting back to Doctor Rick’s suggestion.

Method 1: Solving for parameters

Kurisada, it turned out, already had two ways to do it:

I think that the number of parameters needed can be 2 or 3 because I think that y = x2 – 3x + 2 can be translated to y = x2 – 3x.

As far as I have done, I can only find 2 ways to do it.

The first one is by using the equation (y – a) = (x – b)^2 – 3(x – b) + 2.

The second one is by finding the values of y in the set of values of x:

x: -2  -1  0  . . .

y: 12  6  2  . . .

Then I find two points that have the same gradient with (1,1) and (2, 3).

(This way is similar by doing it with the graph.)

And I can’t figure any other way, and thinking if there is really any other way to do this (and I really want to know the ways).

It appears that these two methods are just ideas, not fully worked out solutions. (That is, in Polya’s terms, they are plans that have not yet been carried out.)

The first method just states what a translation of the function (up a units, right b units) looks like, without specifying how to find a and b.

The second method involves looking through a table of (x, y) pairs to find pairs with the same slope (or, really the same \(\Delta x\) and \(\Delta y\) as in the given points. This is essentially what I did above “with the graph”. In the original discussion both methods were discussed alternately; to keep things simple, I’ll untangle them and follow the discussion of the first method for now, then backtrack to the second after we finish this one.

Half an hour later he added:

Now I’m confused about my own way.

I tried my first way, and I got a = 1, b = -1.

And I got the equation y = x^2 – x + 1. (I checked the graph and this should be correct.)

This answer is what I got above, so he has successfully found the two parameters.

Doctor Rick replied, starting with a comment on the statement, “I think that the number of parameters needed can be 2 or 3 because I think that y = x2 – 3x + 2 can be translated to y = x2 – 3x”:

Here you are giving one example of a translation, namely a vertical translation (down 2 units). The amount of the vertical translation is one parameter. You can also translate horizontally; the amount of the horizontal translation is a second parameter. Those two parameters together fully span the possible translations of the function. This is one way to find the answer to my question; we’ll see how it relates to various other observations we can make.

On the first method, where Kurisada gave the form (y – a) = (x – b)2 – 3(x – b) + 2, he answered,

Ah, good — the a and b here are the two parameters I was talking about above.

He then acknowledged that “a = 1, b = -1, y = x2 – x + 1″ is the correct answer.

Since nothing was said explicitly about how this was obtained, let’s try it. Presumably what was done was to put each (x, y) pair into the form above, to obtain a system of equations in a and b:

$$(1, 1)\rightarrow (1 – a) = (1 – b)^2 – 3(1 – b) + 2\rightarrow 1 – a = 1 – 2b + b^2 – 3 + 3b + 2\rightarrow b^2 + a + b – 1 = 0$$

$$(2, 3)\rightarrow (3 – a) = (2 – b)^2 – 3(2 – b) + 2\rightarrow 3 – a = 4 – 4b + b^2 – 6 + 3b + 2\rightarrow b^2 + a – b – 3 = 0$$

So we are solving the system $$b^2 + a + b – 1 = 0$$ $$b^2 + a – b – 3 = 0$$

We can subtract the second from the first to eliminate a, and we get \(2b + 2 = 0\), so that \(b = -1\); then put that into either equation to obtain \(a = 1\).

Method #2: Directed search

As to the second method, where a list of (x, y) pairs was given in a table,

x: -2  -1  0  . . .
y: 12   6  2  . . .

Doctor Rick initially replied,

I’m not at all sure what you are thinking here. You are finding a set of points on the original curve … ah, perhaps you are thinking, not exactly of the gradient, but of the vector displacement from one of your tabulated points to another, trying to match that to (2-1, 3-1) = (1, 2). In other words, you are looking for two points differing by 1 in x and by 2 in y.

This will work, and can be speeded up by noting the pattern in the y differences. It’s still a “guess and check” or what I’d rather call a “directed search” method; it’s good to think about as we look for a better method, and sometimes a directed search turns out to be the fastest or perhaps even the only method to solve a problem.

By continuing the table and looking for a jump of 2 between successive values of y, we do quickly find this:

x: -2  -1  0  1  2  3
y: 12   6  2  0  0  2

So we know to translate the point (2, 0) to (1, 1). This is essentially what I did with the graph above.

Kurisada’s additional comment on this method did this:

Then I tried my second way:

When x = 2, y = 0.

When x = 3, y = 2.

The gradient of both points above is the same with (1, 1), (2, 3).

But, doesn’t it mean that we add 1 to y and minus 1 to x, in order to get the translation?

But it my first way, a = 1, and thus y – 1 = … . (We minus 1 to y to get the translation.)

So which is correct?

Yes, in translating (2, 0) to (1, 1), we will be shifting up 1 and left 1. But there is a (common) misunderstanding of the relationship between a translation and the equation, which we usually see for horizontal translations, and the same thing happens in the form Kurisada is using for vertical translations.

Doctor Rick answered,

Let me see if I am following your thought this time. You see that the point (2, 0) on the given graph must be translated to (1, 1) on the translated graph. Thus we must subtract 1 from x (2 becomes 1) and add 1 to y (0 becomes 1). That’s what you are saying.

Your first method had the equation (y – a) = (x – b)^2 – 3(x – b) + 2, where a is the vertical translation (up if positive, down if negative) and b is the horizontal translation (right if positive, left if negative). You are apparently thinking that replacing y in the original equation with (y – 1) represents a translation by 1 unit downward, but that’s wrong. If we just make this vertical translation,

y = x2 – 3x + 2 → y – 1 = x2 – 3x + 2

y = x2 – 3x + 3

then you can see that the resulting graph is shifted one unit upward from the original. For more on this, see the recent blog:

Shifting and Stretching Graphs

So both methods in fact give the same equation and represent the same translation.

Method #3: Algebraically matching points

Continuing, Doctor Rick had a new idea:

After working through your work, it occurs to me that another possible method might be to take your “guided search” method and turn it into an algebra problem. What I mean is, we’re looking for a pair of points on the original graph,

(x, x^2 – 3x + 2) and (x+1, (x+1)^2 – 3(x+1) + 2)

and finding the x value such that the difference in the y coordinates is 2:

[(x+1)^2 – 3(x+1) + 2] – [x^2 – 3x + 2] = 2

Once you solve for x, you continue as in method 2. Is that different enough, or is it still your method 2?

I would consider this a different method, though it is built on the same basic idea. Let’s finish it:

The equation simplifies to $$\left[(x+1)^2 – 3(x+1) + 2\right] – \left[x^2 – 3x + 2\right] = 2$$

$$\left[x^2 + 2x + 1 – 3x – 3 + 2\right] – \left[x^2 – 3x + 2\right] = 2$$

$$x^2 + 2x + 1 – 3x – 3 + 2 – x^2 + 3x – 2 = 2$$

$$2x – 2 = 2$$

so \(x = 2\). This means that (2, 0) will translate to (1, 1), and we know the translation we need to do.

Methods #4 and #5: Using the parent function

Kurisada answered,

I was seeking for a totally different method (as in my second method, I directly put in the difference of x and y, that is 1 and – 1 to a and b, so the methods are actually the same at the end).

But now, I’m curious about the two neater ways you stated above. And I think what I need now is these ways. (I want to do this to have a full understanding in how algebra works especially the function, equality, and inequality, because sometimes I find problems that I can’t solve, while I’ve finished reading my books.)

I like the last way you gave me (it’s different enough from my method and it is one that I was seeking to know).

Doctor Rick replied:

I am glad I aroused your curiosity with my mention of two neater variations on your first method, but I would like you to do some further thinking on your own before I give you my ideas in detail. Do you see what I mean when I say that the original equation, y = x2 – 3x + 2, is itself a transformation of a simpler (very basic) function? What is that function, and how would you write general translations of that function? Then, can you write it using another standard form for quadratic functions? In each case there will be two free parameters, but I’m thinking of one in which the parameters do not directly correspond to the horizontal and vertical shifts.

After a little confusion, he expanded his explanation, showing what he meant in his first response by a “family of functions”:

The simplest “parent function” is y = x2. You know how to translate functions in general: replace x with (x – a) to shift the graph a units to the right, and replace y with (y – b) to translate it b units up. This gives the two-parameter family of functions

y – b = (x – a)2                [1]

Expanding the square, we get

y = x2 – 2ax + (a2 + b)

Notice that, by suitable choice of a and b, we can obtain a quadratic with any values we wish for the coefficient of x and for the constant term. Thus another way to express the family of functions obtained by translating y = x2 is

y = x2 + Bx + C               [2]

which is, again, a two-parameter family (changing the coefficient of x2 would introduce a “stretch” in addition to the translation). We could start with either [1] or [2] and proceed the same as in your method 1 to solve the problem, writing an equation for each given point and solving for the two parameters. When I first tried the problem, I chose to start with [2].

There are two parallel, but different, methods here.

Both begin with the fact that we can find a and b such that \(x^2– 2ax + (a^2+ b)\) will be the same as our desired function, \(x^2 – 3x + 2\), for all values of x.  This requires setting the coefficients of each power equal: $$1 = 1$$ $$-2a = -3$$ $$a^2 + b = 2$$

Solving these, we get \(a = \frac{3}{2}\) and \(b = 2 – \frac{9}{2} = -\frac{5}{2}\). This tells us that the given function is \(y = x^2\) translated 3/2 units right and 5/2 units down. (We could have found the same thing by completing the square.)

The same is true of any quadratic of the form \(x^2+Bx + C\). That is, all of these (including our starting function) are translations of the same function.

The point is that in asking for a translation of \(y = x^2 – 3x + 2\) that passes through the given points, they are really asking for a translation of \(y = x^2\) through the two points, so we can find it more quickly, and we can use any form.

So we can do what we did in Method #1 above, where we started with \((y – a) = (x – b)^2 – 3(x – b) + 2\), and instead start with either of the forms [1] and [2].

Kurisada gave it a try, first doing as I did above and finding a and b for the given function, and then using form [1] to solve the problem:

I found that a = 3/2 and b = -9/4 to change it into x2 – 3x.

And I found a = 3/2 and b = -1/4 to change it into x2 – 3x + 2.

I’m not sure about [1], but here what I’ve done:

First, I input (1, 1) and (2, 3) into [1], and I got

a2 – 4a + b = -1

a2 – 2a + b = 0

Resulting a = 1/2 and b = 3/4,

then I did 1/2 – 3/2 = -1 (x shift 1 to left).

And 3/4 – 1/4 = 1 (y shift 1 unit up).

And for the [2], I don’t understand how to apply it. (I tried to apply as above but I think it is wrong.)

On the first bit, Doctor Rick answered,

Good. Thus y = x2 – 3x + 2 results from shifting the graph of y = x2 3/2 unit to the right and 1/4 unit down, putting the vertex of the parabola at (3/2, -1/4). What you do in finding a and b is essentially completing the square. The axis of symmetry is x = 3/2; you can check that the x-intercepts are (1, 0) and (2, 0), which are symmetrical relative to that axis.

Completing the square is an easier way to do the same thing, but this approach fits in with the other techniques being used here.

He continued, on the attempt at using [1]:

You solved correctly for the parameters a and b in the form y = (x – a)2 + b, but I don’t know what you were trying to do after that. You already have the horizontal and vertical shift respectively (from the graph of the parent function y = x2), namely a and b.

OK, I think I see what you’re doing now — you’re finding the shifts from the given function. But you weren’t asked to find the shifts, were you? Doesn’t the problem just ask for the equation of the translated function? That’s y = (x – 1/2)2 + 3/4, which we can expand to y = x2 – x + 1 (the same as the solution we got by other means).

Looking back to the problem statement, I see that it asked for “the equation of the translation,” which could be a little ambiguous. I’m taking it to mean the equation of the translated function, not the amounts of horizontal and vertical shift that constitute the translation (but don’t constitute an equation).

Let’s carry out what Doctor Rick (presumably) intended. We want to find an equation of the form \(y – b = (x – a)^2\), which passes through the points (1, 1) and (2, 3). So, as before, we plug these points into the equation and obtain this system: $$1 – b = (1 – a)^2$$ $$3 – b = (2 – a)^2$$ This is a nonlinear system, which can be a little tricky; I would subtract the two equations, eliminating b, and we find that \(a = \frac{1}{2}\), and then \(b = \frac{3}{4}\). So our answer is \(y – \frac{3}{4} = (x – \frac{1}{2})^2\), which is equivalent to \(y = x^2 – x + 1\).

How about the method starting with [2]?

You solve for B and C similarly to the way you solved for a and b, though it will be somewhat less work. However, B and C aren’t simply related to the horizontal and vertical shifts, so if that is what you need to find, this method will require more work in the end. If all you need is the equation of the translated function, as I believe, then this method is simpler; it gives you that equation in general form directly.

Now Kurisada made a try; after correcting a misstep, it looked like this:

Yes Doctor the question asked me to find the equation. (I didn’t realise to directly use the a and b to find the equation.)

And now I started to understand about (2).

I input (1, 1) and (2, 3) into y = x2 + Bx + C,

resulting in B = -1 and C = 1

So the equation is y = x2 – x + 1

Let’s look at the details: The substitutions give $$1 = 1 + B + C$$ $$3 = 4 + 2B + C,$$ or $$B + C = 0$$ $$2B + C = -1$$ Subtracting the first from the second, we have \(B = -1\), and then the first tells us that \(C = 1\).

Doctor Rick closed the discussion:

In summary, I’d say we’ve solved the same problem in 5 somewhat different ways, though they are based on just 2 fundamentally different approaches, both of which you saw from the start.

Your first method worked directly with the equation of the transformed function. Your second approach started with the observation that if two curves are related to one another by translation, then any two points on one curve have the same coordinate differences, or vector displacement, between them as the corresponding two points on the other curve.

In each case, there was more than one way to implement the basic concept to arrive at the solution.

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