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# 14.1: The Long Rambling Philosophical Introduction

Created by: CK-12

What you’re holding in your hand is much closer to a set of detailed lesson plans than to a traditional textbook. As you read through it, your first reaction may be “Who does he think he is, telling me exactly what to say and when to say it?”

Over a period of time, I have developed a set of in-class assignments, homeworks, and lesson plans, that work for me and for other people who have tried them. If I give you the in-class assignments and the homeworks, but not the lesson plans, you only have $\frac{2}{3}$ of the story, which may not make sense without the other third. So instead, I am giving you everything: the in-class assignments and the homeworks (gathered together in the student book), the detailed explanations of all the concepts (the other student book), and the lesson plans (this document). Once you read them over, you will know exactly what I have done.

What do you do then? You may choose to follow my plan exactly, for a number of reasons—because it worked for me, or because it looks like a good plan to you, or just because you have enough other things to do without planning a lesson that I’ve already planned. On the other hand, you may choose to do something quite different, that incorporates my ideas in some form that I never imagined. This book is not a proscription, in other words, but a resource.

OK, with that out of the way...suppose you decide that you do want to follow my plan, exactly or pretty closely. Here’s what you do.

• Right now, you read this whole introduction—despite the title, it really does contain useful information about these materials.
• Before beginning each new unit, you read my “conceptual explanation” of that unit, so you know what I’m trying to achieve.
• Each day before class, you carefully read over my lesson plan (in this document), and the in-class assignment and homework (in the student book), so you know what I’m doing and why I’m doing it.

A Typical Day in Mr. Felder’s Class (…and why you care)

At the risk of repeating myself, let me emphasize—I’m not trying to insult you by suggesting that my way is the only right way to run a class. But it will help you understand these materials if you understand how I use them.

I begin each day by taking questions on last night’s homework. I answer any and all questions. This may take five minutes, or it may take the entire class period: I don’t stop until everyone is perfectly comfortable with last night’s homework.

Why is that so important? Because, very often, the homework introduces new concepts that the students have never seen in class before. For instance, very early in the first unit, I introduce the idea of “permuting” graphs: for instance, if you add $3$ to any function, the graph moves up by three units. This concept never comes up in class, in any form—it is developed entirely on a homework. So it’s vitally important to debrief them the next day and make sure that they got, not only the right answers, but the point.

After the homework is covered, I begin a new topic. This is almost (almost!) never done in a long lecture. Sometimes it happens in a class discussion; sometimes it happens in a TAPPS exercise (more on that when we do our first one); most often, it happens in an in-class assignment. These assignments should almost always be done in pairs or groups of three, very rarely individually. They generally require pretty high-level thinking. On a good day I can hear three or four heated arguments going on in different groups. Most of my class time is spent moving between different groups and helping them when they are stuck. In general, there is some particular point I want them to get from the exercise, and they will need that point to do the homework—so a lot of my job in class is to make sure that, before they leave, they got the point.

Timing

If you read through this entire document (which I do not recommend at one sitting), you get the illusion that I have everything planned down to the day. If I say, “Do this assignment in class, then do this homework,” they had better get that done in one day, or they will fall irretrievably behind.

Well, suppose you add it all up that way. Every “$1-$ day assignment” (with homework) counts as one day, and what the heck, let’s allocate two days for every test (one day for preparation, using the “Sample Test”—and one day for the actual test). If you add it up that way, you will get a total of $91$ days, or thereabouts. There are $180$ days in the school year.

So what does that mean? Does it mean you will be done in one semester? No, of course not. It means, take your time and do it right.

For one thing, I believe in building in a lot of time for review. Ideally, two weeks before mid-terms and another two weeks before finals. (What I do during this time is cover one topic a day, with the students teaching each class.)

But even leaving that aside, one apparent day’s worth of material will sometimes take you two days to get through. You spend the whole day reviewing last night’s homework and you don’t even get to the new assignment. Or, you get to the end of the class and you realize that most of the groups are only half-way through the in-class assignment. Don’t rush it! It’s much more important to get today’s concept, and really make sure everyone has it, then to rush on to tomorrow. The way I see it, you have three reasonable choices.

1. If most of the class is mostly finished with the in-class assignment, it may make sense to say, “Finish the in-class assignment tonight, and also do the homework.”
2. If most of the class is only half-way done, it may make sense to say, “Finish the in-class assignment tonight, and we will do the homework in class tomorrow.” This puts you a half-day “behind,” which is fine. However, some in-class assignments really cannot be done at home… they require too much group work or help from you. So…
3. Sometimes you just say, “We’ll finish the in-class assignment tomorrow.” This puts you one day “behind,” which is also fine.

Of course, you need to pace yourself. But do it by tests, not by days. There are fifteen tests. If you are going at a clip that will get you through more or less that many tests by the end of the year, you’re doing fine. And even that isn’t exact—some units will naturally take longer than others. Personally, I would much rather skip the unit on Conics (the last unit) entirely, than lose the entire class by trying to rush through Exponents. (However, in real life, I do make it through the entire syllabus.)

Tests

At the end of every unit I have a “Sample Test.” This is for the students’ benefit as much as for yours: it makes a great study guide and/or homework. If you say, “The homework tonight is the sample test. Tomorrow we will go over any questions you have on the sample test and on the topic in general—that will be your last chance to ask me questions! The next day will be our actual test." Thus giving the students a great chance to bone up before the test. Doing this has dramatically improved my classes.

So what about your actual test? Of course, you may (or may not) want to base your test on mine. In that case, however, be careful about timing—some of my “Sample Tests” are actually too long to be a real test. But they are made up of actual questions that I have used on actual tests in the past—and in any case, calling them “Sample Tests” gets students’ attention better than calling them “Review Questions.”

By the way, although I do not generally recommend using my questions exactly—you want to change the numbers at least—it is sometimes OK to use my extra credit exactly. Even if they just did it, it often has enough real learning in it that it is worth giving them a few points if they took the time to look it over and/or ask about it.

## Functions

This is the most important unit in the year, because it introduces many of the major “themes” that will run through the entire class. These themes include:

• What is a function?
• How are functions used to model things in the real world?
• What does it mean for two functions to be “equal”?
• How does a functional description $f(x)=3x$ relate to a graph?

More detailed topics include the vertical line test (the “rule of consistency”), the “dependent” and “independent” variables, domain and range, composite functions, and inverse functions.

Also included in this unit is a quick review of graphing lines. It is assumed that students are mostly familiar with this topic from Algebra I.

The Function Game

This game is an introduction to the idea of a function.

Begin by breaking the students into groups of three. In each group, one student is designated as the leader and another as the recorder. (Do this quickly and arbitrarily: “The shortest person is the leader and the tallest is the recorder”or something similar. Assure them that the roles will rotate.) Go over the instructions (which are in the student packet): walk through a sample session, using the function “add five,” to make sure they understand who does what and what gets written down. In particular, make sure they understand how “add five” can be represented as “$x+5$”: that is, that $x$ is being used to designate the number that comes in. You also want to mention domain in particular, and the idea that if you are doing $\frac{1}{x}$ and someone gives you a $0$, it is “not in your domain.”

Then they can start. Your job is to circle around, keeping them on task, and helping students who are stuck by giving hints: “Do you notice anything in common about all the numbers you’ve gotten back?” or “Why are all negative numbers outside the domain?” or even “Try a $2$ and see what happens.” Also, at $10$ or $15\;\mathrm{minute}$ intervals, instruct them to switch roles.

Toward the end of the class period, interrupt briefly to talk about the word function. What is the leader representing? He is not a number. He is not a variable. He is a process that turns one number into another. That’s all a function is—a mechanical process that takes one number in, and spits a different number back out. Use the analogy of a little machine (you can draw it) with an input and an output. Functions are going to be the main focus of the entire year in Algebra II.

If some groups don’t finish all the problems, that’s OK: as long as they did enough to get the idea. If a group finishes early, tell them to start making up their own functions—challenge them to stump you!

Homework:

“Homework: The Function Game”

“The Real World”

Begin by going over the homework. There are two key points to bring out, which did not come up in class yesterday.

1. If two functions always give the same answer, we say they are equal. This just cannot be stressed enough. Put this on the board:

$3(x-5) = 3x-15$

They’ve all seen this. Most of them know it is the “distributive property.” But do they know what it means? Explain it very carefully. $3(x-5)$ is a function that says, “subtract $5$, then multiply by $3$.” $3x-15$ is a function that says, “multiply by $3$, then subtract $15$.” They are very different processes! But we say they are “equal” because no matter what number you plug in, they yield the same answer. So in the function game, there is no possible question you could ask that would tell you if the person is doing “subtract $5$, then multiply by $3$” or “multiply by $3$, then subtract $15$.” Have the class give you a few numbers, and show how it works—for negative numbers, fractions, zero, anything.

2. The last problem on the homework brings up what I call the “rule of consistency.” It is perfectly OK for a function to give the same answer for different questions. (For instance, $x^2$ turns both $3$ and $-3$ into $9$.) But it is not OK to give different answers to the same question. That is, if a function turns a $3$ into a $9$ once, then it will always turn a $3$ into a $9$.

Once those two points are very clearly made, and all questions answered, you move on to today’s work. Note that there is no “in-class assignment” in the student book: this is a day for interacting with the students.

Remind them that a function is simply a process—any process—that takes one number in, and spits a different number out. Then explain that functions are so important because they model relationships in the real world, where one number depends on a different number. Give them a few examples like the following—no math, just a verbal assertion that one number depends on another:

• The number of toes in class depends on the number of feet in class.
• Which, in turn, depends on the number of people in class.
• The number of points you make in basketball depends on how many baskets you make.
• The amount I pay at the pump depends on the price of gas.

Have them brainstorm in pairs (for $1-2\;\mathrm{minutes}$ at most) to come up with as many other examples as they can. Since this is the first “brainstorming” exercise in class, you may want to take a moment to explain the concept. The goal of brainstorming is quantity, not quality. The object is to come up with as many examples as you can, no matter how silly. But although they may be silly, they must in this case be valid. “The color of your shirt depends on your mood” is not a function, because neither one is a number. “The number of phones in the class depends on the number of computers” is not valid, because it doesn’t. By the end of a few minutes of brainstorming and a bit more talk from you, they should be able to see how easy it is to find numbers that depend on other numbers.

Then—after the intuitive stuff—introduce formal functional notation. Suppose you get two points per basket. If we let $p$ represent the number of points, and $b$ represent the number of baskets, then $p(b)=2b$. If we say $p(c)=2c$ that is not a different function, because they are both ways of expressing the idea that “$p$ doubles whatever you give it” (relate to the function game). So if you give it a $6$, you get $12: p(6)=12$. If you give it a duck, you get two ducks: $\mathrm{p(duck)}=2\;\mathrm{ducks}.$ It doubles whatever you give it.

Key Points to Stress:

• This notation, $p(b)$, does not indicate that $p$ is being multiplied by $b$. It means that $p$ depends on $b$, or (to express the same thing a different way), $p$ is a function of $b$.
• You can plug any numbers you want into this formula. $p(6)=12$ meaning that if you get $6$ baskets, you make $12$ points. $p\left(2\frac{1}{2}\right ) = 5$ is valid mathematically, but in the “real world” you can’t make $2\frac{1}{2}$ baskets. Remind them that $p$ is a function—a process—“double whatever you are given.”
• Also introduce at this point the terminology of the dependent and independent variables.
• Stress clearly defining variables: not “$b$ is baskets,” but “$b$ is the number of baskets you make.” (“Baskets” is not a number.)
• Finally, talk about how we can use this functional notation to ask questions. The question “ How many points do I get if I make $4$ baskets?” is expressed as “What is $p(4)$?” The question “ How many baskets do I need in order to get $50$ points?” is expressed as “$p(b)=50$, solve for $b$.”

For the rest of class—whether it is five minutes or twenty—the class should be making up their own functions. The pattern is this:

1. Think of a situation where one number depends on another. (“Number of toes depends on number of feet.”)
2. Clearly label the variables. ($t=$number of toes, $f=$number of feet.)
3. Write the function that shows how the dependent variable depends on the independent variable $(t(f)=5f)$.
4. Choose an example number to plug in. (If there are $6 \;\mathrm{feet}$, $t(f)=5(6)=30$. $30$ toes.)

Encourage them to think of problems where the relationship is a bit more complicated than a simple multiplication. (“The area of a circle depends on its radius, $A(r)=\pi r^2$.”)

Homework:

“Homework: Functions in the Real World”

When going over this homework the next day, one question that is almost sure to come up is $^\# 4g$: $f(f(x))$. Of course, I’m building up to the idea of composite functions, but there is no need to mention that at this point. Just remind them that $f (anything)= anything^2+2 \ anything +1$. So the answer to $(e)$ is $f(spaghetti)= spaghetti^2+2 \ spaghetti +1$. And the answer to this one is $f(f(x))=f(x)^2+2f(x)+1$. Of course, this can (and should) be simplified, but the point right now is to stress that idea that you can plug anything you want in there.

Algebraic Generalizations

This is the real fun, for me.

Start by telling the students, “Pick a number. Add three. Subtract the number you started with. You are left with...three!” OK, no great shock and surprise. But let’s use algebra to express what we have just discovered. $x+3-x=3$. The key is recognizing what that sentence means. $x$ can be any number. So when we write $x+3-x=3$ we are indeed asserting that if you take any number, add three and then subtract the number, you get three in the end.

Here’s a harder one. Pick a number, add three, multiply by four, subtract twelve, divide by the number you started with. Everyone started with different numbers, but everyone has $4$ in the end. Ask the students to find a generalization to represent that, and see if they can work their way to $\frac{4(x + 3) -12}{x} = 4$. See also if they can guess what number this trick will not work with. Now, have them work on the in-class assignment “Algebraic Generalizations,” in groups of three. Most of the class period should be spent on this. This is hard!!! After the first couple of problems (which very directly echo what you already did in class), most groups will need a lot of help.

Here are some of the answers I’m looking for—I include this to make sure that the purpose of the assignment is clear to teachers.

• In $^\#3$, the object is to get to $2^{x+1}=2\cdot 2^x$ (or, equivalently, $2^x=2 \cdot 2^{x-1})$. Talk through this very slowly with individual groups. “Let’s call this number $x$. So what is this number? $(2^x)$ And how can we represent this number? $(x+1)$," etc... The real goal is to get them to see how, once you have written $2^{x+1}=2 \cdot 2^x$, you have said in one statement that $2^8$ is twice $2^7$, and also that $2^5$ is twice $2^4$, and also that $2^{11}$ is twice $2^{10}$, and so on. It is a “generalization” because it is one statement that represents many separate facts.
• In $^\#4$, the object is to get to $x^ax^b=x^{a+b}$. Again, it will take a lot of hand-holding. It isn’t important for them to do it entirely on their own. It is critically important that by the time they are done, they see how those numbers lead to that generalization and how that generalization leads to those numbers.
• If all that works, they should be able to do $^\# 6$ and come up with something like $(x-1)(x+1)=x^2-1$ pretty much on their own. (Or, of course, $x^2=(x-1)(x+1)+1$, which is a bit more unusual-looking, but just as good.)

Homework:

“Homework: Algebraic Generalizations”

## Graphing

Make sure, on the homework, that they reached something like $(x-a)(x+a)=x^2-a^2$. Of course, someone may have used completely different letters, and someone else may have said that $(x-a)(x+a)+a^2=x^2$. Point out that these are just as good—they look different, but they say the same thing. Finally, show them how they could have arrived at $(x-a)(x+a)=x^2-a^2$ through FOIL.

Once all questions are satisfied, on to graphing. This may actually be two days worth of material. Don’t rush it! And once again, no in-class assignment, but a lot of interaction.

Start by putting the following points on the board: $(6,5)(3,6)(2,9)(5,7)(1,10)(4,8)$. Ask for a general description—if these represent the number of pushups you do each day, what’s the trend? Then graph them, and you can see a definite downward trend with an odd spike in the middle. Moral: we can see things in shapes that we can’t see in numbers.

Now, let’s jump to the idea of graphing functions. Draw a $U-$shape, the equation $f(x)=x^2$, and say, “This drawing is the graph of that function—but what does that mean? What does this drawing actually have to do with that function?” Get to the point where the following ideas have come out. Every time you “do” the function, one number goes in and another comes out. When we graph it, the “in” function is always $x$, and the “out” function is always $y-$ in other words, every time we graph a function, we are always graphing $y=f(x)$. To put it another way, we are graphing all the points that have this particular relationship to each other. (Take as much time as you need on this point.)

Have the students graph $|x|$ (individually at their desks) by plotting points. This is not intended to teach them about absolute value, but to reinforce the ideas I just made about what it means to graph a function.

What can we tell by looking at a graph? Draw the graphs of $x^2$ and $x^3$ on the board. Talk about the things we can tell about these two functions by looking at the graphs. They both have one zero; they are similar on the right, except that $x^3$ rises faster; they are completely different on the left; they both have unlimited domains, but only one has an unlimited range. (Make sure to connect this back to “domain” and “range” from the function game!)

Now, draw this on the board.

Time for another…brainstorming exercise! (Remind them that the object is quantity, not quality!) Each pair of students has to list as many things as possible about this function $f(x)$ by looking at the graph. Key points I want to bring out are:

• The three zeros. (Talk about the word “zeros.”)
• Places where the function is negative and places where it is positive.
• Places where it is increasing and places where it is decreasing. (Talk about slope!)
• What happens for very low and very high values of $x$.
• For the experts, odd symmetry: $f(-x) = -f(x)$.

Talk more about domain and range. This is unrestricted in both. We saw that $x^2$ has an unrestricted domain, but a restricted range. Show graphs of $\sqrt{x}$ and $\frac{1}{x}$ for restricted domains.

Remind them of the “rule of consistency” that we discussed earlier: a function can never take one input and generate two different outputs. Ask them to discuss in pairs, for one minute, how this rule manifests on a graph. Then have class discussion until you have reached the vertical line test.

Come back to the question we started with: why do we graph things? As we demonstrated earlier, graphs enable us to see things visually that are very hard to see in numbers. Draw several different jaggy, shaky graphs, and suggest that they represent the price of gas—ask for verbal descriptions of what each one tells us. Talk again about domain, range, positive, negative.

Throw something straight up into the air and catch it. Tell them there is a function $h(t)$ that represents the height of that object, as a function of time. (Make sure they get this.) Give them one minute to sketch the graph of that function. Then show them that it is an upside-down parabola (you don’t need to use that word). Emphasize that this does not mean the object traveled in an arc shape: it traveled straight up and down. The horizontal axis is time and the vertical axis is height.

If you have a bit of extra time, explain how to generate graphs (and set the window) on the calculator.

Homework:

“Homework: Graphing”

Permutations

The end of the “Graphing” homework sets this topic up. You want to talk for a while about vertical permutations (which tweak the $y-$ value) and horizontal permutations (which tweak the $x-$ value).

With relatively little preamble, hand them the worksheet “Horizontal and Vertical Permutations.” It looks long, but it’s not really a huge deal, and should not take the whole period. Afterwords, when you most-mortem it, you make sure they really get the point. If we know what $y=x^2$ looks like, we don’t have to plot a whole bunch of new points to find what $y=x^2+3$ looks like; we can see from first principles that it will look like the original graph, but moved up three. This should not be a memorized fact: it should be obvious, if they understand what graphing means. Less obviously, $y=(x+3)^2$ will look like the original graph, moved to the left by $3$. The “Conceptual Explanations” contains my second-best explanation of why this is. My very-best explanation is the worksheet itself: plot a bunch of points and see what happens.

If you haven’t already done so, introduce graphing on the calculator, including how to properly set the window. It only takes $5-10 \;\mathrm{minutes}$, but it is necessary for the homework.

Test Preparation

Follow up carefully on the homework from the day before. In $^\#8$, make sure they understand that losing money is negative profit. But mostly talk about $^\#9$. Make sure they understand how, and why, each modification of the original function changed the graph. When we make a statement like, “Adding two to a function moves the graph up by $2$” this is not a new rule to be memorized: it is a common-sense result of the basic idea of graphing a function and should be understood as such.

Ask what they think $-(x+1)^2 -3$ would look like, and then show them how it combines three of the modifications. (The $+1$ moves it to the left, the $-3$ moves it down, and the - in front turns it upside-down.)

Talk about the fact that you can do those generalizations to any function, eg $-|x+1|-3$ (recalling that they graphed absolute value yesterday in class).

Draw some random squiggly $f(x)$ on the board, and have them draw the graph of $f(x)+2$. Then, see if they can do $f(x+2)$.

Go back to the idea of algebraic generalizations. We talked about what it means for two functions to be “equal”—what does that look like, graphically? They should be able to see that it means the two functions have exactly the same graph. But this is an opportunity for you to come back to the main themes. If two functions are equal, they turn every $x$ into the same $y$. So their graphs are the same because a graph is all the $(x,y)$ pairs a function can generate!

I also like to mention at this point that we actually use the $=$ sign to mean some pretty different things. When we say $x+3=5$ we are asking for what value of $x$ is this true? Whereas, when we say $2(x-7)=2x-14$ we are asserting that this is true for all values of $x$. Mention, or ask them to find, statements of equality that are not true for any value of $x$ (eg $x=x+1$).

Now, at this point, you hand out the “sample test.” Tell them this test was actually used for a past class, and although the test you give will be different, this test is a good way of reviewing. I have found this technique—handing out a real test from a previous year, as a review—to be tremendously powerful. But I have to say a word here about how to use it. Sometimes I say, “Work on it in class if you have time, glance at it over tonight if it helps you study; the test is tomorrow.” And sometimes I say, “This is the homework. Tonight, do the sample test, and also look over all the materials we have done so far. Tomorrow, we will go over the sample test and review any questions you have on the material. This will be preparation for the test, which will be the day after tomorrow.” It all depends on timing and on how prepared you think the class is.

Homework:

“Sample Test: Functions I”

I should say a word about $^\#2$ here, just to be clear about what I’m looking for. We have a function $c(s)$. In part (b) I supply $s=20$ and ask for $c$: so the question in function notation is $c(20)$. (Then you plug a $20$ into the formula and go from there.) In part (c), I supply $c=35$ and ask for $s$: so this question in function notation is $c(s)=35$. (Then you set the formula equal to $35$ and solve.) Students have a lot of difficulty with this asymmetry.

Now Give a Test of Your Own on Functions I

You may use something very similar to my test, except changing numbers around. Or you may do something quite different.

I will tell you, as a matter of personal bias, that I feel very strongly about question $^\#7$. Many students will do quite poorly on this question. For instance, they may give you two variables that do not in fact depend on each other (number of guitarists and number of drummers). Or they may give you variables that are in fact constants (let $n$ equal the number of notes in an octave). The answers don’t have to be complicated, although sometimes they are—sometimes a very simple answer gets full credit. (“CDs cost $\12$ apiece, I spend $d \;\mathrm{dollars}$ on $c\;\mathrm{CDs}$I . $d(c)=12c$” is a full credit answer to parts $a-c$.) But they have to show that they can clearly articulate what the variables are and how they depend on each other. If they cannot, spend the time to explain why it is wrong and work them through correct answers. In my opinion, this skill is the best measure of whether someone really understands what a function is. And it is a skill like any other, in the sense that it develops over time and practice.

(Also, I don’t believe in surprises. Tell them in advance that this question will definitely be on the test—the only thing that will change is the topic.)

Lines

This is largely review: if there is one thing the students do remember from Algebra I, it’s that $y=mx+b$ and $m$ is the slope and $b$ is the $y-$intercept. However, we’re going to view this from the viewpoint of linear functions.

Start by giving an example like the following: I have $100$ markers in my desk. Every day, I lose $3$ markers. Talk about the fact that you can write a function $m(d)$ that represents the number of markers I have as a function of day. It is a linear function because it changes by the same amount every day. If I lose three markers one day and four the next, there is still a function $m(d)$, but it is no longer a linear function. (If this is done right, it sets the stage for exponential functions later: linear functions add the same amount every day, exponential functions multiply by the same amount every day. But I wouldn’t mention that yet.)

So, given that it changes by the same amount every day, what do you need to know? Just two things: how much it changes every day $(-3)$, and where it started $(100)$. These are the slope and the $y-$ intercept, respectively. So we can say $y=-3x+100$, but I actually prefer to write the $b$ first: $y=100-3x$. This reads very naturally as, “start with $100$, and then subtract $3$, $x$ times.”

Hammer this point home: a linear function is one that adds the same amount every time. Other examples are: I started with $\100$ and make $\5.50$ each hour. (Money as a function of time.) I start on a $40'$ roof and start piling on bricks that are $\frac{1}{3}'$ each. (Height as a function of number of bricks.)

Then talk more about slope—that slippery concept that doesn’t tell you how high the function is at all, but just how fast it’s going up. With a few quick drawings on the board, show how you can look at a line and guesstimate its slope: positive if it’s going up, negative if it’s going down, zero for horizontal. You can’t necessarily tell the difference between a slope of $3$ and a slope of $5$, but you can immediately see the difference between $3$ and $\frac{1}{2}$. Emphasize that when we say “going up” and “going down” we always mean as you go from left to right: this is a very common source of errors.

Talk about the strict definition of slope. Actually, I always give two definitions. One is: every time $x$ increases by $1$, $y$ increases by the slope. (Again, if the slope is negative, $y$ decreases.) The other is: for any two points on the line, the slope is $\frac{\triangle y}{\triangle x}$ (“rise over run”). Show that this ratio is the same whether you choose two points that are close, or two points that are far apart. Emphasize that this is only true for lines.

Finally, why is it that in $y=mx+b$, the $b$ is the $y-$ intercept? Because the $y-$ intercept is, by definition, the value of $y$ when $x=0$. If you plug $x=0$ into $y=mx+b$ you get $y=b$.

All this may take all day or more than one day. Or, it may go very quickly, since so much of it is review. When you’re done, have them work the in-class assignment “Lines” in groups.

Homework:

“Homework: Graphing Lines”

Composite Functions

OK, it’s getting hard again: this one may take a couple of days. We start with discussion.

There are a number of ways to look at composite functions. It’s important to be able to use all of these ways and to see how they relate.

1. All the way back to the function game. Let one student be the function $4x+6$ and another student be the function $\frac{x}{2}$. You give a number to the first student, who spits out a number at the second student, who spits out a number back at you. For instance, if you give the first student a $3$, his output is an $18$; the second student takes this $18$ and comes out with a $9$. Do this for a while until everyone gets the hang of it. See if anyone realizes that what’s going on is, in the end, the function $2x+3$ is being done to your number.
2. Now, talk about a factory. One box turns garbage into gloop; the next box turns gloop into schlop; the final box turns schlop into food. Each box can be represented by a function that says, “If this much goes in, that much goes out.” The entire factory is a gigantic composite function, where the output of each box is the input of the next, and the composite function says, “If this much garbage goes in, this much food goes out.” (Draw it!)
3. In general, composite functions come up with this variable depends on that variable, which in turn depends on the other variable. The amount of taxes you pay depends on the amount of money you make, which in turn depends on the number of hours you work. Have them come up with a few examples. Be very careful to distinguish composite functions from multivariate functions, eg the number of kids in the class depends on the number of boys, and the number of girls. That is not a function, because those two variables don’t depend on each other.
4. Finally, there is the formalism, $f(g(x))$. Remind them that this is mechanical. If $g(x)=4x+6$ and $f(x)=\frac{x}{2}$, then what is $f(g(x))?$ Well, $\mathrm{f}(anything)=\frac{anything}{2}$. So $f(g(x))=\frac{g(x)}{2}$, which is $\frac{(4x+6)}{2}$ or $2x+3$. Note that this is completely different from $g(f(x))!$ Take a moment to connect this mechanical process with the idea of a composite function that you have already discussed.

Now, have them work through the in-class assignment on “Composite Functions” in groups. Make sure they do well on $^\#4$.

$^\#6$ is a build-up to inverse functions, although you don’t need to mention that. If anyone asks for help, make sure they see that if $h(x)=x-5$, then $h(anything) = anything -5$, so $h(i(x))=i(x)-5$. So $i(x)-5=x$, and we can solve this to find $i(x)=x+5$.

Homework:

“Homework: Composite Functions”

Inverse Functions

This is definitely two days and possibly three.

Just as with composite functions, it is useful to look at this three different ways: in terms of the function game, in terms of real world application, and in terms of the formalism.

1. Ask a student to triple every number you give him and then add $5$. Do a few numbers. Then ask another student to reverse what the first student is doing. This is very easy. You give the first student a $2$, and he gives you an $11$. Then you give the second student an $11$, and he gives you a $2$. Do this a few times until everyone is comfortable with what is going on. Then ask what function the second student is doing. With a little time, everyone should be able to figure this out—he is reversing what the first student did, so he is subtracting five, then dividing by $3$. These two students are “inverses” of each other—they will always reverse what the other one does.

2. Give a few easy functions where people can figure out the inverse. The inverse of $x+2$ is $x-2$ (and vice-versa: it is always symmetrical). The inverse of $x^3$ is $\sqrt[3]{x}$. The key thing to stress is how you test an inverse function. You try a number. For instance…

$10 & \rightarrow x+2 \rightarrow 12 \rightarrow x-2 \rightarrow 10\\-5 & \rightarrow x+2 \rightarrow -3 \rightarrow x-2 \rightarrow -5$

The point is that you take any number and put it into the first function; put the answer in the second function, and you should get back to your original number. Testing inverses in this way is more important than finding them, because it shows that you know what an inverse function means.

3. Ask for the inverse of $x^2$. Trick question: it doesn’t have any! Why not? Because $x^2$ turns $3$ into $9$, and it also turns $-3$ into $9$. It’s allowed to do that, it’s still a function. But an inverse would therefore have to turn $9$ into both $3$ and $-3$, and a function is not allowed to do that—rule of consistency. So $x^2$ is a function with no inverse. See if the class can come up with others. (Some include $y=|x|$ and $y=3$.)

4. Now ask them for the inverse of $10-x$. They will guess $10+x$ or $x-10$; make sure they test! They have to discover for themselves that these don’t work. The answer is $10-x$; it is its own inverse. (It turns $7$ into $3$, and $3$ into $7$.) Ask for other functions that are their own inverses, see if they can think of any. (Other examples include $y=x$, $y=-x$, $y=\frac{20}{x}$.)

5. In practice, inverse functions are used to go backwards, as you might expect. If we have a function that tells us, “If you work this many hours, you will get this much money,” the inverse function tells us, “If you want to make this much money, you have to work this many hours.” It reverses the $x$ and the $y$, the dependent and independent variables. Have the class come up with a couple of examples.

6. Formally, an inverse function is written $f^{-1}(x)$. This does not mean it is an exponent, it is just the way you write “inverse function.” The strict definition is that $f(f^{-1}(x))= x$. This definition utilizes a composite function! It says that if $x$ goes into the inverse function, and then the original function, what comes out is…$x$. This is a hard concept that requires some talking through.

OK, at this point, you have them start working on the in-class exercise “Inverse Functions.” Note that you have not yet given them a way of finding inverse functions, except by noodling around! Let them noodle. Even for something like $y = \frac{2x + 3}{7}$, they should be able to get there, with a bit of hand-holding, by reversing the steps: first multiply by $7$, then subtract $3$, then divide by $2$. If they ask about $^\#11$, make sure they try a few things (such as $\sqrt[x]{2}$) and test them—they will discover they don’t work. Explain that, in fact, we have no inverse function of $2^x$ right now, so we’re going to make one up later in the year and call it a “logarithm.” They can then leave this one blank.

After they have finished noodling their way through the most of the exercises, interrupt the class and say, “Now, I’m going to give you a formal method of finding inverse functions—you will need this for the homework.” The formal way is: first, reverse the $x$ and the $y$, then solve for $y$. For instance, from $y = \frac{2x + 3}{7}$ we first write $x = \frac{2y + 3}{7}$. Then solve for $y$ to get $y = \frac{7x - 3}{2}.$

Homework:

“Homework: Inverse Functions”

But wait! We’re not done with this topic!

What happens the next day is, they come in with questions. Whatever else they did or didn’t get, they got stuck on $^\#10$. (If they got stuck on $^\#9$, point out that it is the same as $^\#2$; make sure they understand why.) $^\#10$ is hard because they cannot figure out how to solve for $y$. This brings us, not to a new conceptual point, but to a very important algebraic trick, which we are going to learn by doing a TAPPS exercise.

TAPPS (Thinking Aloud Pair Problem Solving) is a powerful learning tool, and here’s how it goes. The students are broken into pairs.

One person in each pair is the teacher. His job is to walk through the following solution, step by step, explaining it. For each step, explain two things. Why am I allowed to do that, and why did I want to do that? Your explanation should make perfect sense to a normal Algebra I student. You should never skip steps—go line, by line, by line, explaining each one.

The other person is the student. He also has two jobs. First, whenever the teacher says something that is not perfectly clear, stop him! Even if you understand it, say, “Wait, that didn’t make perfect sense.” Keep pushing until the explanation is completely bullet-proof. Second, keep the teacher talking. If he pauses to think, say “Keep talking. What are you thinking?” The teacher should “think out loud” until he comes up with something.

If the students are stuck on a line, they should raise their hands and ask you.

Take the time to carefully explain the process—we will do other TAPPS exercises. And one more thing—warn them that after everyone has completed the exercise, you will be calling on individuals to explain tricky steps. You will not call for volunteers. After they are done, everyone in the class should be able to answer any question about anything in this derivation. So you will just pick people and ask them questions like, “Why did I do that?”

After they are done, call on individuals and ask questions like “Why did we subtract $2xy$ from both sides?” and “How did we get $y(1-2x)?$” Point out the general strategy is only two steps: get all the $y$ things on one side, then pull out the $y$. Check their answer to the question at the end.

Time for Another Test!

Once again, there is the sample test—you will probably want to assign it as a homework, and tell them to do that and also study everything since the last test. The next day, go over the homework and any questions. Then give the test.

Congratulations, you’re through with your first unit! If things went well, you have laid the groundwork for the entire year. Onward and upward from here!

## Inequalities and Absolute Values

I like to have this unit early in the year, because it introduces another of my main themes: some of these problems require you to think. There is almost no “mechanical” way to solve them. I want to set up that expectation early.

Of course, this is really two very different topics, and both of them contain an element of review as well as some new material. But by the time we get to combining them in problems like $2|3x+4| < 7$, it is new to everybody. Those problems are harder than you might suppose!

Inequalities

This is pretty easy, isn’t it? You can just start by having them work through “Inequalities” with no preamble at all.

After they’ve worked on it for a while, you might want to interrupt the class to talk about it for a while. $^\#1$ is obviously an attempt to get at the old “you have to reverse the inequality when you multiply or divide by a negative number” thing. But stress as loudly as you can, the idea that they shouldn’t just take anyone’s word for it. The question we’re trying to get at is, why do you have to switch the inequality, then and only then? I usually illustrate this point by drawing a number line and showing how, on the left of the zero (in “negative land”) the numbers are going backward (an observation that every second grader notices, but that they have forgotten by high school). So you can visually show how $2 < 3$ becomes $-2 > -3$ when you move it to the other side.

Also, tell them how important it is to distinguish carefully between ANDs and ORs—this will become a major issue later. I have two pet peeves on this topic.

The first pet peeve is people who memorize a facile rule (such as “less-than problems become AND and greater than problems become OR”) without having the slightest idea what they are doing. I go out of my way to create problems that frustrate such rules (which isn’t hard to do). They have to see and understand what these conjunctions mean.

The second pet peeve is $x > \pm 4$. This is, for all intents and purposes, meaningless. I want them to realize that on $^\#9$, and then I warn them that I will always take off points if they answer any question this way. (This comes up in the context of $|x| > 4$ and is a good example of how wrong you go if you answer mechanically instead of thinking.) $x$ can equal $\pm 4$, if you know what that means (shorthand for $x=4$ or $x=-4$), but it cannot be greater than, or less than, $\pm$ anything.

Homework:

“Homework: Inequalities”

Inequality Word Problems

The difficulty with this is that there is really nothing to say about it at all: it’s just something I want them to see. So there is this homework assignment, which you can give at any time, before or after anything else. Stick it in when you find yourself, at the end of a day, not quite ready to give out the next homework, or something like that.

Homework: "Inequality Word Problems"

In one problem, they have to make up an inequality word problem. You might think that, just because they have made up so many word problems and functions by now, they would knock this one down easily—but it ain’t so. I’ve had students who could create functions all day long (by this time) who could not create a good inequality to save their lives. They create scenarios like, “I eat three bowls of cereal a day, how many do I eat in a week?” There is nothing unequal there. Just like everything else, this takes practice. But I do want them to see that inequality relationships are all around us.

Absolute Value Equations

Now it gets tough. But once again, little or no preamble is needed: just have them start working on “Absolute Value Equations.”

Here’s the thing. Problems $1-9$ really contain all the math: all the concepts they need to get. And, for the most part, they will get them right—although you need to check this before they go on any further.

But when it comes to the more complicated-looking problems in the second half of the assignment, they panic. They stop thinking, revert to rules, and start getting wrong answers. If they are diligently checking, they will realize that their answer to $^\#12$ doesn’t work. But they may need you to point out that this is because it is analogous to $^\#6$ and has no answer.

So, around this time, I spend a lot of time insisting, “Think, think, think!” The way to think it through is this. Once you have solved for the absolute value, go back to the kind of thinking you did in the first page. For instance, when you have $|x+3|=-1$, cover up the $x+3$ and ask yourself the question like this: “The absolute value of something is $-1$. What is the something?” The answer, of course, is “nothing.” Think, and you will get it right. Plug and chug, and you will get it wrong.

OK, if $|x+3|=7$ has two answers, and $|x+3|=0$ has one, and $|x+3|=-4$ has none, then what about $|x-2|=2x-10?$ The answer is, you don’t know until you try. You begin by splitting it the same way you did before: $x-2=2x-10$ or $x-2=-(2x-10)$. Find both answers. But then check them: even if you did the math right, they may not work! Don’t tell them this up front, but make sure to discuss this with them toward the end of class, in the context of $^\#13$; they will need to know this for the homework.

Homework:

“Homework: Absolute Value Equations”

Absolute Value Inequalities

They are going to work on the assignment “Absolute Value Inequalities” in class. You may want to begin by reminding them that they have already been solving absolute value inequalities. On the previous assignment they turned, “the absolute value of my number is less than $7$ into an inequality and solved it, by trying a bunch of numbers. These are no different. We are going to use the same sort of thinking process as before: confronted with $|3x-1|>10$ we will say, “OK, the absolute value of something is greater than $10$. What could the something be? <think,think> OK, the something must be greater than $10$ (like $11,12,13$) or else less than $-10$ (like $-11,-12,-13$).” So we write $3x-1>10$ or $3x-1<-10$ and go from there.

Gee, why are you making it so hard? My Algebra I teacher taught me that if it’s greater than, just make it an “or” and if it’s less than, just make it an “and.”

OK, let’s try a slight variation: $|3x-1|>-10$. Now what? “OK, the absolute value of something is greater than $-10$. What could the something be? <think, think> OK, the something can be…anything! The absolute value of anything is greater than $-10$, so any $x-$ value will work!

That’s not what my Algebra I teacher taught me.

Fine, then, let’s try that same problem your way. Then, let’s test our answers, by plugging into the original inequality to see which answer works.

You get the idea? The whole point of this unit (to me) is to—very early in the year—establish a pattern that the only way to solve math problems is by thinking about them. This unit is great for that.

Homework:

“Homework: Absolute Value Inequalities”

Graphing Inequalities and Absolute Values

This is a two-day topic. possibly three.

Start by putting the function $y=x^2-1$ on the board. Now, distinguish between two different kinds of questions.

1. Solve (or graph the solution of) $x^2-1 < 0$. This should remind the students of problems we did in the last unit, where we asked “For what $x-$ values is this function negative?” The answer is $-1 < x < 1$; it could be graphed on a number line.
2. Graph $y. This is a completely different sort of question: it is not asking “For what $x-$values is this true?” It is asking “For what $(x,y)$ pairs is this true?” The answer cannot live on a number line: it must be a shaded region on a two-dimensional graph. Which region? Well, for every point on this curve, the $y-$ value is equal to $x^2-1$. So if you go up from there, then $y$ is greater than...but if you go down from there, then $y$ is less than... So you shade below it.

It is important to be able to solve both types of problems, but it is even more important, I think, to distinguish between them. If you answer the first type of question with a shaded area, or the second type on a number line, then you aren’t just wrong—you’re farther than wrong—you’re not even thinking about what the question is asking. ($2+2=5''$ is wrong, but “$2+2=$George Washington” is worse.)

With that behind you, get them started on the assignment “Graphing Inequalities and Absolute Values.” They should get mostly or entirely finished in class, and they can finish it up and also do the homework that evening.

Homework:

“Homework: Graphing Inequalities and Absolute Values”

If they come in the next day asking about $^\#4$, by the way, just tell them to turn it into $y< -2|x|$ and then it is basically like the other ones.

Second day, no worksheet. After going over the homework, stressing the ways we permute graphs, warn them that a problem they cannot solve by permuting is coming up. Challenge them—first person with the correct shape is the winner, no calculators allowed—and then put on the board $y=x+|x|$. Give them a couple of minutes to plot points. Then let someone who got it right put it on the board—both the points and the resultant shape.

Now, you point out that this shape is really a combination of two different lines: $y=2x$ on the right, and $y=0$ on the left. This odd two-part shape is predictable, without plotting points, if you understand absolute values in a different way. This is our lead-in to the piecewise definition of the absolute value:

$|\text{x}| = \begin{cases} \text{x} & \text{x} \ge 0 \\ -\text{x} & \text{x} < 0\end{cases}$

This takes a whole lot of explaining: it is just one of those things that students find difficult. Here are a few ways to explain it (use all of them).

1. Just try numbers. If $x=3$, then $|x|=3$, so $|x|=x$. Same for $x=4, x=52,$ and even $x=0$. But if $x=-3$, then $|x|=3$, so $|x| \neq x$ (they are not the same)! Instead, $|x|=-x$. Why, because $-x$ in this case is $-(-3)$, which is $3$, which is indeed $|x|$.

But how can $|x|=-x$ when $|x|$ is never negative? Well, that brings us to…

2. Putting a - sign in front of a number does not make it negative: it switches the sign. It makes positive numbers negative and negative numbers positive. So you can read that piecewise definition as “if $x$ is negative, then the absolute value switches the sign.”

3. Finally, come back to the graph of good old $y=|x|$. Point out that it is, indeed, the graph of $y=x$ on the right, and the graph of $y=-x$ on the left.

Now, how does all this relate to our original problem? When $x<0$, we replace $|x|$ with $-x$ so our function becomes $y=x-x=0$. When $x \ge 0$, we replace $|x|$ with $x$ so our function becomes $y=x+x=2x$. That’s why the graph came out the way it did.

Why is this important? It’s an important way to understand what absolute value means. But it’s also our first look at piecewise functions (one of the only looks we will get) so take a brief timeout to talk about why piecewise functions are so important. Throw an object into the air and let it drop, and talk about the function $h(t)$. We previously discussed this function only during the flight. But to get more general, you have to break it into three different functions: $h=3$ before you throw it (assuming it was in your hand $3'$ above the ground); $h=16-t^2$ or something like that during the flight; and $h=0$ after it hits the ground. Do a few more examples to get the idea across that piecewise functions come up all the time because conditions change all the time.

OK, back to our friend the absolute value. The students should now graph $y=\frac{x}{|x|}$ on their own (individually, not in groups), not by plotting points, but by breaking it down into three regions: $x<0, x=0,$ and $x>0$. (It is different in all three.) Get the right graph on the board.

Now, hopefully, you have at least $10-15 \;\mathrm{minutes}$ left of class, because now comes the hardest thing of all. You’re going to graph $|x|+|y|=4$. Since $x$ is under the absolute value, we have to break it into two pieces—the left and the right—just as we have been doing. Since $y$ is under the absolute value, we also have to break it vertically. So what we wind up doing is looking at each quadrant separately. For instance, in the second quadrant, $x<0$ (so we replace $|x|$ with -$x$) and $y>0$ (so we replace $|y|$ with $y$). So we have $y-x=4$, which we then put into $y=mx+b$ format and graph, but only in the second quadrant. You do all four quadrants separately.

Explain this whole process—how to divide it up into the four quadrants and how to rewrite the equation in the second quadrant. Then, set them going in groups to work the problem. Walk around and help. By the end of the class, most of them should have a diamond shape.

After they are all done, you may want to mention to them that this exact problem is worked out in the, “Conceptual Explanations” at the very end of this chapter. So they can see it again, with explanations.

Homework:

Graph $|x|-2|y|<4$. This requires looking at each quadrant as a separate inequality and graphing them all in the appropriate places!

Time for Another Test!

Once again, there is the sample test—you will probably want to assign it as a homework, and tell them to do that and also study everything since the last test. The next day, go over the homework and any questions. Then give the test.

## 2 Equations and 2 Unknowns

(aka Simultaneous Equations)

Like all our topics so far, this unit reviews something the students covered in Algebra I—but it goes deeper.

Introduction to Simultaneous Equations

Begin by having them work their way through the assignment, “Distance, Rate, and Time” in pairs. Most of it should be pretty easy, including getting to the general relation $d=rt$. It should not take much time.

Until the last question, that is. Let them work on this for a while. Some will get all the way, some will not get very far at all. But after they’ve been at it for a while, tell them to stop working and pull them back to a classwide discussion. Show them how to set up $d=rt$ for each case, and make sure they understand. Maybe come up with another problem or two along the same lines, including one where the times are the same and the distances are different. (A train leaves Chicago and a train leaves New York, when do they crash…) Get them comfortable with setting up the two equations—we’re not really focused on solving them.

Toward the end, see if they remember that there are three ways of solving these equations. Two of them, Substitution and Elimination, will be covered tomorrow. Tonight, on homework, we are going to solve by graphing. Your big job is to drive home the point that a graph represents all the points where a particular relationship is true; therefore, the place where the two graphs intersect is the point where both relationships are true. Also talk about the fact that graphing is not $100\%$ accurate (you sort of eyeball a point and say, “It looks like around this”) and how to check your answer (plug it back into both equations).

Homework:

“Homework: Simultaneous Equations by Graphing”

Simultaneous Equations

At the start of the second day, explain the two techniques of substitution and elimination. This is review, so they should get it with just a few examples.

Then have them do the assignment, “Simultaneous Equations.”

At the end of class, they can finish anything they haven't completed for homework, and you can have them do the "homework" the next day in class.

Homework:

“Homework: Simultaneous Equations”

The “Generic” Simultaneous Equations

This should be done in class as another TAPPS exercise. Remind them of the ground rules, and especially of the fact that you will be asking them questions afterwards to make sure they got it.

Time for Another Test!

Once again, there is the sample test. If everyone finishes the TAPPS exercise early, and they all seem pretty comfortable with the material, you may not want to say, “Do the sample test tonight and we’ll go over it tomorrow and then have the test the next day,”— because otherwise, tomorrow may be pretty boring! Instead, it may be OK (depending on the class), to just say, “Now work in class on the sample test, use it to help you study tonight, and we will have a real test tomorrow.”

There are really three separate pieces of this unit: factoring, solving quadratic equations, and graphing quadratic functions. The first piece is vital and important, but small. Nonetheless, you may want to add a small quiz between that section and the next. I have included two sample tests—the first on factoring and solving quadratic equations and the second on graphing.

Multiplying Binomials

Sounds trivial, doesn’t it? But this is one of the most important days in the year.

What they do know, from Algebra I, is how to FOIL. This takes two seconds of review and you’re done. However, there are two points to make which their Algebra I teacher never taught them.

1. When we say $(x+3)(x+4)=x^2+7x+12$, we are asserting the equality of two functions—that is, if I plug any number into $(x+3)(x+4)$, and plug that same number into $x^2+7x+12$, it should come out the same.
2. FOIL leaves you high and dry if you have to multiply $(x+2)(x+y+3)$. The real algorithm for multiplying polynomials is to multiply everything on the left by everything on the right. Walk through an example of this on the board. Show them how FOIL is just a special case of this rule when both things are binomials.

At this point, they can start working in pairs on the exercise, “Multiplying Binomials.” They should have no problem with the first few. As you are walking around, your main job is to make sure that they are doing $^\#5$ correctly. They should not be multiplying these out explicitly (so that $(x+4)(x+4)$ becomes $x^2+4x+4x+16$) and then combining the middle terms. They should instead be using the formula they just developed, $(x+a)^2=x^2+2ax+a^2$, going straight to the right answer. A lot of them will find this very confusing. I always explain it this way: $x$ and $a$ are both placeholders that could represent anything. So when we say:

$(x+a)^2=x^2+2ax+a^2$

What we’re really saying is:

(something+something_else)$^2=$something$^2 + 2$(something)(something_else) $+$ something_else$^2$

Maybe walk them through the first one as an example.

One point of this exercise is to get them to the point where they can see immediately, with no in-between steps, that $(x+4)^2=x^2+8x+16$. Some of them will think that this new, confusing method may be faster, but they can just go right on doing it the “old way” with FOIL. I always explain that, in a few days, we’ll be learning a technique called “completing the square” that involves reversing this formula, which cannot possibly be done with FOIL. Therefore, they need to know the formula.

Another point is to get our three formulae on the table: $(x+a)^2, (x-a)^2$, and $x^2-a^2$. There are very few things I ask the class to memorize during the year, but these three formulae should all be committed to memory.

But the larger point is to give them a new understanding and appreciation for what variables do—to understand that $x$ and $a$ represent anything, so that once you have a formula for $(x+a)^2$ you can use that formula directly to find $(2y+6z)^2$.

Homework:

“Homework: Multiplying Binomials”

Factoring

Begin class by reminding them of what they already know: factoring means turning $x^2+7x+12$ into $(x+3)(x+4)$. Then ask—how would we check that? There are two ways. First, we can multiply it back (using FOIL for instance). Second, we can try a number (since we are making the claim that these two functions are “equal,” remember?). Stress this very heavily: they have to know both ways of checking. Why? Because if you don’t know both ways of checking, then you don’t really understand what factoring is, even if you get the right answer.

OK, on to…how do you do it? There are three steps to factoring.

1. Pull out common terms. This is always the first step!
2. Use the formulae from yesterday. For instance, given $x^2-9$, you recognize it as the difference between two squares. Given $x^2-6x+9$, you recognize it as $(x-3)^2$.
3. When all else fails, plain old-fashioned factoring. Do a few examples on the board, just to refresh their memory. $x^2+7x+12, \ x^2-7x+12, \ x^2+x-12, \ x^2-x-12$ are good starting examples, and give you an opportunity to talk about what effect negative numbers have, both in the middle term (not much) and in the last term (lots). The thing I always stress is to start with the last term—find all the pairs of numbers that multiply to give you the last term, and then see if any of them add to give you the middle term. The real test is when you are faced with something like $x^2+4x+8$; it should not take long to determine that it cannot be factored!

Now they are ready to start on the “Factoring” assignment during class.

Homework:

“Homework: Factah Alla Dese Heah Spressions"

As I mentioned, this might be a good place to break and have a quiz. Or it might not. What do I know? Anyway, on to quadratics…

Get them started on the assignment, “Introduction to Quadratic Equations” with little or no preamble. Then, after a few minutes—after everyone has gotten through $^\#5$ —stop them.

Make sure they all got the right answers to numbers $2$ and $3$ and that they understand them. If $xy=0$ then either $x$, or $y$, must equal zero. There is no other way for it to happen. On the other hand, if $xy=1$, that doesn’t tell you much—either one of them could be anything (except zero).

Now, show how this relates to quadratic equations. They do remember how to solve quadratic equations by factoring. $x^2+x-12=0, (x+4)(x-3)=0, x=-4$ or $x=3$. But that last step is taken as a random leap, “because they told me so.” The thing I want them to realize is that, when they write $(x+4)(x-3)=0$, they are in fact asserting that, “these two numbers multiply to give zero,” so one of them has to be zero. This helps reinforce the idea of the previous lesson, that $x$ and $y$ can mean anything: $(x+4)(x-3)=0$ is in fact a special case of $xy=0$.

The acid test is what do you do with $(x+5)(x+3)=3?$ The ones who don’t get it will turn it into $x+5=3, x+3=3$. And get two wrong answers. Instead you have to multiply it out, then get everything on one side so the other side is $0$, and then factor.

Now they can keep going. Many of them will need help with $^\#6$ —talk them through it if they need help, but make as much of it as possible come from them. This is a very standard sort of “why we need quadratic equations” type of problem.

The last four problems are a sneaky glimpse ahead at completing the square. For $^\#11$, many students will say $x=3$; remind them that it can also be $-3$. For ,$^\#12$, this is yet another good example of the “$x$ can be anything” rule, and should remind them in some ways of the work we did with absolute values: $\mathrm{something}^2 =9$, so $\mathrm{something}= \pm 3$. $^\#13$ is obviously, $^\#12$ rewritten, and $^\#14$ can be turned into $^\#13$ by adding $16$ to both sides.

Homework:

Completing the Square

When you’re going over the homework, talk for a while about the throwing-a-ball-into-the-air scenario. It will come up again, and I really want people to understand it. The particular point I try to make is how the math reflects the reality. You have a function $h(t)$ where if you plug in any $t$ at all, you will get an $h$. You’re using it backward, specifying $h$ and asking for $t$ (as in, “when will the ball hit the ground?”). What kind of answers would you expect? Well, suppose you throw the ball $16 \;\mathrm{ft}$ in the air. If you ask, “When will it be at $20\;\mathrm{ft?}$” you would expect to get no answer at all. If you ask, “When will it be at $5 \;\mathrm{ft?}$” you would expect two answers—one on the way up, and one on the way down. If you ask, “When will it be at $16 \;\mathrm{ft?}$” you would expect exactly one answer. In all three cases, the math gives you exactly what you expect.

On the other hand, suppose you ask “When will it be at $-3 \;\mathrm{ft?}$” (That is, under the ground.) You might expect no answer at all, since the ball never is under the ground. But the math doesn’t know that—it thinks the ball is following the same function forever. So you get two answers. One is after the ball hits the ground. The other is before it left—a negative time! This is where you have to use common sense to find the “real” answer, as distinct from the answer the math gave you.

I spend a good half-period, at least, talking this through. I think it is an incredibly important point about the way we use math to model the world. See http://www.ncsu.edu/felder-public/kenny/papers/physicist.html for an exercise you can use just on this.

Anyway, onward…the assignment “Completing the Square” pretty much speaks for itself. Probably the only preamble you need is to point out that many quadratic equations, which do have solutions, cannot be factored. So we are going to learn another technique, which has the advantage that it can always be used. (Factoring is still easier and faster when it works.)

Now you can just get them started on it, and then wander around and help. Just make sure that before the class is done, everyone gets the technique. You may also want to point out to them that they already did this on yesterday’s assignment.

On $^\#4$ make sure they get two answers, not just one!

Homework:

“Homework: Completing the Square”

The hard ones that you will get questions on the next day are $^\#9$ and $^\#10$. Note that, on $^\#9$, I am not looking for the discriminant and the quadratic formula and stuff; just the obvious fact, based on completing the square, that if $c < 0$ we have no real answers, if $c=0$ we have one, and if $c>0$ we have two. $^\#10$ is worth looking at closely if there are questions, because it leads to the next day.

Begin by reminding them of what we did with simultaneous equations. First, we learned how to solve them (using substitution or elimination). Then we used those exact same techniques to solve the generic version—that is, simultaneous equations where all the numbers were replaced by letters. This, in turn, gave us a formula that could instantly be used to solve any pair of simultaneous equations.

Now we are going to do that same thing with quadratic equations. The “generic” quadratic equation is, of course, $ax^2+bx+c=0$. Now, we have learned two different ways of solving such equations. The “generic” version is hard to solve by factoring (although it is possible); we are going to do it by completing the square.

Make sure they look over my example of completing the square; this might be a good opportunity for a quick TAPPS exercise. There are other examples in the “Conceptual Explanations” so you could do two TAPPS exercises—that way everyone gets a chance to be the teacher.

Then have them work through the sheet. They should derive the quadratic formula and then use it.

By the time they are done, they should have two things. They should have the quadratic formula and a bit of practice using it—so now we have three different techniques for solving quadratic equations. They should also have derived the formula. I always warn them that I will ask for this derivation on the next test: it is not enough to know the formula (although that too is good), you have to be able to derive it.

At the end of class, you may want to talk for just a couple of minutes about the discriminant, in reference to $^\#11$. It should be fairly obvious by that point to most of them.

Homework:

Time for Another Test!

As always, there is the sample test, which may or may not be assigned as a homework. Then there is the test—on multiplying polynomials, on factoring, and mostly on solving quadratic equations.

Once again, there is the sample test—you will probably want to assign it as a homework, and tell them to do that and also study everything since the last test. The next day, go over the homework and any questions. Then give the test. Make it shorter than my sample ☺

OK, we’re done solving quadratic equations—we already have three techniques and that’s enough. But—one thing I say a million times throughout my class—you never really understand a function until you graph it.

So, they can do the exercise “Graphing Quadratic Functions.” It doesn’t require any buildup, they can do it right now. Note that we are not introducing any of the formal machinery of parabolas (focus, directrix, etc)—all that will come much later, in the unit on conics. We are graphing both horizontal and vertical parabolas, the same way as in the first unit on functions—by taking an initial starting point ($y=x^2$ or $x=y^2$) and moving it up and down and left and right and stretching it and turning it upside-down. None of this should require a calculator.

It might be worth mentioning that a horizontal parabola is not a function. But we can still talk about it and graph it.

Homework:

Finish the in-class assignment. That was a long one, wasn’t it?

The beginning of the “Graphing Quadratic Functions II” exercise is review of yesterday. After letting them work on it together, you may want to interrupt and have them do the thing in the middle as a TAPPS exercise. The key things you need to ask them about are how this is the same as, and different from, the way we completed the square before. For instance, we used to add nine to both sides (because, let’s face it, we had two sides). Now we only have one side, so we add $9$ to it, and subtract $9$ from it, at the same time. This gives us what we want (the perfect square) without changing the function.

Homework:

“Homework: Graphing Quadratic Functions II” You will get questions the next day about $^\#8$ (which is really a line) and $^\#11$ (which they just flat can’t graph at this point). These lead nicely into $^\#12$. If it has an $x^2$ but no $y^2$, it’s a vertical parabola. If it has a $y^2$ but no $x^2$, it’s a horizontal parabola.

Solving Problems by Graphing Quadratic Functions

Now, at long last, we see a use for all this graphing we’ve been doing.

In the throwing-a-ball scenario, we have an $h(t)$ that can be used to answer two kinds of questions. “I know the time, but what is the height?” (easy, plug in) and “I know the height, what is the time?” (harder, requires solving a quadratic equation). But there is a third kind of question, very important in the real world, which is: “How high does it go?” Now we don’t know the time or the height! But if we graph it, and find the vertex, we can find both.

Now they can work a while on the in-class assignment. Many of them will get stuck dead on $^\#3$. This is where you have to pull back and lecture a bit more. Help them draw it, and set up the function $A(x)$. But more importantly, talk about what that function means. You plug in any $x$ (length) and you get back an $A$ (area). So, if the graph looks like this $\cap$ what does that tell us? Well, at the peak, that is the highest $A$ ever gets on our graph—that is, the highest the area ever gets. Find the vertex, and you will find the $x$ that maximizes $A!$

This is worth a lot of time to make sure people really get it. It goes back to week $1$, and the idea of graphing a function. On one level, it’s incredibly abstract—we are drawing an upside-down parabola that somehow represents the “possibility spaces” for a bunch of rectangles. But if you understand the idea of graphing a function, it is really very simple. Every point on that parabola pairs an $x$ (length) with an $A$ (area). Every point represents one farm that our farmer could create. It’s obvious, looking at it, that the point at the top represents the one with the highest area.

This is one of those cases where the combined in-class assignment and the homework could easily take two days instead of one. Let it take that, if it does. Make up more problems, if you have to. But don’t let them get away with thinking, “I understand everything else, I just don’t get the word problems.”

Homework:

“Homework: Solving Problems by Graphing Quadratic Functions”

For some reason, this is one of the hardest topics in the course. It shouldn’t be hard. There is nothing hard about it. But students get incredibly tied in knots on this, by trying to take short cuts. The hardest part is convincing them that they have to think about it graphically.

So, begin by simply putting these two problems on the board.

$& x^2-3x-4 > 0\\& x^2-3x+3 > 0$

Allow them to work in pairs or groups. Offer a piece of candy or a bit of extra credit or something equivalent to anyone who can find the answer to both problems. Give them time to really work it. Almost no one will get it right and that’s the point. It’s very hard to think about a problem like this algebraically. It’s very easy if you think about it the right way: by graphing.

So, we’re going to graph both functions. But strangely enough, we’re going to do it without completing the square or finding the vertex. In each case, we’re only going to ask two questions: what are the zeros of the function, and which direction does it open in? We only need these two questions to answer the inequality.

In the first case, by factoring, we find two zeros: $4,$ and $-1$. In the second case, we find with the quadratic formula that there are no zeros. Both graphs open up. (Why? Because the coefficient of the $x^2$ term is positive.) So the graphs look something like this.

What are the vertices, exactly? We don’t know. If we wanted to know that, we would have to complete the square, just as we did before.

But if all we want to know is where each graph is positive, we now have it. The graph of the first function should make it clear that all numbers to the right of $4$ work, as do the numbers to the left of $-1$, but the numbers in between don’t work. (Quick review: how can we write that answer with inequalities? With set notation?) The graph of the second function makes it clear that all numbers work.

Check this by trying numbers in the original inequalities.

Now, challenge them to find a quadratic inequality that is in the form $f(x) > 0$ where $f(x)$ is a quadratic function and the solution is nothing works. The key is, of course, it has to be an upside-down quadratic.

Homework:

Time for Another Test!

Congratulations, kids! We are done with our entire unit on quadratic equations—probably the biggest unit in the course. (Certainly the only think I remember from my own Algebra II course.)

## Exponents

After all the incredibly new stuff we’ve been doing, it’s a nice break to get back to something with a large element of review in it.

But it is also a problem. Many of the kids know already that $2^3 2^5=2^8$. A fair number of them even know that $2^{-7}=\frac{1}{2^7}$. But they don’t know why. It’s vital to keep reminding them that it isn’t enough to know it, they have to know why—and this will indeed be reflected on the test.

Rules of Exponents

Yes, a lot of this assignment was already done, verbatim, in the unit on Functions. But there are a lot of reasons for bringing it back (just as we did in Quadratics). First, they (and you) can discover that they have gotten better at finding generalizations. But more importantly, when we did this in functions, we were only interested in the process of finding generalizations. Now we are focused on creating, memorizing, and using the three rules of exponents.

$2d$ is really pushing them another step toward “the way mathematicians think”—seeing that $2(2^x)=2^{x+1}$ is really just a special case of $2^a 2^b =2^{a+b}$, where $a=1$.

In $^\#3$, I really want to get at the idea that $\frac{3^{12}}{3^{10}} =3^2$, and $\frac{3^{10}}{3^{12}}= \frac{1}{3^2}$ . In other words, the whole thing can and should be done without negative exponents. Why? Because we haven’t yet defined what they mean—and why.

Homework:

“Homework: Rules of Exponents”

Extending the Idea of Exponents

This is one of my favorite class discussions. You’re going to talk almost the entire class and just give out an assignment toward the end.

So far, we have only talked about exponents in the context of positive integers. The base can be anything: for instance, we can find $(-3)^4$ or $(\frac{1}{2})^3$. But when we say that $2^3$ means $2 \cdot 2 \cdot 2$, that definition is really only meaningful if the exponent is a positive integer. We can’t multiply $2$ by itself, “$-3$ times” or “$\frac{1}{2}$ times.” Or “$0$ times” for that matter.

So, let’s plop ourselves down in an imaginary point in history where exponents are only defined for positive integers. We are the king’s mathematicians. The king has just walked in and demanded that we come up with some sort of definition for what $2^{-3}$ means. “Zero and negative numbers have rights too,” he growls. “They must be treated equally, and given equal rights to be in the exponent.”

So we start with a brainstorming exercise. The object is to come up with as many possible things as you can think of, for $2^{-3}$ to mean. As always, this should be done in groups of $2$ or $3$, and remind them that the object is quantity, not quality: let’s get creative. If you can’t think of more than two or three definitions, you’re not trying hard enough.

By the way, somewhere in class there is a smart-aleck who knows the right answer and therefore won’t plan. “It’s $\frac{1}{2^3}$,” he insists proudly. “Why are we doing this?” To which you reply: “The people in the class are coming up with dozens of things it could mean. Can you give them a good argument as to why it should mean that, instead of all the others?” And he weakly answers, “Well, my Algebra I teacher told me…” and you’ve won. The point, you explain, is not to parrot what your teacher told you, but to understand several things. The first is that our old definition of an exponent (“multiply by itself this many times”) just doesn’t apply here, and so we need quite literally a new definition. The second is that there are a ton of definitions we could choose, and it frankly seems arbitrary which one we pick. The third is that there really is a good reason for choosing one and only one definition. But if he doesn’t know what that is, how about if he gets with the program and brainstorms? End of discussion.

OK, so after a few minutes, you start collecting ideas from the groups. $2^{-3}$ means…$2^3$, only negative (so it’s $-8$). It means $2$ divided by itself three times,$2/2/2$ (so it’s either $2$ or $\frac{1}{2}$, depending on how you parenthesize it). And so on. With a bunch of ideas on the board, you say, now we have to choose one. How do we do that? That is, what criteria do we use to decide that one definition is better than the others? (Silence)

The answer is—the definition should be as consistent as possible with the one we already have. Of course it won’t mean the same thing. But it should behave mathematically consistently with the rules we have: for instance, it should still obey our three laws of exponents. That sort of consistency is going to be the guideline that we use to choose a definition for the king.

And hey, what exactly do negative numbers mean anyway? One way to look at them is, they are what happens when you take the positive numbers and keep going down. That is, if you go from $5$ to $4$ to $3$ and just keep going, you eventually get to $0$ and then negative numbers. This alone is a very powerful way of looking at negative numbers. You can use this to see, for instance, why positive-times-negative-equals-negative and why negative-times-negative-equals-positive.

$3 \cdot 5 & = 15 && 3 \cdot -5 = -15 \\2 \cdot 5 & = 10 && 2 \cdot -5 = -10 \\1 \cdot 5 & = 5 && 1 \cdot -5 = -5 \\0 \cdot 5 & = 0 && 0 \cdot -5 = 0 \\-1 \cdot 5 & = -5 && -1 \cdot -5 = 5 \\-2 \cdot 5 & = -10 && -2 \cdot -5 = 10$

OK, you may or may not want to get into this, but I think it’s cool, and it does help pave the way for where we’re going. (It also helps reinforce the idea that even the rules you learned in second grade have reasons.) The numbers I’ve written on the left show what happens to “multiply-by $-5$” as you count down. Clearly, looking at the positive numbers, the answers are going down by $5$ every time. So if that trend continues as we dip into negative numbers, then we will get $-5$ and then $-10$ on the bottom: negative times positive equals negative.

On the right, we see what happens to “multiply-by $-5$” as you count down. Since we already know (just proved) that positive-times-negative-equals-negative, we know that $3 \cdot -5=-15$ and so on. But what is happening to these answers as we count down? They are going up by $5$. So, continuing the trend, we find that $-1\cdot -5=5$ and so on.

As I mentioned, you might want to skip that. The essential point to get across is that we need a new definition that will cover negative exponents, and that we are going to get there by looking for consistency with the positive ones. Then they are ready for the in-class assignment: it shouldn’t take long. Do make sure to give them $10 \;\mathrm{minutes}$ for it, though—you want them to finish it in class, and have time to ask you questions, so you know they are ready for the homework.

Homework:

“Homework: Extending the Idea of Exponents”

Fractional Exponents

Start by reminding them of where we are in the big picture. We started with nothing, but the idea that exponents mean “multiply by itself a bunch of times”—in other words, $7^4$ means $7 \cdot 7 \cdot 7 \cdot 7$. We went from there to the rules of exponents $-x^a x^b =x^{a+b}$ and so on—by common sense. Then we said, OK, our definition only works if the exponent is a positive integer. So we found new definitions for zero and negative exponents, but extending down from the positive ones.

Now, we don’t have a definition for fractional exponents. Just as with negative numbers, there are lots of definitions we could make up, but we want to choose one carefully. And we can’t get there using the same trick we used before (you can’t just count and “keep going” and end up at the fractions). But we still have our rules of exponents. So we’re going to see what sort of definition of fractional exponents allows us to keep our rules of exponents.

From there, you just let them start working. I can summarize everything on the assignment in two lines.

1. The rules of exponents say that ${(x^\frac{1}{2})}^2=x$. So whatever $x^\frac{1}{2}$ is, we know that when we square it, we get $x$. Which means, by definition, that it must be $\sqrt{x}$ . Similarly, $x^\frac{1}{3}= \sqrt[3]{x}$ and so on.
2. The rules of exponents say that ${(x^\frac{1}{3})}^2=x^\frac{2}{3}$. Since we now know that $x^\frac{1}{3}= \sqrt[3]{x}$, that means that $x^\frac{2}{3}= \left (\sqrt[3]{x} \right )^2$. So there you have it.

Thirty seconds, written that way. A whole class period to try to get the students to arrive at that place on their own; once there, many of them will still require a lot of help to see the point. Toward the end, you may just call the class’s attention to the board and write out the answers. But by the time they leave, you want them to have the following rule: for fractional exponents, the denominator is a root and the numerator is an exponent. And they should have some sense, at least, that this rule followed from the rules of exponents.

There is one more thing I really want them to begin to get. If a problem ends up with $\sqrt{25}$ , you shouldn’t leave it like that. You should call it $5$. But if a problem ends up with $\sqrt{2}$ , you should leave it like that: don’t type it into the calculator and round off. This is also worth explicitly mentioning toward the end.

Homework:

“Homework: Fractional Exponents”

Mostly this is practicing what they learned in class. The inverse functions are a good exercise: it forces them to review an old topic, but also forces them to practice the current topic. For instance, to find the inverse function of $y=x^\frac{2}{3}$, you write:

$x=y^\frac{2}{3}=\sqrt[3]{y^2} \cdot x^3=y^2 \cdot y=\sqrt{x^3} =x^{\frac{3}{2}}.$

After you go through that exercise a few times, you start to see the pattern that the inverse function actually inverts the exponent. The extra fun comes when you realize that $x^0$ has no inverse function, just as this rule would predict.

At the end of the homework, they do some graphs—just by plotting points. You will want to make sure they have figured out the shapes correctly, because this paves the way for the next topic. When going over the homework the next day, sketch the shapes quickly and point out that, on the graph of $2^x$, every time you move on to the right, $y$ doubles. On the graph of $(\frac{1}{2})^x$, every time you move one to the right, $y$ drops in half.

“Real Life” Exponential Curves

This is another one of those topics where the in-class exercise and the homework may take a total of two days, combined, instead of just one. This is a difficult and important topic.

We begin with a lecture something like the following:

Earlier this year, we talked about “linear functions”: they add a certain amount every time. For instance, if you gain $\5$ every hour, then the graph of your money vs. time will be a line: every hour, the total will add $5$. The amount you gain each hour ($5$ in this case) is the slope.

Can a line also subtract every day? Sure! The rule isn't different because adding is the same as subtracting a negative number. So if Mr. Felder is losing ten hairs a day, and you graph his hairs vs. time, the graph will be a line going down. The total subtracts $10$ every day, but another way of saying that is, it adds $-10$ every day. The slope is $-10$. This is still a linear function.

So why am I telling you all this? Because “exponential functions” are very similar, except that they multiply by the same thing every time. And, just as linear functions can subtract (by adding negative numbers), exponential functions can divide (by multiplying by fractions: for instance, multiplying by $\frac{1}{3}$ is the same as dividing by $3$). The amount you multiply by is called…well, come to think of it, it doesn’t have a cool name like “slope.” I guess we could call it the “base.”

Then they can begin to work on the assignment. They will make it through the table all right. But when it comes to finding the formula for the $n^{\mathrm{th}}$ day, many will fall down. Here is a way to help them. Go back to the table and say: “On day $3$, let’s not write $4''$—even though it is $4 \;\mathrm{pennies}$. It is $2$ times the previous amount, so let’s just write that: $2 \times 2$. On day $4$, it’s $2$ times that amount, or $2 \times 2 \times 2$. On day $5$, it’s $2$ times that amount, or $2 \times 2 \times 2 \times 2$. This is getting tedious... what’s a shorter way we can write that?” Once they have expressed every answer in powers of $2$, they should be able to see the $2^{n-1}$ generalization. If they get the wrong generalization, step them through to the next paragraph, where they test to see if they got the right answer for day $30$.

You go through the same thing on the compound interest, only harder. A lot of hand-holding. If you end one year with $x$ then the bank gives you $.06x$ so you now have a total of $x+.06x,$ which is, in fact, $1.06x$. So, hey, your money is multiplying by $1.06$ every year! Which means if you started with $\1000$ then the next year you had $\1000 \cdot 1.06$. And the next year, you multiplied that by $1.06$, so then you had $\1000 \times 1.06 \times 1.06$. And the year after that...

Toward the end of class, put that formula, $\1000 \times 1.06^n$, on the board. Explain to them that they can read it this way: “Just looking at it, we can see that it is saying you have $\1000$ multiplied by $1.06, n$ times.” This is always the way to think about exponential functions—you are multiplying by something a bunch of times.

The assignment is also meant to bring out one other point that you want to mention explicitly at the end. When we developed our definitions of negative and fractional exponents, we wanted them to follow the rules of exponents and so on. But now they are coming up in a much more practical context and we have a new need. We want $x^{2\frac{1}{2}}$ to be bigger than $x^2$ and smaller than $x^3$, right? After all, after $2 \frac{1}{2}$ years, you certainly expect to have more money than you had at the beginning of the year! It isn’t obvious at all that our definition, $x^{\frac{5}{2}}=\sqrt{x^5}$ , will have that property: and if it doesn’t, it’s useless in the real world, even if it makes mathematicians happy. Fortunately, it does work out exactly that way.

Homework:

“Homework: ‘Real Life’ Exponential Curve"

Time for Another Test!

The sample test will serve as a good reminder of all the topics we’ve covered here. It will also alert them that knowing why $x^{\frac{1}{2}}$ is defined the way it is really does count. And it will give them a bit more practice (much-needed) with compound interest.

## Logarithms

I talk to a surprising number of math teachers who are really uncomfortable with logs. There’s something about this topic that just makes people squeamish in Algebra, in the same way that “proving a series converges” makes people squeamish in Calculus.

It doesn’t have to be hard. It is not intrinsically more complicated than a radical. When you see $\sqrt[3]{x}$ you are seeing a mathematical question: “What number, raised to the $3^{\mathrm{rd}}$ power, gives me $x?$” When you see $\log_3 x$ you are seeing a question which is quite similar: “$3$, raised to what power, gives me $x?$” I say this about a hundred times a day during this section. My students may forget the rules of logs and they may forget what a common log is and they will almost certainly forget e, but none of them will forget that $\log_2 8$ means the question “$2$ to what power is $8?$” You may want to show them the “Few Quick Examples” at the beginning of the Conceptual Explanations chapter to drive the point home.

It is possible to take any arbitrary logarithm on a standard scientific or graphing calculator. I deliberately never mention this fact to my students, until the entire unit (including the test) is over. Faced with $\log_ 2 8$ I want them to think it through and realize that the answer is $3$ because $2^3=8$. The good news is none of them will figure out how to do that problem on the calculator, if you don’t tell them.

Introduction to Logarithms

This is a pretty short and self-explanatory exercise. There isn’t anything you need to say before it. But you do need to do some talking after the assignment. Introduce the word “log” and explain it, as I explained it above: $\log_2 8$ means “$2$ to what power is $8?$” Also discuss the fact that the log is always the inverse of the exponential function.

After they have done the assignment, and heard your explanation of the word log, then they are ready for the homework. It wouldn’t hurt if that happens in the middle of the class, so they can get started on the homework in class and finish it up at home. The in-class exercise is short, the homework is long.

Homework:

“Homework: Logs”

When going over the homework the next day, $^\#20$ can be explained two ways. First, $5$ to what power is $5^4?$ When asked that way, it’s easy, isn’t it? You don’t have to find what $5^4$ is, to see that the answer is $4!$ But there is also another way to explain it, which gets back to the idea of $5^x$ and $\log_5 x$ being inverse functions. The first function turns $4$ into $5^4$. So the second one has to reverse this process, and turn $5^4$ back into $4$. This way is harder to understand, but it makes it a lot easier to see why $^\#21$ also has to be $4$.

Then, there is the graph—as always, make sure they get the right general shape. Point out that the most salient feature of this graph is that it grows…incredibly…slowly as you go farther out to the right. (Every time $x$ doubles, the graph just goes up by $1$.) This is a lot of what makes logs useful, as we will see.

Properties of Logarithms

This is very standard stuff. Using the in-class exercise in groups of $2$ or $3$, they should be able to find—in some cases with a bit of help from you—some rules of logarithms. In the homework, they practice using those rules.

One thing I don’t do in the text is formally prove the rules. However, I have been known to “throw in” the proofs sometimes in class, either for a group that finishes early, or for the whole class if enough people are interested. One of the proofs is provided as an example in the “Conceptual Explanations” along with guidelines for the other two.

But what I really care about is giving them an intuitive grasp of why the rules work, rather than the proof. The intuitive grasp is what comes from the exercise, from realizing that the logarithm is essentially a counter. Once you see that $\log_2 8$ is asking how many $2s$ there are in $8$, then it’s obvious that $\log_2 (8 \cdot 16)$ will add up all the $2s$ in $8$, and in $16$.

Homework:

“Homework: Properties of Logarithms”

So What Are Logarithms Good For, Anyway?

As always, things get harder when we get into word problems. There are a couple of concepts I want them to take away.

First—logs are used in a wide variety of real world situations.

Second—logs are used because they compress scales. In other words, because they grow so slowly, we use logarithmic scales whenever we want to work with a function that, by itself, grows too quickly. Or, to put it another way, we use logarithms whenever something varies so much that you don’t care about the exact number; you are more interested in the order of magnitude (what the power of $10$ is). Don’t say all this before they start working, but hopefully they will come up with something like this on $^\#6$.

Homework:

“Homework: What Are Logarithms Good For, Anyway?”

In addition to following up on the in-class work, the homework here also introduces the common and natural logs. It’s a bit of a weak connection, but I had to stick them somewhere.

Time for Another Test!

The sample test is actually pretty important here. It pulls together a lot of ideas that have been covered pretty quickly.

It also introduces one new idea in the last problem. We haven’t done this kind of equation before. You may want to spend a couple of minutes talking about it, and then put something like it on your actual test. I don’t think it’s worth a whole class period—just $5-10 \;\mathrm{minutes}$. But it is another good opportunity to remind them of the rules of logs. $\ln(3)+\ln(x)$ is $\ln(3x)$ so we end up with $3x=21$. You may then want them to try solving a few more equations. I’m very fond of:

$\ln(x+3)+\ln(x-3)=\ln(55)$

That’s worth ten minutes all by itself. It becomes $x^2=64$. So the answer to that is $x=8$. But oops, it isn’t: it’s actually $x=\pm 8$. But oops, that isn’t right either—because $-8$ is not a valid solution to our original problem, since you can’t take $\ln$ of a negative number. So the answer is $8$ after all!

The extra credit is just a pun. The answer is log cabin, or better yet, natural log cabin. Who says math can’t be fun?

According to my reckoning, you are now approximately halfway through the curriculum. Mid-terms are approaching. If there are a couple of weeks before mid-terms, I would not recommend going on to radicals—spend a couple of weeks reviewing. Each topic (each test, really) can stand a whole day of review. It may be the most important time in the whole class!

## Rational Expressions

We’ve talked about the word “rational”—it doesn’t mean “sane,” it means a “ratio” or, in other words, a fraction. A rational expression is just a fraction with variables.

This section is unique, perhaps, because it introduces practically no new skills. They have to be able to factor; they have to know the rules of exponents; they have to be able to work with fractions; and they even have to be able to do long division. There is nothing new in any of that. It’s just putting it all together to simplify and work with rational expressions.

Part of the benefit of this unit is that there are always a few kids in class—maybe more than a few—who have a lingering, secret fraction-phobia. They are hoping that no one will ever notice because the calculator will always rescue them. You can spot these people because they always answer everything—including, “What is $2$ divided by $3?$”—in decimals. But this unit will flush them out. You can’t get through rational expressions unless you know how to do fractions and your calculator will not help you. (I always point this out, very explicitly, several times.) In the “Conceptual Explanations” I begin each section by working plain-old-number-fraction problems (simplifying them, multiplying them, adding them, and so on); tell them they can look there if they want a quick review.

Because of the nature of this unit—no new concepts and fraction phobia—it has fewer “creative thinking” types of problems, and more “drill and practice” than any other unit. It gets boring for you, but don’t let them see that. For a few students at least, this has the potential to break down a barrier that they have been struggling with since the third grade.

Rational Expressions

Begin by explaining what rational expressions are and making the points I made above—we are going to put together some of our old skills in a new way, which will require us to be good at fractions. So, we’re going to start by reviewing how to work with fractions.

Then pair the students up. We’re going to do a sort of do-it-yourself TAPPS exercise. One partner is the student, one is the teacher. The teacher’s job is to add $\frac{1}{2}+\frac{1}{3}$—not just to come up with the answer, but to walk through the process, explaining what he is doing and why he is doing it at every step, all on paper. The students should ask for clarification of any unclear points. By the time they are done, they should have a written, step-by-step instruction guide for adding fractions.

Then they switch roles. The former student becomes the new teacher and gives two lessons: how to multiply $\left (\frac{1}{2} \right ) \left (\frac{1}{3} \right )$ and how to divide $\frac{\frac{1}{2}}{\frac{1}{3}}$. Once again, they should wind up with a step-by-step guide.

The original teacher takes over again and shows how to simplify a fraction.

Finally, you give a brief lesson on multiplying fractions—they’ve already done that, but the key point to emphasize here is that you can cancel before you multiply. For instance, if you want to multiply $\frac{7}{8}$ times $\frac{10}{21}$, you could say:

$\frac{7}{8} \times \frac{10}{21}=\frac{70}{168}$

and then try to simplify that. But it’s a lot easier to simplify before you multiply. The $\frac{7}{21}$ becomes $\frac{1}{3}$, the $\frac{10}{8}$ becomes $\frac{5}{4}$, so we have:

$\frac{\overset{1}{\cancel{7}}}{\underset{4}{\cancel{8}}} \cdot \frac{\overset{5}{\cancel{10}}} {\underset{3}{\cancel{21}}}=\frac{5}{12}$

Of course, you can only do this trick—canceling across different fractions—when you are multiplying. Never when you are adding, subtracting, or dividing!

So why are we going through all this? Because, even though they know how to do it with numbers, they are going to get confused when it comes to doing the exact same thing with variables. So whenever they ask a question (“What do I do next?” or “Do I need a common denominator here?” or something similar), you refer them back to their own notes on how to handle fractions. I have had a lot of students come into the test and immediately write on the top of it:

$\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$

They did this so they would have a “template” to follow when adding rational expressions—it’s a very smart move.

Other than basic fraction manipulation, there is only one other big thing to know about rational expressions—always factor first. Factoring shows you what you can cancel (especially when multiplying), and how to find the least common denominator (when adding or subtracting).

So, walk through some sample problems for them on the blackboard. Put up the problem, and ask them what the first step is…and what the second step is…and so on, until you have something like this on the blackboard.

$\frac{1}{x}+\frac{2}{y}=\frac{y}{xy}+\frac{2x}{xy}=\frac{y+2x}{xy}$

Emphasize over and over that these are the same steps you would take to add $\frac{1}{2}+\frac{2}{3}$. However, point out that ultimately just as we did in the first week of class, we have asserted that two functions are equal. That means they should come out exactly the same for any $x$ and $y$. So have everyone choose an $x-$ value and a $y-$ value, and plug them into both $\frac{1}{x}+\frac{2}{y}$ and $\frac{y+2x}{xy}$, and make sure they come out with the same number.

Now walk through something harder on the blackboard, like this:

$\frac{x}{x^2+5x+6} & - \frac{2x}{x^3-9x} && \text{The original problem} \\\frac{x}{(x+3)(x+2)} & - \frac{2x}{x(x+3)(x-3)} && \text{Always factor first!}\\\frac{x}{(x+3)(x+2)} & - \frac{2}{(x+3)(x-3)} && \text{Simplify (cancel the x'' terms on the right)}.\\\frac{x(x-3)}{(x+3)(x+2)(x-3)} & - \frac{2(x+2)}{(x+3)(x-3)(x+2)} && \text{Get a common denominator. This step}\\&& &\text{requires a lot of talking through. You have to}\\ && &\text{explain where the common denominator came}\\&& &\text{from, and how you can always find a common}\\ && & \text{denominator once you have factored.}\\& \frac{(x^2-3x)-(2x+4)}{(x+2)(x+3)(x-3)} && \text{Now that we have a common denominator, we}\\&& &\text{can combine. This step is a very common}\\&& &\text{place to make errors-by forgetting}\\&& &\text{to parenthesize the} \ (2x+4) \ \text{on the right, students}\\&& &\text{wind up adding the} \ 4 \ \text{instead of subtracting it.}\\& \frac{x^2-5x-4}{(x+2)(x+3)(x-3)}&& \text{Done! Of course, we could multiply the}\\&& &\text{bottom through, and many students want to. I}\\&& &\text{don't mind, but I don't recommend it-there}\\&& &\text{are advantages to leaving it factored.}$

Once again, have them try numbers (on their calculators) to confirm that $\frac{x}{x^2+5x+6}-\frac{2x}{x^3-9x}$ gives the same answer as $\frac{x^2-5x-4}{(x+2)(x+3)(x-3)}$ for any $x$. But they should also remember (from week $1$) how to find the domain of a function—and in this case, the two are not quite the same. The function we ended up with excludes $x=-2, x=-3,$ and $x=3$. The original function excludes all of these, but also $x=0$. So in that one case, the two are not identical. For all other cases, they should be.

Whew! OK, you’ve been lecturing all day. If there are $10 \;\mathrm{minutes}$ left, they can begin the exercise. They should work individually (not in groups or pairs), but they can ask each other for help.

Homework:

Finish the in-class exercise and do “Homework—Rational Expressions”

Rational Equations

After you have answered all the questions on the previous homework, they can just get started on this assignment immediately—it should explain itself pretty well.

However, after about $5 \;\mathrm{minutes}$—when everyone has gotten past the first two problems—pull them back and talk to the whole class for a moment, just to make sure they get the point. The point is that if the denominators are the same, then the numerators must be the same ($^\#1$); and if the denominators are not the same, then you make them the same ($^\#2$). It’s pretty straightforward with these two problems, but it may be deceptively easy. The real thing to make sure they “get” is that, having established these two principals with these easy problems, they are now going to apply them in much more complicated ones.

Then they can get back to it, and you just float around and help. In $^\#3$, the only trick is remembering that if $x^2=25$, then $x= \pm 5$ (not just $5$). Numbers $4$ is straightforward. Give them time to struggle with $^\#5$ before pointing out that they should factor-and-simplify first—always factor first! $^\#6$ is really what all this is working towards: to solve rational equations in general, you must be able to solve quadratic equations!

Homework:

“Homework: Rational Expressions and Equations”

Polynomial Division

Today at least half the class perhaps the whole class will consist of a lecture and a demonstration—you have to show them how to do this. The lecture goes something like this.

We’re going to talk about everybody’s favorite topic…long division!

Before we do that, I have to start by pointing out some very important cases where you don’t have to use long division. For instance, suppose you have this:

$\frac{36x^3+8x^2+5x+10}{2x}$

That problem doesn’t require any hard work—you should be able to divide it on sight. (Have them do this.) You should have gotten:

$18x^2+4x+2\frac{1}{2}+\frac{5}{x}$

$\frac{x^3-6x^2+5x}{x^2-5x}$

That one isn’t quite as easy. What do we do first? (factor!) Oh yes, let’s do that! So it is $\frac{x(x-5)(x-1)}{x(x-5)}$ . Oh, look…it’s just $x-1!$ Once again, no long division necessary.

OK, but suppose we had this?

$\frac{6x^3-8x^2+4x-2}{2x-4}$

How can we simplify it? This is where we’re going to have to use…long division.

So, let’s start the same way we started with fractions: by remembering how to do this with numbers. Everyone, at your seats, work out the following problem on paper.

$\frac{4327}{11}$

(Here you pause briefly while they work it out on paper—then you work it out on the blackboard.) OK, you should have gotten 393 with a remainder of $4$. So the actual answer is $393\frac{4}{11}$. How could we check that? That’s right…we would want to make sure that $393\frac{4}{11}$ times $11$ gives us back $4327$. Because that’s what multiplication is—it’s division, backward.

Now, let’s go back to that original problem. OK, kids...I’m going to leave my number long division over here on the blackboard, and I’m going to work this rational expressions long division next to it on the blackboard, so you can see that all the steps are the same.

We’ll start here:

$& \overset{\quad \qquad \ {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}} {{2x-4}\big |{6x^3 - 8x^2 + 4x -2}}$

From there, develop the whole thing on the blackboard, step by step. (I do this exact problem in the “Conceptual Explanations.” I would not suggest you tell them to look it up at this point; instead, I would recommend that you go through it on the blackboard, explaining steps as you go, and continually reinforcing the analogy to what you did with numbers.) In the end you conclude that $\frac{6x^3-8x^2+4x-2}{2x-4}$ is $3x^2+2x+6$ with a remainder of $22$: or, to put it another way, $3x^2+2x+6+\frac{22}{2x+4}$.

So then you ask: OK, how could we test that? Hopefully they will come up with one answer, then you say “Good, how else?” and they come up with the other one. One way is to plug a number—any number—into $\frac{6x^3-8x^2+4x-2}{2x-4}$, and the same number into $3x^2+2x+6+\frac{22}{2x-4}$, and make sure you get the same thing. The other way is to multiply back $3x^2+2x+6+\frac{22}{2x-4}$ by $2x-4$ and make sure you get back to $6x^3-8x^2+4x-2$. Make sure they understand both ways. If you don’t understand the first way, you don’t understand function equality; if you don’t understand the second way, you don’t understand what division is!

Finally, you ask: how could we have made all that a bit easier? The answer, of course, is that we should have divided the top and bottom by $2$ before we did anything else. This brings us back to our cardinal rule of rational expressions: always factor first!

If there is still time left in class, let them get started on the homework.

Homework:

“Dividing Polynomials”

When going over it, see how many people did $^\#1$ the “hard way.” Remind them that if the bottom is only one term, you can just do the whole thing quickly and painlessly!

Optional Exercise

If you have extra time—if some students get way ahead and you want to give them an extra assignment, or if you want to spend more time on this topic—here is a good exercise that brings things together. Suppose you want to solve the equation $6x^3-5x^2-41x-30=0$. You can use the “Solver” on the calculator and it will find one answer: most likely, $x=-1$. (This requires a $5-\;\mathrm{minute}$ introduction to the “Solver.”) How do you find the other answers?

Well, we recall from our study of quadratic equations that if $x=-1$ is an answer, then the original function must be expressible as $(x+1)$(something). How do you find the something? With synthetic, or long, division! (Maybe have them try it both ways.) What you end up with, after you divide, is $6x^2-11x-30$. You can factor that the “old-fashioned” way (which takes a bit of time) and you get $(2x+3)(3x-10)$, which gives you the other two roots.

Time for Another Test!

And we’re done with yet another unit.

On $^\#6$ of the sample test, stress that they will get no credit without showing work. They can check their answer either way—multiplying back or trying a number—but they have to show their work.

The extra credit is a good problem that I like to get in somewhere. I generally give one point for the obvious pairs $(0,0)$ and $(2,2)$, and two points for the equation $xy=x+y$. But what I really want to see is if they remember how to solve that for $y$ and get $y=\frac{x}{x-1}$. Finally, from that, they should be able to see that $x=1$ has no pairing number (which is obvious if you think about it: nothing plus one gives you the same thing times one!).

Well, this is easier, isn’t it? They know what a radical is. But they’re going to go places with them that they have definitely never been before…

This looks long, but a lot of it is very fast. You don’t have to do much setup, except to remind them what a square root is. I would explain it by analogy to the way we explained logs. $\log_2 8$ asks the question, “$2$ to what power is $8?$” Well, $\sqrt{9}$ also asks a question: “What squared is $9?$

But then, there is an important distinction—one that I like to make right away and then repeat several times. The question, “What square is $9?$” actually has two answers. So if we defined square root as the answer to that question, square root would not be a function $-9$ would go in, and both $3$ and $-3$ would come out (the old “rule of consistency” from day $1$). We somewhat arbitrarily designate the $\sqrt{}$ symbol to mean the positive answer, so that it is a function. So if you see $x^2=9$ you should properly answer $x=\pm 3$. But if you see $x=\sqrt{9}$ then you should answer only $x=3$. This is a subtle distinction, but I really want them to get it, and to see that it is nothing inherent in the math—just a definition of the square root designed to make it single-valued. This is why if you see $x^2=2$ you have to answer $x=\pm \sqrt{2}$ to get both answers.

The assignment starts with a couple of word problems, just to set up the idea that radicals really are useful (which is not obvious). After everyone is done with that part, you may want to ask them to make up their own problems that require square roots as answers (they are not allowed to repeat $^\#1$). Get them to realize that we square things all the time, and that’s why we need square roots all the time, whenever we want to get back. (We’ll be returning to this theme a lot.)

Then there are problems with simplifying radicals. For many of them, they have seen this before—they know how to turn $\sqrt{8}$ into $2\sqrt{2}$. But they don’t realize that they are allowed to do that because of the general rule that $\sqrt{ab}=\sqrt{a} \sqrt{b}$. So it’s important for them to get that generalization, but it’s also important for them to see that it allows them to simplify radicals. And it’s equally important for them to see that $\sqrt{a+b}$ is not $\sqrt{a}+\sqrt{b}$.

The final question is a trap, of course—many of them will answer $x^4$. But they should know they can test their answer by squaring back, and oops, $(x^4)^2$ is not $x^{16}$.

There are two ways to look at this problem correctly. One is that $\sqrt{x^{16}}$ asks a question, “What number, squared, is $x^{16}?$" The rules of exponents are enough to answer this with $x^8$. The other way to look at it is to remember that raising something to the $\frac{1}{2}$ power is the same as taking a square root. So $\sqrt{x^{16}}$ is the same as $(x^{16})^\frac{1}{2}$, which again by the rules of exponents is $x^8$. (I prefer the first way.)

Homework:

When going over the homework the next day, make sure to talk about the last few problems (the inverse functions). Make sure they tested them! There are several points that you want to make sure they got.

• $x^2$ has no perfect inverse. $\sqrt{x}$ works only if we confine ourselves to positive numbers. On the other hand, $\sqrt[3]{x}$ is a perfect inverse of $x^3$. I always take a moment here to talk about what $\sqrt[3]{x}$ is, and also to make sure they understand that it is not the same thing as $3\sqrt{x}$, so it’s vitally important to be careful in how you write it.
• $x^3$ and $3^x$ are completely different functions. This is why exponents really have two different inverses, logs and radicals. Our last unit was on one of them, this unit will be on the other.

Yeah, it’s sort of a grab bag—a miscellaneous compilation of word problems, review from yesterday, and so on. You can just get them started working on the assignment after you’re done going over yesterday’s homework.

Toward the end, however, they are going to start running into trouble. This is when you introduce rationalizing the denominator. You may want to bring the whole class together to see who can figure out how to rationalize $\frac{1}{\sqrt{3}+1}$. It’s a great opportunity to review our rules of multiplying binomials: $(a+b)^2=a^2+2ab+b^2$, which is why multiplying by $\sqrt{3}+1$ doesn’t work; $(a-b)^2=a^2-b^2$, which is why multiplying by $\sqrt{3}-1$ does.

But please be very careful here, because this particular topic has a very subtle danger. A lot of teachers communicate the idea that denominators should always be rationalized, “just because”—because I said so, or because somehow $\frac{\sqrt{2}}{2}$ is “simpler” than $\frac{1}{\sqrt{2}}$. This is one of the best ways to convince students that math just doesn’t make sense.

What I’m trying to do with this exercise is demonstrate a real practical benefit of rationalizing the denominator, which is that it helps you add and subtract fractions. It’s difficult or impossible to come up with a common denominator without doing this first!

And of course, we have the, “you never understand a function until you’ve graphed it” question. Talk a bit about the graph after they get it. They should be able to see that the domain and range are both $\ge 0$ and why this must be so. They should see that for large values it grows very slowly (like a log), but without the drastic behavior that the log shows in the $0 \le x \le 1$ range.

Homework:

If you read over the assignment carefully, I think it’s pretty self-explanatory. Encourage the students to read it carefully as they go (not just skip to the equations).

In my experience, most the trouble with this section comes from trying to make easy problems, hard (that is, squaring both sides when you don’t need to); or from trying to make hard problems, easy (neglecting to square both sides when you should). Make sure they are clear on the distinction—if there is a variable under the radical, you will need to square; otherwise, you won’t.

Also, it can’t hurt to say this about a hundred times: whenever you square both sides, you have to check your answers—they may not work even if you did all your math right! Make sure they understand when this rule applies, and also why squaring both sides can introduce false answers. (I work through that explanation pretty carefully in the “Conceptual Explanations.”)

Homework:

Time for Another Test!

Not much to say here, except that the real point of the extra credit is to see if they realize that the behavior will be very similar on the right, but it will extend down to the left as well.

## Imaginary Numbers

This is an interesting unit in several ways both good and bad.

The good news is it’s fun. It’s like a game, and I always try to present it that way.

The bad news is it’s incredibly abstract. It’s abstract because it’s hard to understand these numbers-that-aren’t-numbers, and it’s also abstract because to save my life, I can’t come up with any good explanation of why imaginary numbers are useful. Of course, they are useful—invaluable even—but how can I explain that to an Algebra II student? Here are a few things I always say (several times).

1. These numbers are indeed useful, and they are used in the real world, even if I can’t do a great job of explaining why to you right now.
2. Nothing in the real world is imaginary. That is, you will never have $i$ tomatoes, or measure a brick that is $5i \;\mathrm{feet}$ long, or wait for $3i+2 \;\mathrm{seconds}$. So why are these useful? Because there are very often problems where the problem is real, and the answer is real, but in between, as you get from the problem to the answer, you have to use imaginary numbers. Repeat this several times. You have a problem, or real-world situation, which (of course) involves all real numbers. You do a bunch of math, which includes imaginary numbers. In the end, you wind up with the answer, which (of course) involves all real numbers again. But it would have been difficult or impossible to find that answer, if you didn’t have imaginary numbers.
3. One example is electrical engineering. In an electric circuit you have resistors, capacitors, and inductors. They all act very differently in the circuit. When you model the circuit mathematically to determine how it will behave (what current will flow through it), the inductor has an inductance and the capacitor has a capacitance and the resistor has a resistance; they are all very different, which makes the math really hairy. However, you can define a complex quantity called impedance, which makes resistors, capacitors, and inductors all look mathematically the same. The disadvantage is that you are now working with a complex number instead of all real numbers. The advantage is that resistors, capacitors, and inductors now look the same in the equations, which makes life a whole lot simpler. So this is a good example of how you use complex numbers to make the math easier. (As a side note, electrical engineers call the imaginary number $j$ whereas everyone else on the planet calls it $i$. I think kids like that bit of trivia.)
4. Imaginary numbers are also used in many other applications such as quantum mechanics.

It’s all very hand-wavy, and I admit that up front, and I don’t hold the kids responsible for it. But I want them to know that this is really useful, and at the same time, I want to explain why we aren’t going to have any “real world” problems in this unit.

This lecture, by the way, usually comes toward the end of day $1$ or during day $2-$ not at the very beginning of day $1$. In the beginning, I prefer to treat it as a game—“What if there were a square root of $-1?$ Just suppose, what if there were?” With one class I went so far in treating it as a game that it was day $3$ before they realized I wasn’t making the whole thing up.

Oh yeah, one more thing. The calculators will do imaginary numbers for them. I never tell them this. If they figure it out, more power to them. But I literally don’t tell them until after the test that the calculator can be used to solve imaginary numbers! I want them to be able to do these things on their own.

Introduction to Imaginary Numbers

This is a fun day or possibly two days. The first exercise is something that, in theory, they could walk all the way through on their own. But it sets up all the major themes in imaginary numbers.

In practice, of course, some groups will have problems and will need help at various points. But beyond that, almost no groups will see the point of what they have done even if they get it right. So a number of times in class, you are going to interrupt them and pull them back together into a classwide discussion, and talk about what they have just done. The ideal time to do this is after everyone in the class has reached a certain point—for instance, after they have all done $^\#2$ (or struggled with it in vain), you pull them back and talk about $^\#2$. All my suggested interruptions are described below.

Before you start, remind them that the equation $x^2=-1$ has no answer and talk about why. Then explain that we are going to pretend it has an answer. The answer is, of course, an “imaginary” number, so we will call it $i$. The definition of $i$ is therefore, $i=\sqrt{-1}$ or, equivalently, $i^2=-1$.

There are two ways to play this. One is to go into the whole “why $i$ is useful” spiel that I spelled out above. The other approach, which is the one I take is to treat it as a science fiction exercise. I always start by telling the class that in good science fiction, you start with some premise: “What if time travel were possible?” or “What if there were a man who could fly?” or something like that. Then you have to follow that premise rigorously, exploring all the ramifications of that one false assumption. So that is what we are going to do with our imaginary number. We are going to start with one false premise: “What if you could square something and get $-1?$” And we are going to follow that premise logically using all the rules of math and see where it would lead us.

Then they get started. And in $^\#2$, they get stopped in their tracks. So you give them a minute to struggle, and then walk it through on the board like this.

• $-i$ means $-1 \cdot i$ (* Stress that this is not anything unusual about $i$, it is a characteristic of $-1$. We could just as easily say $-2$ means $-1 \cdot 2$, and so on. So we are treating $i$ just like any other number.)
• So $i(-i)$ is $i \cdot -1 \cdot i$.
• But we can rearrange that as $i \cdot i \cdot -1$. (You can always rearrange multiplication any way you want.)
• But $i \cdot i$ is $-1$, by definition. So we have $-1 \cdot -1$, so the final answer is $1$.

The reason to walk through this is to get across the idea of what I meant about a science fiction exercise. Everything we just did was simply following the rules of math—except the last step, where we multiplied $i \cdot i$ and got $-1$. So it illustrates the basic way we are going to work: assume that all the rules of math work just like they always did and that $i^2=-1$.

The next few are similar. Many of them will successfully get $\sqrt{-25}$ on their own. But you will have to point out what it means. So, after you are confident that they have all gotten past that problem (or gotten stuck on it), call the class back and talk a bit. Point out that we started out by just defining a square root of $-1$. But in doing so, we have actually found a way to take the square root of any negative number! There are two ways to see this answer. One is (since we just came off our unit on radicals) to write $\sqrt{-25} =\sqrt{25 \cdot -1} = \sqrt{-25} \sqrt{-1}=5i$. The other—which I prefer—is to say, $\sqrt{-25}$ is asking the question, “What number squared is $-25?$ The answer is $5i$. How do you know? Try it! Square $5i$ and see what you get!”

Now remind them of the subtle definition of square root as the positive answer. If you see the problem $x^2=-25$ you should answer $x= \pm 5i$ (take a moment to make sure they all got the right answer to $^\#4$, so they see why $(-5i)^2$ gives $-25$). On the other hand,$\sqrt{-25}$ is just $5i$.

Next we move on to the cycle of powers. Again, they should be able to do this largely on their own. If a group needs a hint, remind them that if we made a similar table with powers of $2$ ($2^1, 2^2, 2^3$, and so on), we would get from each term to the next one by multiplying by $2$. So they should be able to figure out that in this case, you get from each term to the next by multiplying by $i$, and they should be able to do the multiplication. They will see for themselves that there is a cycle of fours. So then you can ask the whole class what $i^{40}$ must be, and then $i^{41}$ and so on, and get them to see the general algorithm of looking for the nearest power of $4$. (I have tried mentioning that we are actually doing modulo $4$ arithmetic and I have stopped doing this—it just confuses things. I do, however, generally mention that the powers of $-1$ go in a cycle of $2$, alternating between $1$ and $-1$, so this is just kind of like that.)

$^\#13$ is my favorite “gotcha” just to see who falls into the trap and says it’s $9-16$.

After they do $^\#18$, remind them that this is very analogous to the way we got square roots out of the denominator. And this is not a coincidence—$i$ is a square root, after all, that we are getting out of the denominator! You may want to introduce the term, “complex conjugate” even at this stage, but the real discussion of complex numbers will come later.

Homework:

“Homework: Imaginary Numbers”

When going over this homework the next day, make sure they got the point. Our “pattern of fours” can be walked backward as well as forward. It correctly predicts that $i^0=1$, which it should anyway, of course,since$\mathrm{anything}^0=1$. It correctly predicts that $i^{-1} >= -i$, which is less obvious—but remind them that, just yesterday, they showed in class that $\frac{1}{i}$ simplifies to $-i!$

Complex Numbers

The first thing you need to do is define a complex number. A complex number is a combination of real and imaginary numbers. It is written in the form $a+\mathrm{bi}$, where $a$ and $b$ are both real numbers. Hence, there is a “real part” $(a)$ and an “imaginary part” $(\mathrm{bi})$. For instance, in $3+4i$, the real part is $3$ and the imaginary part is $4i$.

At this point, I like to try to put this in context by talking about all the different kinds of numbers we have seen. We started with counting numbers: $1, 2, 3, 4,$ and so on. If you are counting pebbles, these are the only numbers you will ever need.

Then you add zero and negative numbers. Are negative numbers real things? Can they be the answers to real questions? Well, sure…depending on the question. If the question is, “How many pebbles do you have?” or “How many feet long is this stick?” then the answer can never be $-2$: negative numbers are just not valid in these situations. But if the answer is, “What is the temperature outside?” or “How much money is this company worth?” then the answer can be negative. This may seem like an obvious point, but I’m building up to something, so make sure it’s clear—we have invented new numbers for certain situations, which are completely meaningless in other situations. I also stress that we have gone from the counting numbers to a more general set, the integers, which includes the counting numbers plus other stuff.

Then we add fractions and the same thing applies. If the question is, “How many pebbles do you have?” or “How many live cows are on this farm?” the answer can never be a fraction. But if the question is, “How many feet long is this stick?” a fraction may be the answer. So again, we have a new set—the rational numbers—and our old set (integers) is a subset of it. And once again, these new numbers are meaningful for some real life questions and not for others. I always mention that “rational numbers” (“rational” not meaning “sane,” but meaning rather a “ratio”) are always expressable as the ratio of two integers, such as $\frac{1}{2}$ or $\frac{-22}{7}$. So since we have already defined the integers, we can use them to help define our larger set, the rational numbers.

But some numbers are not rational—they cannot be expressed as the ratio of two integers. These are the irrational numbers. Examples are $\pi, e,$ and $\sqrt{2}$ (or the square root of any other number that is not a perfect square).

If we add those to our collection—put the rational and irrationals together—we now have all the real numbers. You can draw a number line, going infinitely off in both directions, and that is a visual representation of the real numbers.

And now, finally, we have expanded our set even further, to the complex numbers, $a+\mathrm{bi}$. Just as we piggybacked the definition of rational numbers on top of our definition of integer, we are piggybacking our definition of complex numbers on top of our definition of real numbers. The complex numbers are, as far as I know, the broadest set—it doesn’t get any more general than this.

All that may sound unnecessary, and of course, it is. But some students really get into it. I have had students draw the whole thing into a big Venn diagram—which I did not ask them to do. (*It is a good extra credit assignment, though.) My own diagram is at the very end of this unit in the “Conceptual Explanations,” under the heading “The World of Numbers.” Many students like seeing all of math put into one big structure. And it helps make the point that complex numbers—just like each other generalization—are valid answers to some questions, but not to others. In other words, as I said before, you will never measure a brick that is $5i \;\mathrm{inches}$ long. (It never hurts to keep saying this.)

The complex numbers are completely general—any number in the world can be expressed as $a+\mathrm{bi}$. This is not obvious! There are plenty of things you can write that don’t look like $a+\mathrm{bi}$. One example is $\frac{1}{i}$, which does not look like $a+\mathrm{bi}$, but can be put into that form, as we have already seen. Other examples are $2^i$ and $\ln(i)$, which we are not going to mess with, but they are worth pointing out as other examples of numbers that don’t look like $a+\mathrm{bi}$, but take-my-word-for-it you can make them if you want to. And then there is $\sqrt{i}$ , which we are going to tackle tomorrow.

That is probably all the setup you need. They can do the in-class exercise on Complex Numbers and see for themselves that whether you add, subtract, multiply, or divide them, you get back to a complex number. Also make sure they get the point about what it means for two complex numbers to be equal: this will be very important as we move on. One other thing I like to mention at some point (doesn’t have to be now) is that there are no inequalities with imaginary numbers. You cannot meaningfully say that $1>i$ or that $1. Because they cannot be graphed on a number line, they don’t really have “sizes”—they can be equal or not, but they cannot be greater than or less than each other.

Homework:

“Homework: Complex Numbers”

When going over this homework, make a special point of talking about $^\#12$. This helps reinforce the most important point of the year, about generalizations. Once you have found what happens to $(a+\mathrm{bi})$ when you multiply it by its complex conjugate you have a general formula, which can be used to multiply any complex number by its complex conjugate without actually going through the work. Show them how $^\#6, 8,$ and $10$ can all be solved using this formula. Also, remind them that $a$ and $b$ are by definition real—so the answer $a^2+b^2$ is also real. That is, whenever you multiply a number by its complex conjugate you get a real answer. This is why there is no possible answer to $^\#14$.

Me, Myself, and the Square Root of $i$

This is arguably the most advanced and difficult thing we do all year. But I like it because it contains absolutely nothing they haven’t already done. It’s not here because it’s terribly important to know $\sqrt{i}$ , or even because it’s terribly important to know that all numbers can be written in $a+\mathrm{bi}$ format. It is here because it reinforces certain skills—squaring out a binomial (always a good thing to practice), working with variables and numbers together, setting two complex numbers equal by setting the real part on the left equal to the real part on the right and ditto for the imaginary parts, and solving simultaneous equations.

Explain the problem we’re going to solve, hand it out, and let them go. Hopefully, by the end of class, they have all reached the point where they know that $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$ and $-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i$ are the two answers and have tested them.

Note that right after this in the workbook comes a more advanced version of the same thing, where they find $\sqrt[3]{-1}$ (all three answers: $-1,\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{1}{2}-\frac{\sqrt{3}}{2}i$ ). I tried using this for the whole class, but it was just a bit too difficult. However, it may be appropriate for some very advanced students—either as an alternative to the $\sqrt{i}$ exercise or as an extra credit follow-up to it.

Homework:

They should finish the worksheet if they haven’t done so, including $^\#7$. Then they should also do the “Homework: Quadratic Equations and Complex Numbers.”

It’s a good opportunity to review quadratic equations and to bring in something new! (It’s also a pretty short homework.)

When going over the homework, make sure they did $^\#3$ by completing the square—again, it’s just a good review, and they can see how the complex answers emerge either way you do it. $^\#4$ is back to the discriminant, of course: if $b^2-4ac < 0$ then you will have two complex roots. The answer to $^\#6$ is no. The only way to have only one root is if that root is $0$. (OK, $0$ is technically complex…but that’s obviously not what the question meant, right?)

The fun is seeing if anyone got $^\#5$. The answer, of course, is that the two roots are complex conjugates of each other—real part the same, imaginary part different sign. This is obvious if you rewrite the quadratic formula like this:

$x = \frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$

and realize that the part on the left is always real, and the part on the right is where you get your $i$ from.

Time for Another Test!

Not much to say here, except that you may want to reuse this extra credit on your own test—if they learn it from the sample (by asking you) and then get it right on your test, they learned something valuable.

## Matrices

The EOC spends an inordinate amount of time on problems like this:

The Kind of Problem I Don’t Bother Too Much With

This matrix shows McDonald’s sales for a three-day period.

Big Macs Fries Coke
Monday $\1,000$ $\500$ $\2,000$
Tuesday $\1,500$ $\700$ $\2,700$
Wednesday $\800$ $\800$ $\1,500$

What were their total sales on Monday? What were their total sales of Big Macs? On which day did they make the most profit? etc etc…

I guess the object is to make it appear that “matrices are useful,” but it is really deceptive. Of course, matrices are useful, but not because they give you a convenient way to organize tabular data and then add columns or look things up.

So I don’t spend much time on this kind of thing. I start with what a matrix is (which is sort of like that). I develop the rules for adding matrices, subtracting them, multiplying a matrix by a constant, and setting two matrices equal to each other—all of which are very obvious and should not be presented as a mystery, but rather just as something obvious.

Then comes the big two days of magic in which we learn to multiply matrices. I use a “gradebook” application, which gives an example of why you would want to do this strange operation—it makes a lot of sense up to the point where you are multiplying an arbitrary-dimensions matrix by a column matrix; although, it gets a bit strained when you expand the second matrix. No matter. They need to get the mechanics of how you multiply matrices and just practice them.

After that bit of magic, the rest should follow logically. The definition of $[I]$, the definition of an inverse matrix and how you find one, and (the final hoorah) the way you use matrices to solve linear equations, should all be logical and consistent, based on the one magic trick, which is multiplying them.

There is one other thing I need to address, which is calculators. There is a day that I set aside to teach them explicitly how to do matrices on the calculator. But that day is after we have learned to multiply matrices, and perhaps even had a quiz on it. Even after that, I stress doing things by hand, and give them problems that will force them to do so (by using variables).But I do love showing them that you can solve five equations with five unknowns quickly and easily by using matrices and a calculator!

Introduction to Matrices

Tell them to get into groups and work on “Introduction to Matrices.” I think it is very self-explanatory. You may want to make the analogy at some point that setting two matrices equal to each other is kind of like setting two complex numbers equal to each other: for “this” to equal “that,” all their respective parts must be equal.

Homework:

“Homework: Introduction to Matrices”

Multiplying Matrices I (Row × Column)

Once again, you may just want to get them started on the assignment and let it speak for itself.

However, toward the end of class—after they have all struggled through or gotten stuck—pull them back and do some blackboard talk. Start by pointing out that we are doing two different things here. Multiplying a matrix times a constant is both easy and intuitive. If you can add two matrices, then you can add $[A]+[A]$ and thus figure out what $2[A]$ has to be.

On the other hand, multiplying two matrices (a row times a column) is not easy or intuitive. Show how to multiply a row matrix by a column matrix; do a few examples, and talk about how it applies to the gradebook example. In other words, make sure they get it.

What I do, about a hundred times, is try to get them to visualize the row floating up in the air and twisting around so that it lines up with the column. I use my hands, I use sticks, anything to get them to visually see this row floating up and twisting to line up with that column. This visualization is not essential today, but if they get it today, it will really help them tomorrow, when we start multiplying full matrices! As I mentioned earlier—this is a radical departure from my normal philosophy—I am more concerned that they get the mechanics here (how to do the multiplication) than any sort of logic behind it. They should be able to see, for instance, that if the row and column do not have the same number of elements then the matrix multiplication is illegal.

Oh yeah, one more thing—I always stress that when you multiply two matrices, the product is a matrix. In the case of a row times a column, it is a $1 \times 1$ matrix, but it is still a matrix, not a number.

Homework:

“Homework: Multiplying Matrices I”

Multiplying Matrices II (The Full Monty)

This time, you’re going to have to lecture. You are going to have to explain, on the board, how to multiply matrices. Probably a good $20 \;\mathrm{minutes}$ (half the class) dedicated to showing them that this row goes over here to this column, and then we go down to the next row and so on. Get them to work problems at their desks, make sure they are cool with it. You can also refer them to the “Conceptual Explanations” to see a problem worked out in a whole lot of detail.

Two things to stress:

1. Keep doing the visualization of a row (in the first matrix) floating up and twisting to get next to a column (in the second matrix). If the two do not line up—that is, they have different numbers of elements—then the multiplication is illegal.
2. Matrix multiplication does not commute. If you switch the order, you may turn a legal multiplication into an illegal one. Or, you may still have a legal multiplication, but with a different answer. $AB$ and $BA$ are completely different things with matrices.

You may never get to the in-class assignment at all. If you don’t, that’s OK, just skip it! However, note that the in-class assignment is built on one particular application, which is showing how Professor Snape can do just one matrix multiplication to get the final grades for all his students. This exercise is one of the few applications I have for matrix multiplication.

Homework:

“Homework: Multiplying Matrices II”

$^\#4$ is important for a couple of reasons. First, of course, by using variables, it forces them to do the work manually even if they have figured out how to do it on a calculator. More importantly, it continues to hammer home that message about what variables are—you can solve this leaving $x, y,$ and $z$ generic, and then you can plug in numbers for them if you want.

$^\#5$ and $^\#7$ set up the identity matrix, and $^\#6$ sets up using matrices to solve linear equations. You don’t need to mention any of that now, but you may want to refer back to them later. I don’t want them to think of $[I]$ as being defined as “a diagonal row of 1s.” I want them to know that it is defined by the property $AI=IA=A$, and to see how that definition leads to the diagonal row of 1s. $^\#7$ is the key to that.

This might be a good time to insert a small quiz—maybe on everything so far, or maybe just on multiplying matrices. The rest of the unit depends heavily on that skill!

Identity and Inverse Matrices

This may, in fact, be two days masquerading as one—it depends on the class. They can work through the sheet on their own, but as you are circulating and helping, make sure they are really reading it and getting the point! As I said earlier, they need to know that $[I]$ is defined by the property $AI=IA=A$, and to see how that definition leads to the diagonal row of $1\;\mathrm{s}$. They need to know that $A^{-1}$ is defined by the property $AA^{-1}=A^{-1}A=I$, and to see how they can find the inverse of a matrix directly from this definition. That may all be too much for one day.

I also always mention that only a square matrix can have an $[I]$. The reason is that the definition requires I to work commutatively: $AI$ and $IA$ both have to give $A$. You can play around very quickly to find that a $2 \times 3$ matrix cannot possibly have an $[I]$ with this requirement. And of course, a non-square matrix has no inverse, since it has no $[I]$ and the inverse is defined in terms of $[I]!$

Homework:

“Homework: The Identity and Inverse Matrices”

Inverse of the Generic 2x2 Matrix

This is one of those things that should be easy, but it isn’t. It should be easy because they have already been doing it with numbers, and it’s just the same with letters. But hey, that’s what Algebra II is about, right?

They should definitely work in groups here. Make sure they understand what they are doing. A clear sign that they don’t understand what they are doing, even a little, is that they wind up solving for $a$, or solving for $w$ in terms of $x$, or something like that. They need to understand that the object is to solve for $w, x, y,$ and $z$ in terms of $a, b, c,$ and $d$. Only by doing this can they come up with a generic solution to the inverse of a $2 \times 2$ matrix, which can then be used quickly and easily to find the inverse of any $2 \times 2$ matrix. If they don’t understand that, they just don’t have any idea what we’re doing—it’s important to get them to understand the problem instead of just focusing on solving it.

The answer, by the way, is $A^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d && -b \\ -c && a\end{bmatrix}.$ Once they have it in this form help them understand how to use it—the numbers in this diagonal switch places, the numbers in that diagonal change signs. It’s not hard to memorize that way.

Use Matrices for Transformation

This is a fun day. No new math, just a cool application of the math we’ve seen.

The in-class assignment pretty well speaks for itself. It’s worth mentioning that despite the very simplified and silly nature of this specific assignment, the underlying message—that matrices are used to transform images in computer graphics—is absolutely true.

Some students (good students) may question why the last column $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ is necessary in specifying Harpoona’s initial condition. The reason is this representation is not simply a list of all the corner points in a shape. Yes, each column represents a point. But the matrix is a set of instructions to the computer, to draw lines from this point to that point. Without the last column, Harpoona would be missing her hypotenuse.

Homework:“Homework: Using Matrices for Transformation”. Here we see a matrix that rotates any object by $30^o$,counter-clockwise. The inverse matrix, of course, rotates clockwise. Hopefully, they will discover all that for themselves. As well as, realize why an inverse matrix always does the exact opposite of the original matrix, since if you do them one after the other, you end up back where you started.

Matrices on the Calculator

This starts with a lecture. You have to show them how to do matrices on the calculator. They should be able to…

• Enter several matrices at once
• Subtract
• Multiply
• Find inverses

All of this is explained step-by-step in the “Conceptual Explanations.”

There are two things I stress. First, whenever I enter a matrix, I always check it. For instance, after I enter matrix $[A]$, I go back to the home screen and go $[A][Enter]$, and the calculator displays matrix $[A]$ to me, so I can make sure I typed it right. One small mistype will ruin a whole problem and it’s really easy to do!

Second, the calculator is very smart about interpreting equations. After you enter three matrices, you can just type $[A][B]^{-1}-[C]$ and it will multiply $[A]$ by the inverse of $[B]$ and then subtract $[C]$.

Homework:

“Homework: Calculators” Depending on how things go, they may be able to finish this in class and have no homework.

Solving Linear Equations

Just let them start this assignment, it should explain itself.

This is the coolest thing in our whole unit on matrices. It is also the most dangerous.

The cool thing is you can solve real-world problems very quickly, thanks to matrices—so matrices (and matrix multiplication and the inverse matrix and so on) prove their worth.

If you are given:

$&2x - 5y + z = 1 \\&6x - y + 2 = 4 \\&4x - 10y + 2z = 2$

you plug into your calculator $[A]=\begin{bmatrix} 2 & -5 & 1 \\ 6 & -1 & 2 \\ 4 & -10 & 2 \end{bmatrix},$ $[B]=\begin{bmatrix} 1 \\ 4 \\ 2\end{bmatrix}$, ask for $A^{-1}B$, and the answer pops out! They should be able to do this process quickly and mechanically.

But the quick, mechanical nature of the process is also its great danger. I want them to see the logic of it. I want them to see exactly why the equation $\begin{bmatrix} 2 & -5 & 1 \\ 6 & -1 & 2 \\ 4 & -10 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} \begin{bmatrix} 1 \\ 4 \\ 2\end{bmatrix}$ is exactly like those three separate equations up there. I want them to be able to solve like this:

$AX &=B\\A^{-1}AX & =A^{-1}B\\IX &=A^{-1}B\\X &=A^{-1}B$

to see why it comes out as it does. If they see that this is all perfectly logical, and they know how to do it, the day is a big success—and in fact, this sort of justifies the whole unit on matrices.

As a final point, mention what happens if the equations were unsolvable: matrix A will have no inverse, so the equation won’t work. (You get an error on the calculator.)

Homework:

“Homework: Solving Linear Equations”

Time for Another Test!

And, time to conclude our unit on matrices.

## Modeling Data with Functions

This unit is really two different topics joined together by a somewhat weak thread. The first topic is direct and inverse variation. The second topic is regression on a calculator.

The weak thread that connects them is that they both involve starting with data and finding a function that models that data. (At least, the second topic is clearly about that, and the first topic is about that if you approach it the way I do…)

Direct and Inverse Variation

This is one of those days where you want to get them working right away (on the “Direct Variation” assignment), let them finish the assignment, and then do $10-15 \;\mathrm{minutes}$ of talking afterwards. You want to make sure they got the point of what they did.

Direct variation is, of course, just another kind of function—an independent variable, and a dependent variable, and consistency, and so on. But the aspect of direct variation that I always stress is that when the independent number doubles, the dependent number doubles. If one triples, the other triples. If one is cut in half, the other is cut in half…and so on. They are in a word, proportional.

This is not the same thing as saying, “When one goes up, the other goes up.” Of course that is true whenever you have direct variation. But that statement is also true of $\ln(x)$, $\sqrt{x}, x^2, 2^x, x+3,$ and a lot of other functions: they do “when one goes up, the other goes up,” but not “when one doubles, the other doubles,” so they are not direct variation. The only function that has that property is $f(x)=kx$, where $k$ is any constant. (Point out that $k$ could be $\frac{1}{2}$, or it could be $-2$, or any other constant—not just a positive integer.)

In $^\#3$ they arrived at this point. They should see that the equation $y=kx$ has the property we want because if you replace $x$ with $2x$ then $y=k(2x)$, which is the same thing as $2kx$, which is twice what it used to be. So if $x$ doubles, $y$ doubles.

They should also see that direct variation always graphs as a line. And not just any line, but a line through the origin.

But the thing I most want them to see is that there are many, many situations where things vary in this way. In other words, $^\#4$ is the most important problem on the assignment. You may want to ask them to tell the whole class what they came up with, and then you throw in a few more just to make the point of how common this is. The amount of time you spend waiting in line varies directly with the number of people in front of you; the amount you pay varies directly with the number of meals you order; the weight of your french fries measured in grams varies directly with the weight of your french fries measured in pounds; and both of these, in turn, vary directly with the number of fries; and so on, and so on, and so on.

Homework:

“Homework: Inverse Variation”

That’s right, no homework on direct variation—it’s time to develop the second one. They should be able to do it pretty well on their own, in analogy to what happened in class. But you will spend a fair amount of the next day debriefing them on inverse variation just as you did on direct. The defining property is that when the independent variable doubles, the dependent variable chops in half. Again, it is true to say, “When one goes up, the other goes down”—but it is not enough. $\frac{1}{x^2}$ has that property, and so does $10-x$, and neither of those is inverse variation.

Examples are a bit harder to think of off the top of your head. But there is an easy and systematic way to find them. I always warn the students that I will ask for an example of inverse variation on the test, and then (now that I have their attention) I explain to them how to do it. Inverse variation is $y=\frac{k}{x}$ (graphs as a hyperbola). This equation can be rewritten as $xy=k$. This is useful for two reasons. First, it gives you the ability to spot inverse variation—if the product is always roughly the same, it’s inverse. Second, it gives you the ability to generate inverse variation problems, by determining whenever two things multiply to give a third thing, and then holding that third thing constant.

For instance: the number of test questions I have to grade is the number of students, times the number of questions on the test. That’s obvious, right?

If I want to turn that into a direct variation problem, I hold one of the two multiplying variables constant. What do I mean “hold it constant?” I mean, pick a number. For instance, suppose there are twenty students in my class. Now the dependent variable (number of questions I have to grade) varies directly with the independent variable (number of questions I put on the test).

On the other hand, if I want to turn that exact same scenario into an inverse variation problem, I hold the big variable constant. For instance, suppose I know that I am only capable of grading $200$ problems in a night. So I have to decide how many questions to put on the test, based on how many students I have. You see? Double the number of students and the number of questions on the test drops in half.

Algebraically, if I call $t$ the number of questions on the test, $s$ the number of students, and $g$ the number of questions I have to grade, then $g=ts$. In the first case, I set $s=20$ so I had the direct variation equation $g=20t$. In the second case, I set $g=200$ so I had the inverse variation equation $t=\frac{200}{s}$.

I explain all that to my class, slowly and carefully. They need to know it because the actual test I give them will have a question where they have to make up an inverse variation problem, and I always tell them so. That question, if nothing else, will come as no surprise at all.

Homework:

“Homework: Direct and Inverse Variation”

This is on the long side, and introduces a few new ideas: the idea of being proportional to the square (or square root) of a variable, and the idea of dependence on multiple variables. So you should ideally hand it out in the middle of class, so they have time to work on it before the homework—and be prepared to spend a lot of time going over it the next day.

Calculator Regression

Lecture time. Talk about the importance of regression again, and about Kepler. Walk them, step-by-step, through a few regressions on the calculator. (Detailed instructions for a $TI-83$ are included in the “Conceptual Explanations” for this section.)

My advice right now is to look at the points first (which may require resetting the window!) and then categorize them according to concavity—although I don’t use that word. I talk about three kinds of increasing functions:

• Logs and square roots increase more and more slowly
• Parabolas and exponential functions increase more and more quickly

Similarly, of course, for decreasing functions. Choose an appropriate kind of regression, and let the calculator do the rest.

After a few examples, hand out the homework.

Homework:

“Homework: Calculator Regression”

Time for Another Test!

Question $8c$ is not something I would put on a real test, for a couple of reasons. First, it would be pretty hard to grade; second, and more importantly, it is unlike anything we’ve seen on a homework. But I might use it for an extra credit, and in any case, it can’t hurt them to see it and discuss it on a sample.

## Analytic Geometry (or) Conic Sections

The last topic! And it’s a big one. Here is the overall plan.

We start with a day on distance—finding the distance between two points, finding the distance from a point to a line—this will form a basis for the whole unit.

Then we do our shapes: circles, parabolas, ellipses, and hyperbolas. For each shape, there are really two things the students need to learn. One is the geometric definition of the shape. The other is what I call the machinery associated with the shape—the standard equation and what it represents. I cover these two things separately and then connect them at the end.

Distance

Before I hand this out, I tell them a bit about where we’re going in terms of the whole unit. We’re going to do “analytic geometry”—that is, linking geometry with algebra. We’re going to graph a bunch of shapes. And everything we’re going to do is built upon one simple idea: the idea of distance.

Let’s start with distance on a number line. (Draw a number line on the board.) Here’s $4$ and here’s $10$. What’s the distance between them? Right, $6$. You don’t need to do any math, you can just count—$1, 2, 3, 4, 5, 6$. There we are.

How about the distance from $4$ to $1$? Right, $3$. $4$ to $0$? Right, $4$. One more: what is the distance from $4$ to $-5$? Again, just count…$1,2,3,4,5,6,7,8,9$. The distance is nine.

So, what’s going on here? In each case, we’re counting from $4$ to some other number: to $10$, to $1$, to $0$, then to $-5$. What happened mathematically? We subtracted. This is a point that looks incredibly obvious, but really it isn’t, so it’s worth repeating: if you subtract two numbers, you get the distance between them. $10-4$ is $6$. $4-1$ is $3$. And the last one? $4-(-5)=4+5=9$. So even that works. Remember that we saw, by counting, that the distance from $4$ to $-5$ is $9$. Now we see that it works mathematically because subtracting a negative is like adding a positive.

Oh yeah…what if we had subtracted the other way? You know, $4-9$ or $-5-4$. We would have gotten the right answers only negative. But the distance would still be positive because distance is always positive.

So, based on all that, at your seats, write down a formula for the distance from $a$ to $b$ on a number line: go! (give them thirty seconds). What did you get? $|a-b|?$ Good! $|b-a|?$ Also good! They are the same thing. $a-b$ and $b-a$ are not the same thing, but when you take the absolute value, then they are.

Now, let’s get two-dimensional here. We’ll start with the easy case, which is when the points line up. In that case, we can use the same rule, right? For instance, let’s look at $(4,3)$ and $(10,3)$. How far apart are they? Same as before—$6$. We can just count, or we can just subtract because the $y$-coordinates are the same. (Show them this visually!) Similarly, suppose we take $(-2,5)$ and $(-2,-8)$. Since the $x$-coordinates are the same we can just count again, or just subtract the $y$-coordinates and get a distance of $13$.

Now, what if neither coordinate is the same? Then it’s a bit trickier. But we’re not going to use any magic “distance formula”—if you ever memorized one, throw it out. All we need is what we’ve already done. Let’s look at $(-2,1)$ and $(4,9)$. (Draw it!) To find that distance, we’re going to find the distance across and the distance up. So draw in this other point at $(4,1)$. Now, draw a triangle with the distance we want over here, the distance across here, and the distance up here. These two sides are easy, because they are just what we have already been doing, right? So this is $6$ and this is $8$. So how do we find this third side, which is the distance we wanted? Right, the Pythagorean Theorem! So it comes out as $10$.

The moral of the story is—whenever you need to find a distance use the Pythagorean Theorem.

OK, one more thing before you start the assignment. That was the distance between two points. How about the distance from a point to a line? For instance, what is the distance from you to the nearest street? The answer, of course, is—it depends on where on the street you want to get. But when we say, distance from you to the street, we mean the shortest distance. (Do a few drawings to make sure they get the idea of shortest distance from a point to a line. If the line is vertical or horizontal, then we are back to just counting. If it’s diagonal, life gets much more complicated, and we’re not going to get into it. Except I sometimes assign, as an extra credit assignment, “find the distance from the arbitrary point $(x,y)$ to the arbitrary line $y=mx+b$. It’s complicated and difficult, but I usually have one or two kids take me up on it. For the rest of the class, just promise to stick with horizontal and vertical lines and counting.)

After all that is said, they are ready to start on the assignment.

Homework:

“Homework: Distance”

Circles

Our first shape. (Sometimes I have started with parabolas first, but I think this is simpler.)

Here’s how you’re going to start—and this will be the same for every shape. Don’t tell them what the shape is; instead, tell them this. A bunch of points are getting together to form a very exclusive club. The membership requirement for the club is this: you must be exactly $5\;\mathrm{units}$ away from the origin. Any point that fulfills this requirement is in the club; any point that is either too close or too far is not in the club. Give them a piece of graph paper and have them draw all the points in the club. You come around and look at their work. If they are stuck, point out that there are a few very obvious points on the $x$-axis. The point is that, before any math happens at all, every individual group should have convinced themselves that this club forms a circle.

Then you step back and say—in Geometry class you used circles all the time, but you may never have formally defined what a circle is. We now have a formal definition: a circle is all the points in a plane that are the same distance from a given point. That distance is called what? (Someone will come up with “radius.”) And that given point is called what? (Someone may or may not come up with “center.”) The center plays a very interesting role in this story. It is the most important part, the only point that is key to the definition of the circle—but it is not itself part of the circle, not itself a member of the club. (The origin is not $5\;\mathrm{units}$ away from the origin.) This is worth stressing even though it’s obvious because it will help set up less obvious ideas later (such as the focus of a parabola).

Point out, also, that—as predicted—the definition of a circle is based entirely on the idea of distance. So in order to take our general geometric definition and turn it into math, we will need to mathematically understand distance. Which we do.

At this point, they should be ready for the in-class assignment. They may need some help here, but with just a little nudging, they should be able to see how we can take the geometric definition of a circle leads very directly to the equation of a circle, $(x-h)^2+(y-k)^2=r^2$ where $(h,k)$ is the center and $r$ is the radius. This formula should be in their notes, etc. Once we have this formula, we can use it to immediately graph things like $(x-2)^2+(y+3)^2=25$—to go from the equation to the graph and vice-versa.

Once you have explained this and they get it, they are ready for the homework. They now know how to graph a circle in standard form, but what if the circle doesn’t come in standard form? The answer is to complete the square—twice, once for $x$ and once for $y$. If it works out that the homework gets done in class (this day or the next day), you may want to do a brief TAPPS exercise with my little completing-the-square demo in the middle of the homework. However, they should also be able to follow it on their own at home.

Homework:

“Homework: Circles”

It may take a fair amount of debriefing afterward and even a few more practice problems before you are confident that they “get” the circle thing. They need to be able to take an equation for a circle in non-standard form, put it in standard form and then graph it. And they need to still see how that form comes directly from the Pythagorean Theorem and the definition of a circle.

There is one other fact that I always slip into the conversation somewhere, which is: how can you look at an equation, such as $2x^2+3x+2y^2+5y+7=0$, and even tell if it is a circle? Answer: it has both an $x^2$ and a $y^2$ term, and they have the same coefficient. If there is no $x^2$ or $y^2$ term, you have a line. If one exists, but not the other, you have a parabola. By the time we’re done with this unit we will have filled out all the other possible cases, and I expect them to be able to look at any equation and recognize immediately what shape it will be.

Parabolas, Day 1

Once again, when you start, don’t tell them we’re doing parabolas! Tell them we’re going to create another club. This time the requirement for membership is: you must be exactly the same distance from the point $(0,3)$ that you are from the line $y=-3$. For instance, the point $(3,3)$ is not part of our club—it is $3\;\mathrm{units}$ away from $(0,3)$ and six units away from $y=-3$.

Now, let them work in groups on “All the Points Equidistant from a Point and a Line” to see if they can find the shape from just that. If they need a hint, tell them there is one extremely obvious point and two somewhat obvious points. After that they have to tinker around.

When all or most groups have it, go through it on the blackboard something like this. The extremely obvious point is the origin. The “somewhat” obvious points are $(-6,3)$ and $(6,3)$. Show why all those work.

Now, can any point below the $x$-axis work? Clearly not. Any point below the $x$-axis is “obviously” (meaning, after you show them for a minute) closer to the line than to the point.

So, let’s start working up from the origin. The origin was in the club. As we move up, we are getting closer to the point and farther away from the line. So how can we maintain equality? The only way is to move farther away from the point by moving out. In this way, you sketch in the parabola.

Now, you introduce the terminology. We’re already old friends with the vertex of a parabola. This point up here is called the focus. This line down here is the directrix. The focus and directrix are kind of like the center of a circle, in the sense that they are central to the definition of what a parabola is, but they are not themselves part of the parabola. The vertex, on the other hand, is a part of the parabola, but is not a part of the definition.

The directrix, of course, is a horizontal line: but what if it isn’t? What if the directrix is vertical? Then we have a horizontal parabola. Of course it isn’t a function, but it’s still a shape we can graph and talk about, and we have seen them a few times before. If you have time, work through $x=4y^2-8y$.

Homework:

“Homework: Vertical and Horizontal Parabolas”

Parabolas, Day 2

What good are parabolas? We’ve already seen some use for graphing parabolas, in terms of modeling certain kinds of behavior. If I throw a ball into the air, not straight up, its path through the air is a parabola. But here is another cool thing: if parallel lines come into a parabola they all bounce to the focus.

So telescopes are made by creating parabolic mirrors—all the incoming light is concentrated at the focus. Pretty cool, huh?

We are already somewhat familiar with parabola machinery. We recall that the equation for a vertical parabola is $y=a(x-h)^2+k$ and the equation for a horizontal parabola is $x=a(y-k)^2+h$. The vertex in either case is at $(h,k)$. We also recall that if $a$ is positive, it opens up (vertical) or to the right (horizontal); if $a$ is negative, it opens down or to the left. This is all old news.

Thus far, whenever we have graphed parabolas, we have found the vertex, determined whether they open up or down or right or left, and then gone “swoosh.” If we were a little more sophisticated, we remembered that the width of the parabola was determined by $a$; so $y=2x^2$ is narrower than $y=x^2$, which is narrower than $y=\frac{1}{2}x^2$. But how can we use $a$ to actually draw the width accurately? That is the question for today. And the answer starts out with one fact, which may seem quite unrelated: the distance from the vertex to the focus is $\frac{1}{4a}$ . I’m going to present this as a magical fact for the moment—we will never prove it, though we will demonstrate it for a few examples. But that one little fact is the only thing I am going to ask you to “take my word for”—everything else is going to flow logically from that.

(Go back to your drawing of our parabola with focus $(0,3)$ and directrix $y=-3$, and point out the distance from vertex to focus—label it $\frac{1}{4a}$.) Now, what is the distance from the vertex to the directrix? Someone should get this: it must also be $\frac{1}{4a}$. Why? Because the vertex is part of the parabola, so by definition, it must be the same distance from the directrix that it is from the focus. Label that.

Now, let’s start at this point here (point to $(-6,3)$) and go down to the directrix. How long is that? Again, someone should get it:$\frac{1}{4a}+\frac{1}{4a}$ , which is $\frac{2}{4a}$ or $\frac{1}{2a}$. Now, how far over is it, from this same point to the focus? Well, again, this is on the parabola, so it must be the same distance to the focus that it is to the directrix: $\frac{1}{2a}$.

If you extend that line all the way to the other side, you have two of those:$\frac{1}{2a}+\frac{1}{2a}=\frac{2}{2a}$, or $\frac{1}{a}$, running between $(6,3)$ and $(-6,3)$. This line is called the latus rectum—Latin for “straight line”—it is always a line that touches the parabola at two points, runs parallel to the directrix, and goes through the focus.

At this point, the blackboard looks something like this:

The point should be clear. Starting with the one magical fact that the distance from the focus to the vertex is $\frac{1}{4a}$, we can get everything else, including the length of the latus rectum, $\frac{1}{a}$. And that gives us a way of doing exactly what we wanted, which is using $a$ to more accurately draw a parabola.

At this point, you do a sample problem on the board, soup to nuts. Say, $x=\frac{1}{2}(y+3)^2-4$. They should be able to see quickly that it is horizontal, opens to the right, with vertex at $(-4,-3)$—that’s all review. But now we can also say that the distance from the focus to the vertex is $\frac{1}{4a}$, which in this case (since $a=\frac{1}{2}$) is $\frac{1}{2}$. So where is the focus? Draw the parabola quickly on the board. They should be able to see that the focus is to the right of the vertex, so that puts us at $(-3\frac{1}{2},-3)$. The directrix is to the left, at $x = -4 \frac{1}{2}$. (I always warn them at this point, a very common way to miss points on the test is to say that the directrix is $(-4\frac{1}{2},-3)$. That’s a point—the directrix is a line!)

How wide is the parabola? When we graphed them before, we had no way of determining that: but now we have the latus rectum to help us out. The length of the latus rectum is $\frac{1}{a}$, which is $2$. So you can draw that in around the focus—going up one and down one—and then draw the parabola more accurately starting from the focus and touching the latus rectum on both sides.

Whew! Got all that? Good then, you’re ready for the homework! (You may want to mention that a different sample problem is worked in the “Conceptual Explanations.”)

Homework:

“Homework: Parabolas and the Latus Rectum”

This one will take a lot of debriefing afterwards. Let’s talk about $^{\#}3$ for a minute. The first thing they should have done is draw it—that can’t be stressed enough—always draw it! Drawing the vertex and focus, they should be able to see that the parabola opens to the right. Since we know the vertex, we can write immediately $x=a(y-6)^2+5$. Many of them will have gotten that far and then gotten stuck. Show them that the distance from the vertex to the focus is $2$, and we know that this distance must be $\frac{1}{4a}$, so $\frac{1}{4a}=2$. We can solve this to find $a=\frac{1}{8}$, which fills in the final piece of the puzzle.

On $^{\#}5$, again, before doing anything else, they should draw it! They should see that there is one vertical and one horizontal parabola that fits this definition.

But how can they find them? You do the vertical, they can do the horizontal. Knowing the vertex, we know that the vertical one will look like $y=a(x+2)^2+5$. How can we find a this time, by using the information that it “contains the point $(0,1)$”? Well, we have to go back to our last unit, when we found the equation for a parabola containing certain points! Remember, we said that if it contains a point, then that point must make the equation true! So we can plug in $(0,1)$ to the equation, and solve for a. $1=a(0+2)^2+5$, $4a+5=1$, $a=-1$. The negative $a$ tells us that the parabola will open down—and our drawing already told us that, so that’s a good reality check.

Parabolas: From Definition to Equation

OK, where are we? We started with the geometric definition of a parabola. Then we jumped straight to the machinery, and we never attempted to connect the two. But that is what we’re going to do now.

Remember what we did with circles? We started with our geometric definition. We picked an arbitrary point on the circle, called it $(x,y)$, and wrote an equation that said, “You, Mr. $(x,y)$, are exactly $5\;\mathrm{units}$ away from the origin.” That equation became the equation for the circle.

Now we’re going to do the same thing with a parabola. We’re going to write an equation that says, “You, Mr. $(x,y)$, are the same distance from the focus that you are from the directrix.” In doing so, we will write the equation for a parabola, based on the geometric definition. And we will discover, along the way, that the distance from the focus to the vertex really is $\frac{1}{4a}$.

That’s really all the setup this assignment needs. But they will need a lot of help doing it. Let them go at it, in groups, and walk around and give hints when necessary. The answers we are looking for are:

1. $\sqrt{x^2 + y^2}$. As always, hint at this by pushing them toward the Pythagorean triangle.
2. $y+4$. The way I always hint at this is by saying, “Try numbers. Suppose instead of $(x,y)$ this were $(3,10)$. Now, how about $(10,3)$?” and so on, until they see that they are just adding $4$ to the $y$-coordinate. Then remind them of the rule, from day one of this unit—to find distances, subtract. In this case, subtract $-4$.
3. $\sqrt{x^2 + y^2}=y+4$. This is the key step! By asserting that $d1=d2$, we are writing the definition equation for the parabola.
4. Good algebra exercise! Square both sides, the $y^2$ terms cancel, and you’re left with $x^2=8y+16$. Solve for $y$, and you end up with $y=\frac{1}{8}x^2-2$. Some of them will have difficulty seeing that this is the final form—point out that we can rewrite it as $y=\frac{1}{8}(x-0)^2-2$, if that helps. So the vertex is $(0,-2)$ and the distance from focus to vertex is $2$, just as it should be.

Warn them that this will be on the test!

Which brings me to…

Time for Another Test!

Our first test on conics. I go back and forth as to whether I should give them a bunch of free information at the top of the test—but it’s probably a good idea to give them a chart, similar to the one on top of my sample.

Ellipses

Only two shapes left! But these two are doozies. Expect to spend at least a couple of days on each—they get a major test all to themselves.

In terms of teaching order, both shapes are going to follow the same pattern that we set with parabolas. First, the geometry. Then, the machinery. And finally, at the end, the connection between the two.

So, as always, don’t start by telling them the shape. Let them do the assignment “Distance to this point plus distance to that point is constant” in groups, and help them out until they get the shape themselves. A good hint is that there are two pretty easy points to find on the $x$-axis and two harder points to find on the $y$-axis. As always, keep wandering and hinting until most groups have drawn something like an ellipse. Then you lecture.

The lecture starts by pointing out what we have. We have two points called the foci. (One “focus,” two “foci.”) They are the defining points of the ellipse, but they are not part of the ellipse. And we also have a distance, which is part of the definition.

Because the foci were horizontally across from each other, we have a horizontal ellipse. If they were vertically lined up, we would have a vertical ellipse. You can also do diagonal ellipses, but we’re not going to do that here.

Let’s talk more about the geometry. One way you can draw a circle is to thumbtack a piece of string to a piece of cardboard, and tie the other end of the string to a pen. Keeping the string taut, you pull all the way around, and you end up with a circle. Note how you are using the geometric definition of a circle, to draw one: the thumbtack is the center, and the piece of string is the radius.

Now that we have our geometric definition of an ellipse, can anyone think of a way to draw one of those? (Probably not) Here’s what you do. Take a piece of string and thumbtack both ends down in a piece of cardboard, be sure the string is not taut. Then, using your pen, pull the string taut.

Now, pull the pen around, keeping the string taut. You see what this does? While the string is taut, the distance from the pen to the left thumbtack, plus the distance from the pen to the right thumbtack, is always a constant—namely, the length of the string. So this gives you an ellipse. I think most people can picture this if they close their eyes. Sometimes I assign them to do this at home.

OK, so, what good are ellipses? The best example I have is orbits. The Earth, for instance, is traveling in an ellipse, with the sun at one of the two foci. The moon’s orbit around the Earth, or even a satellite’s orbit around the Earth are all ellipses.

Another cool ellipse thing, which a lot of people have seen in a museum, is that if you are in an elliptical room and one person stands at each focus, you can hear each other whisper. Just as a parabola collects all incoming parallel lines at the focus, an ellipse bounces everything from one focus straight to the other focus.

OK, on to the machinery. Here is the equation for a horizontal ellipse centered at the origin.

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Here is a drawing of a horizontal ellipse.

There are three numbers in this drawing. $a$ is the distance from the center to the far edge (right or left). If you double this, you get the horizontal length of the entire ellipse—this length, $2a$ of course, is called the major axis.

$b$ is the distance from the center to the top or bottom. If you double this, you get the vertical length of the entire ellipse—this length, $2b$ of course, is called the minor axis.

$c$ is the distance from the center to either focus.

$a$ and $b$ appear in our equation. $c$ does not. However, the three numbers bear the following relationship to each other: $a^2=b^2+c^2$. Note also that $a$ is always the biggest of the three!

A few more points. If the center is not at the origin—if it is at, say, $(h,k)$, what do you think that does to our equation? They should all be able to guess that we replace $x^2$ with $(x-h)^2$ and $y^2$ with $(y-k)^2$.

Second, a vertical ellipse looks the same, but with $a$ and $b$ reversed:

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

Let’s see how all that looks in an actual problem—walk through this on the blackboard to demonstrate. Suppose we want to graph this:

$x^2+9y^2-4x+54y+49=0$

First, how can we recognize it as an ellipse? Because it has both an $x^2$ and a $y^2$ term, and the coefficients are different. So we complete the square twice, sort of like we did with circles. But with circles, we always divided by the coefficient right away—this time, we pull it out from the $x$ and $y$ parts of the equation separately.

$x^2 +9y^2-4x+54y+49 & =0 && \text{The original problem} \\x^2-4x+9y^2 +54y & =-49 && \text{Group the} \ x \ \text{and} \ y \ \text{parts} \\ (x^2-4x)+9(y^2+6y) & =-49 && \text{Factor out the coefficients. In this case, there}\\&& &\text{is no} \ x \ \text{coefficient, so we just have to do} \ y. \ \text{In}\\&& & \text{general, we must do both}. \\(x^2-4x+4)+9(y^2+6y+9) & =-49+4+81 && \text{Complete the square, twice} \\(x-2)^2+9(y+3)^2 & =36 && \text{Finish completing that square} \\\frac{(x-2)^2}{36} + \frac{(y+3)^2}{4} & =1 && \text{Divide by} \ 36. \ \text{This is because we need a} \ 1 \ \text{on}\\& &&\text{the right, to be in our standard form!}$

OK, get all that? Now, what do we have?

First of all, what is the center? That’s easy: $(2,-3)$.

Now, here is a harder question: does it open vertically or horizontally? That is, does this look like $\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$, or like $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$? The way to tell is by remembering that $a$ is always bigger than $b$. So in this case, the $36$ must be $a^2$ and the $4$ must be $b^2$, so it is horizontal. We can see that the major axis $(2a)$ will be $12$ long, stretching from $(-4,-3)$ to $(8,-3)$. (Draw all this as you’re doing it.) And the minor axis will be $4$ long, stretching from $(2,-5)$ to $(2,-1)$.

And where are the foci? Since this is a horizontal ellipse, they are to the left and right of the center. By how much? By $c$. What is $c$? Well, $a^2=b^2+c^2$. So $c^2=32$, and $c= \sqrt{32} = 4 \sqrt{2}$ , or somewhere around $5 \frac{1}{2}$. (They should be able to do that without a calculator: $32$ is somewhere between $25$ and $36$, so $\sqrt{32}$ is around $5 \frac{1}{2}$.) So the foci are at more or less $\left (-3\frac{1}{2},-3 \right )$ and $\left (7\frac{1}{2},-3 \right )$. We’re done!

Homework:

“Homework: Ellipses”

There are two things here that may throw them for a loop.

One is the fractions in the denominator, and the necessity of (for instance) turning $25y^2$ into $\frac{y^2}{\frac{1}{25}}$. You may have to explain very carefully why we do that (standard form allows for a number on the bottom, but not on the top), and how we do that.

The other is number $7$. Some students will quickly and carelessly assume that $94.5$ is $a$ and $91.4$ is $b$. So you want to draw this very carefully on the board when going over the homework. Show them where the $94.5$ and $91.4$ are and remind them of where $a, b$, and $c$ are. Get them to see from the drawing that $94.5+91.4$ is the major axis and is therefore, $2a$. And that $a-91.4$ is $c$, so we can find $c$, and finally, we can use $a^2=b^2+c^2$ to find $b$. This is a really hard problem, but it’s worth taking a lot of time on, because it really drives home the importance of visually being able to see an ellipse in your head, and knowing where $a,b$, and c are in that picture.

Oh, yeah…number 6 may confuse some of them too. Remind them again to draw it first, and that they can plug in $(0,0)$ and get a true equation.

OK, now we’re at a bit of a fork in the road. The next step is to connect the geometry of the ellipse with the machinery. This is a really great problem because it brings together a lot of ideas, including some of the work we did forever ago in radical equations. It is also really hard. So you can decide what to do based on how much time you have left and how well you think they are following you. You may want to have them go through the exercise in class (expect to take a day). Or, you may want to make photocopies of the completely-worked-out version (which I have thoughtfully included here in the teacher’s guide), and have them look it over as a TAPPS exercise, or just ask them to look it over. Or you could skip this entirely, or make it an extra credit.

The Ellipse: From Definition to Equation

Here is the geometric definition of an ellipse. There are two points called the “foci”: in this case, $(-3,0)$ and $(3,0)$. A point is on the ellipse if the sum of its distances to both foci is a certain constant: in this case, I’ll use $10$. Note that the foci define the ellipse, but are not part of it.

The point $(x,y)$ represents any point on the ellipse. $d1$ is its distance from the first focus and $d2$ to the second. So the ellipse is defined geometrically by the relationship: $d1+d2=10$.

To calculate $d1$ and $d2$, we use the Pythagorean Theorem as always: drop a straight line down from $(x,y)$ to create the right triangles. Please verify this result for yourself! You should find that $d1=\sqrt{(x+3)^2+y^2}$ and $d2=\sqrt{(x-3)^2+y^2}$. So the equation becomes:

$\sqrt{(x+3)^2+y^2} + \sqrt{(x-3)^2+y^2} = 10 && \text{This defines our ellipse}$

The goal now is to simplify it. We did problems like this earlier in the year (radical equations, the “harder” variety that have two radicals). The way you do it is by isolating the square root and then squaring both sides. In this case, there are two square roots, so we will need to go through that process twice.

$& \sqrt{(x+3)^2+y^2} = 10 - \sqrt{(x-3)^2+y^2} && \text{Isolate a radical} \\& (x+3)^2 + y^2 = 100 - 20 \sqrt{(x-3)^2+y^2} + (x-3)^2+y^2 && \text{Square both sides} \\& (x^2+6x+9)+y^2=100-20 \sqrt{(x-3)^2 + y^2} + (x^2-6x+9)+y^2 && \text{Multiply out the squares} \\& 12x=100-20 \sqrt{(x-3)^2+y^2} && \text{Cancel combine like terms} \\& \sqrt{(x-3)^2+y^2} =5-\frac{3}{5} x && \text{Rearrange, divide by}\ 20 \\& (x-3)^2+y^2=25-6x+ \frac{9}{25} x^2 && \text{Square both sides again} \\& (x^2-6x+9)+y^2=25-6x+ \frac{9}{25} x^2 && \text{Multiply out the square} \\& \frac{16}{25} x^2+y^2=16 && \text{Combine like terms} \\& \frac{x^2}{25} + \frac{y^2}{16} = 1 && \text{Divide by}\ 16$

…and we’re done! Now, according to the “machinery” of ellipses, what should that equation look like? Horizontal or vertical? Where should the center be? What are $a, b$, and $c$? Does all that match the picture we started with?

Hyperbolas

The good news about hyperbolas is they are a lot like ellipses—a lot of what has already been learned will come in handy here. The bad news about hyperbolas is they are a lot like ellipses—so all the little differences can be very confusing.

We will start as always with the geometric definition. Let them do the assignment “Distance to this point minus distance to that point is constant” in groups, and help them out until they get the shape themselves. There are two pretty easy points to find on the $x-$axis, but from there they just sort of have to noodle around like we did with parabolas, asking…what happens as I move inside? What happens as I move outside? As always, keep wandering and hinting until most groups have drawn something like a hyperbola. Then you lecture.

The lecture starts by pointing out what we have. We have two points, once again called the foci. They are the defining points of the hyperbola, but they are not part of the hyperbola. And we also once again have a distance, which is part of the definition.

Because the foci were horizontally across from each other, we have a horizontal hyperbola. If they were vertically lined up, we would have a vertical hyperbola. You can also do diagonal hyperbolas—anyone remember where we have seen one of those? That’s right, inverse variation! That was a hyperbola, just like these. But we’re not going to talk about those in this unit, just the horizontal and vertical ones.

Incidentally, a hyperbola is not two back-to-back parabolas. It looks like it, but these shapes are actually different from parabolic shapes, as we will see.

OK, so, what good are hyperbolas? The analogy continues…orbits! Suppose a comet is heading toward the sun. (Draw.) If it has a low energy—that is, a low velocity—it gets trapped by the sun and winds up orbiting around the sun in an elliptical orbit. But if it has high energy (high velocity) it zooms around the sun and then zooms away forever. Its path in this case is half a hyperbola.

Another cool use is in submarine detection. A submarine sends out a pulse. Two receiving stations get the pulse. They don’t know what direction it came from or when it was sent, but they do know that one station received it exactly two seconds before the other one. This enables them to say that the distance from the sub to this station, minus the distance to this other station is such-and-such. And this, in turn, locates the sub on a hyperbola.

OK, on to the machinery. Here is the equation for a horizontal hyperbola, centered at the origin.

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Looks familiar, doesn’t it? But the plus has changed to a minus and that makes all the difference in the world.

Here is a drawing of a horizontal hyperbola.

Let’s be very careful in seeing how this is, and is not, like an ellipse.

$a$ is defined in a very similar way. It goes from the center to the edges. In this case, the edges are called the “vertices” (in analogy to parabolas). The distance from one vertex to the other ($2a$, of course) is called the transverse axis.

$c$ is defined in a very similar way: it goes from the center to either focus.

$b$ is perpendicular to the other two, just as before. But it goes from the center to…well, to a sort of strange point in the middle of nowhere. We’re going to use this point. The distance from the top point to the bottom point $(2b)$ is called the conjugate axis.

But here is one major difference. In an ellipse, the foci are inside; in a hyperbola, they are outside. So you can see looking at an ellipse that $a>c$, and you can see looking at a hyperbola that $c>a$. Hence, our equation relating the three shapes is going to be different. Instead of $a^2=b^2+c^2$, we have $c^2=a^2+b^2$. This reflects the fact that $c$ is the biggest one in this case.

Once again, the class should be able to see that if the center is $(h,k)$ instead of the origin, we replace $x^2$ with $(x-h)^2$ and $y^2$ with $(y-k)^2$.

How about a vertical hyperbola? That looks like this:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

The way we tell vertical from horizontal is completely different. In an ellipse, we can tell by assuming that $a>b$. In a hyperbola, we have no such guarantee; either $a$ or $b$ could be the greater, or they could even be the same. Instead, we look at which one comes first. If we are doing an $x^2-y^2$ thing, it’s horizontal; if we are doing a $y^2-x^2$ thing, it’s vertical. This is a very common source of errors…be careful!

There should be no need to go through the whole completing-the-square rigmarole on the board—just tell them it is exactly like with ellipses, including making sure you have a $1$ on the right, and there is nothing multiplied by $x^2$ or $y^2$. But the part that is different is the graphing. So let’s just pick it up there. Suppose you want to graph:

$\frac{(x-2)^2}{36} - \frac{(y+3)^2}{4} = 1$

First of all, what is the center? That’s easy: $(2,-3)$.

Now, here is a harder question: does it open vertically or horizontally? We can answer this question without even looking at the numbers on the bottom! The $x^2$ is positive and the $y^2$ is negative, so this is horizontal.

As we did with ellipses, we will then find $a, b$, and $c$. $a=6$ and $b=2$. We will find c with the hyperbola equation $c^2=a^2+b^2$ (different from the ellipse equation!) and get $c=\sqrt{40}$, which is $2\sqrt{10}$ or somewhere just above $6$ (again, because $40$ is just above $36$).

Now, it’s drawing time. We start at the center. We go out horizontally by $6$ to find the vertices and by a little more than 6 to find the foci. We go out vertically by $2$ to find the endpoints of the conjugate axis, those weird little points in space. Then what?

Here’s what you do. Draw a rectangle, going through the vertices and the endpoints of the conjugate axis. Then, draw diagonal lines through the corners of that rectangle. Those diagonal lines are going to serve as asymptotes or guides: they are not part of the hyperbola, but they help us draw it. Why? Because as it moves out, the hyperbola gets closer and closer to the asymptotes, but never quite reaches them. So once you have drawn your asymptotes, you have a guide for drawing in your hyperbola.

Note that I draw the box and the asymptotes in dotted lines, indicating that they are not really part of the hyperbola.

It’s worth talking about what an asymptote is since it is such an important concept in Calculus—the line that the curve gets closer and closer and closer to, without ever quite reaching. It’s also worth pointing out that this shows that a hyperbola is not two back-to-back parabolas, since parabolas do not display asymptotic behavior.

Finally, I always mention the comet again. Remember that if a comet comes in with high energy, it swoops around the sun and then flies away again. Now, if there were no sun—if there were nothing in the universe, but our comet—the comet would travel in a straight line. And clearly, when the comet is very, very far away from the sun (either before or after its journey through our solar system), the effect of the sun is very small, so the comet travels almost in a straight line. That straight line is the asymptote. The farther away from the sun the comet gets, the closer it gets to that straight line.

Homework:

“Homework: Hyperbolas”

The only really unusual thing here is that I ask for the equation of one of the asymptotes. This is just a quick review of the skill of finding the equation for a line, given that you already know two points on the line.

You will note that I have absolutely nothing here about going from the geometry of the hyperbola to the equation. The reason is that it is exactly like the ellipse. You may want to do it, or you may not. If you do it, you shouldn’t need to hand them anything—just say, “By analogy to what we did with the ellipse, do this.”

Now that you have done all the shapes, the one vital skill that cuts across all of it is looking at an equation $(ax^2+by^2+cx+dy+e=0)$ and telling what shape it is. This is done entirely by looking at the coefficients of the squared terms ($a$ and $b$), and you should refer them to the chart at the end of the “Conceptual Explanations.”

Time for Our Very Last Test!

With luck, you have two or three weeks left for review after this, before the final test. Congratulations, you made it through!

## Date Created:

Feb 23, 2012

Apr 29, 2014
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