Tag Archives: quadratic equations

The Sum of a Parabola and a Line

For the past two years, my algebra students have determined that the product of two lines is a parabola, which instantly provides a visual of the solutions and the line of symmetry.  For the past year, they’ve determined that squaring a line is likewise a parabola, and can be moved up and down the line of symmetry, which is instantly visible as the line’s x-intercept. In this way, I have been able to build understanding from lines to quadratics without just saying hey, presto! here’s a parabola. I introduce them to adding and subtracting functions, and from there, it’s a reasonable step to multiplying functions.

Typically, I’ve moved from this to binomial multiplication, introducing the third form of the quadratic we deal with in early high-level math, the standard form. (The otherwise estimable Stewart refers to the vertex form as standard form, to which I say sir! you must reconsider, except, well, he’s dead.)

At some point in teaching this, you come to the “- b over 2a” (-b2a) issue. That is, teachers who like to build on existing knowledge towards each new step are a bit stuck when it comes to finding the vertex in a standard form equation.

(For non-mathies, the standard form of an equation is ax2+bx+c and the vertex form is a(x-h)2+k.  The parameters “a” “b”, and “c” are often just referred to by letter. Vertex form, we’re more likely to talk about the x and y values of the vertex, just like  when we talk about lines in the form y=mx+b, we don’t say “m” and “b” but rather “slope” and “y-intercept”. But teachers, at least, often talk about teaching different aspects of standard form operations by parameters: a>1, a<0, to say nothing of the quadratic formula.  So the way to find the vertex of a parabola in standard form is to take the “a” and “b” term and use the algorithm -b2a to find the line of symmetry,  which is the x-value of the vertex. Then”plug it in”, or evaluate, the x-value in the quadratic equation to find the y-value for the vertex.)

The only way I’ve found until now of building on existing knowledge to establish it is setting standard form equal to vertex form to establish that the “h” of vertex form is equal to the -b2a of standard form, something only the top kids really understand and don’t often enjoy. (they’re much more interested by pre-calc.)

Last year, I was putting together a worksheet on adding and subtracting lines, and on impulse I added a few that involved adding a simple parabola with its vertex at the origin with a line, mainly to add a bit of challenge for the top kids. I could see that adding a line and a parabola doesn’t provide the instant visual “hook” that multiplying or squaring lines does.

sumparabolaline

It’s obvious that the y-intercept of the sum will be the same as the y-intercept of the line. One can logically ascertain that in this particular case, the right side of the y-axis will only increase—adding two positives. The left side, therefore, as x approaches negative infinity is where the action is. But not too much action, since the parabola’s y is galloping towards positive infinity at a faster clip than the line’s is trotting towards negative infinity. So for a brief interval, the negative of the line will offset a bit of the positive of the parabola, but eventually the parabola’s growth will drown out the line’s decline.

All logically there to construe, but far less obvious at a glance.

This year, I decided to explore the relationship further, because deciphering standard form is where my weakest kids tend to check out. They’ve held on through binomial multiplication, to hang on, at least temporarily, to the linear term so that (x+3)2 doesn’t become x2 + 9. They’ve mastered factoring quadratics, to their shock. They understand how to graph parabolas in two forms. And suddenly this bizarre algorithm that has to be remembered, then calculated, then more calculations to find “y”, whatever that is. Can you say “cognitive load“, boys and girls? Before you know it, they’re using the quadratic formula for linear equations and other bad, bad things that happen when it’s all kerfluzzled in their noggins. That’s when you realize that paralysis isn’t the worst thing that can happen.

Could I break the process down into discrete steps that told a story?  Build on this notion of modifying the parent function ax2 with a line to shift it left or right? Find Raylene a new kidney now that her third husband discovered her affair with the yoga instructor and will no longer give her one of his?

My  first thought was to wonder if the slope of the line had any relationship to the graph’s location. My second thought was yes, you dweeb, “b” is the slope of the added line and b’s fingerprints are all over the line of symmetry. No, no, the other half of my brain, the English major, protested. I know that. But is there some way I can get the kids to think of “b” as a slope, or to link slope to the process in a meaningful way?

(This next part is probably incredibly obvious to actual mathematicians, but in my own defense I ran it by three teachers who actually studied advanced math, and they were like hey, wow. I didn’t know that.)

What information does standard form give? The y-intercept, or “c”. What information do we want that it doesn’t readily provide? The vertex. Factors would be nice, but they aren’t guaranteed. I always want the vertex. So if I graph the resulting parabola of the sum of, say,  x2 and 6x + 5, how might the slope be relevant?

The obvious relationship to wonder about first is the slope between the y-intercept, which I have, and the vertex, which I want. Start by finding the slope between these two points. And right at that point I realize hey,  by golly, that’s the rate of change(!).

sumparabolalineslope

The slope–that is, by golly, the rate of change(!)–is 3. The line of symmetry is -3. The vertex is exactly 9 units below the y-intercept, or the product of the rate of change and the line of symmetry. Heavens. That’s interesting. Does it always happen? Let’s assume for now a=1.

Sum Slope from y-int
to vertex
Line of
Symmetry
units from y-int to
y-value of vertex
Vertex
x2 – 4x – 12 -2 x=2 -4 (2,-16)
x2 – 10x + 9 -5 x=5 -25 (5,-16)
x2 – 2x – 3 -1 x=1 -1 (-1,-4)
x2 +6x + 8 3 x=-3 -9 (-3,-1)

Hmm. So according to this, if I were trying to get the vertex for x2 +12x + 15, then I should assume that the slope–that is, by golly, the rate of change(!)– from the vertex to the y-intercept is 6. That would make the line of symmetry is x=-6. The y-value of the vertex should be 36 units down from 15, or -21. So the vertex should be at (-6,-21). And indeed it is. How about that?

So what happens if a is some other value than 1? I know the line of symmetry will change, of course, but what about the slope–that is, by golly, the rate of change(!). Is it affected by changes in a?

Sum Slope from y-int
to vertex
Line of
Symmetry
units from y-int to
y-value of vertex
Vertex
2x2 – 8x – 5 -4 x=2 (-4/2) -8 (2,-3)
-x2 +2x + 4 1 x=1 (-1/-1) 1 (1,5)
-2x2 +14x +7 7 x=3.5 (-7/-2) 24.5 (49/2) (3.5,31.5)
4x2 +8x -15 4 x=-1 (-4/4) -4 (-1,-19)

Here’s a Desmos application that I created to demonstrate it.  The slope–that is, by golly, the rate of change(!)–from the vertex to the y-intercept is always half of the slope of the line added to the parabola–that is, half of “b”. The rate of change is not affected by the stretch factor, or a. The line of symmetry, however, is affected by the stretch, which makes sense once you realize that what we’re really calculating is the horizontal distance (the run) from the vertex to the y-axis. The stretch would affect how quickly the vertex is reached. So the vertex y-value is always going to be the rise for the number of iterations the run went through to get from the y-axis to the line of symmetry, or the rate of change multiplied by the line of symmetry x-value.

sumparabolathenut

Mathematically, these are the exact steps used to complete the square but considerably less abstract. You’re finding the “run” to the line of symmetry and the “rise” up or down to the vertex.

Up to now, I’ve been describing my own discovery? How to explain this to the kids? As is always the case in a new lesson, I keep it pretty flexible and don’t overplan. I created a quick activity sheet.sumparabolalinehandout

The goal here was just to get things started. Notice the last question on the back: “Do you notice any patterns?” I was fully prepared for the answer to be “No”, which is good, because it was. We then developed the table similar to the first one above, and they quickly caught on to the pattern when a=1.

I was a bit worried about moving to other a values. However,  the class eventually grasped the basic relationship. The slope from the vertex to the y-intercept was always related to the slope of the line added  to the parabola. But the line of symmetry, the distance from the y-axis, would be influenced by the stretch. This made intuitive sense to most of the kids. They certainly screwed up negatives now and again, but who doesn’t.

Good math thinking throughout. I heard a lot of discussions, saw graphs where kids were clearly thinking through the spatial relationship. Many kids realized that when a=1, a negative b means the slope of the line from the y-intercept to the vertex is also negative, which means the vertex must be to the right of the y-intercept. A positive “b” means the slope is positive which means the vertex is to the left. Then they realize that the sign of “a” will flip that relationship around. he students start to see the “b” value as an indicator. That is, by making bx+c its own unit, they realize how important the slope of the added line is, and how essential it is to the end result.

All that and, you might have noticed, they get an early peek at rate of change concepts.

Definitely no worse than my usual -b2a  lesson and the weak kids did much, much better. This was just the first run; the next time I teach algebra 2 I’ll get more ambitious.

So I can now build on students’ existing knowledge to decipher and graph a standard form equation rather than just provide an algorithm or go through the algebra. On the other hand, the last tether holding my quadratics unit to the earth of typical algebra 2 practice has been severed; it’s now wandering around in the stratosphere.

I don’t mean the basics aren’t covered. I teach binomial multiplication, factoring, projectile motion, the quadratic formula, complex numbers, and so on. But the framework differs considerably from my colleagues’.

But if anyone is thinking that I’m dumbing this down, recall that my students are learning that functions can be combined, added, subtracted, multiplied. They’re learning that rate of change is linked directly to the slope of the line added to  the parabola, and that the original parabola’s stretch doesn’t influence the rate of change–but does impact the line of symmetry. And the weaker kids aren’t getting lost in algorithms that have no meaning.

I could argue about this, but maybe another day. For now, I’m interested in what to call this method, and who else is using it.

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Binomial Multiplication and Factoring Trinomials with The Rectangle

I was reading about Joseph Nebus’s factoring method….

(okay, a brief note. Early in his writeup, Nebus (Joe? Joseph?) writes: “It’s a method for factoring quadratic expressions into binomial expressions, and I must admit, it’s not very good. It’s cumbersome and totally useless once one knows the quadratic equation.”

Many, many math teachers have expelled much breath on the uselessness of factoring, as the skill is completely nullified by the quadratic formula (which I think what he means here). But when they make this comment, they are thinking as mathematicians, not as teachers. Mathematicians work with math to solve problems. Teachers teach math so their kids can demonstrate their knowledge on tests*—not just state tests, but college admissions tests and placement tests. And on these tests, the questions are designed for either factoring or the quadratic formula—and far more the former than the latter. All students must learn to factor trinomials if they are to escape remediation. The quadratic formula is optional. And if the test pragmatism isn’t enough of an argument, please know that students with limited integer operations skills will do better with factoring than the formula because they rarely have squares memorized and please, please believe me when I say that they will ALWAYS subtract 4ac from b squared, even if c is negative. End brief note.)

There are teachers who think this is a science, and teachers who know it’s an art based largely on the audience. And which kind of teacher you are is a religion, or an expression of personality (which I often think is the same thing).

So when I say that the method Nebus describes sounds extremely convoluted, I am simply a Jehovah’s Witness expressing doubt about the utility of the Amish rumspringa.

But many math teachers aren’t even aware of the box/diamond method, and many others who do use it don’t teach it in a fully integrated manner. So for the teachers of the artist mindset looking to find the right method for certain audiences, here’s an outline of the method.

I got the approach from CPM. I don’t know if it originated with CPM, so apologies if the original idea goes back earlier. CPM’s curriculum is insanely irritating: text heavy, lousy examples and insufficient practice. But in many cases, its approach to a topic provides a beautiful, fully integrated, and consistent framework that I steal without shame.

Factoring out common terms

I always introduce the generic rectangle when introducing or reviewing simple factoring (pulling out common terms). The area model uses the fact that the rectangle’s area is both the product of the length and width and the sum of the individual areas. You break up the side of the rectangle into as many different segments as there are terms.

So 8x + 18 is the sum of two areas, both created by a product of length and width. One side is used for both areas, so it must be a factor common to both areas. In other words, what is the greatest common factor of both terms?

Once you find the GCF, work backwards. What do I multiply by 2 in order to get 8x? Most students do well on this, but if they struggle, I show them how to divide in order to find the answer.

I don’t stress its use here, as I do during binomial multiplication; I just want students to be familiar with my use of it. At the same time, I always find a few students who struggle with factoring common terms and find the approach helpful.

Binomial Multiplication

I do not mention FOIL, although most of my students have learned it at one time or another. While I don’t require my students to use any particular method for tests, I require them to use the area model method for binomial multiplication at least for a day or two so they understand the underpinnings of the factoring method.

So obviously, binomial multiplication is the opposite of factoring; the terms go on the outside of the box and generate the individual areas. (x+2) is a segment of length of x and 2, (x+3) a segment of x and 3. I always point out that the lengths don’t need to be accurate or drawn to scale.

I demonstrate this method several times, up front. I explain the area concept again and how multiplication of length times width for each smaller rectangle is the same as the area of the larger rectangle. I don’t really expect my students to be able to repeat it back to me. What I expect, or hope, they will think is “Oh, okay, that makes sense”. Because from this point on, when they think of this method, I want them to remember that the method made sense to them, even if they don’t remember the specifics. That’s also why I don’t write any of this down—most of my students will toss any documentation, anyway. I work a variety of examples (at this point, a=1), picking students to walk through the process.

The kids have a handout with 20-30 problems (this is one of the few topics that kuta software doesn’t have a good handout for), but I don’t have my usual handout online. The original problems would all be a=1, b and c of all signs, because I want them to work dozens of problems and see the pattern, if they are able to. Then, on day 2, 3, and 4, I introduce difference of two squares (what happens to b?), a>1, and 2×3 or 3×3 polynomial multiplication—which works really well with this model, as the kids just make a bigger rectangle.

I wish I could say that this method eliminates the problem of (x+2)(x+3) = x2 + 6. Alas. However, when a student makes the mistake and I scowl and draw the rectangle, with no other explanation, 9 out of 10 kids making the mistake go “Oh, yeah.” That’s the win, such as it is.

Factoring Trinomials

So after a week or so of multiplication, I point out something interesting about the completed rectangle:

This is particularly interesting when we consider the two “middle” terms that add up to bx. We now know that they add to bx and multiply to the same product as ax2 and c.

Interesting, yes, but also useful. I remind the kids that distribution is the inverse of factoring, that distribution converts a product into a sum, while factoring turns a sum into a product. So if they are faced with a quadratic equation in ax2 + bx + c—say, for example, x2 + 9x + 14—how could they turn this sum back into a product?

I ask the kids, if I’d multiplied two binomials to get x2 + 9x + 14, what would have been in the box?

Factoring trinomials is the task of finding the numbers for the other half of the rectangle.

And thanks to the properties of the generic rectangle, we know that the terms we are looking for add to 9x and multiply to 14x2, the product of the first kittycorner.

So we use the “diamond” as a visual tool to help find those terms.

No matter what method a teacher chooses to teach, factoring comes down to that question: What do I multiply to get ac that I add to get b?

I teach the students to write out the factors in pairs, starting with 1 and the number itself (otherwise they tend to forget) and working up from there. Remember that I teach students who will often have a tough time remembering all the factors of 24, and pause on each term to remember the pair.

So once you find the terms that meet the requirements, you put them in the box. It doesn’t matter which goes where. I repeat that phrase a lot. I sometimes wonder if I should create a rule for where the terms go, just so I won’t get the question any more.

It’s worth stressing to your students that, while you’ve found the missing terms, you still have one more step! They’ll still forget, and this will bite them back when they start factoring trinomials in which a>1.

The last step involves finding the GCF for each row. This is where I get the payoff for introducing “single row” factoring much earlier. The students are familiar with the process; they’ve seen me explain that the outside terms are the GCF for a month or more, even if they didn’t use it themselves.

Again, I work five or six problems with the class as a group each day. The kids have a page of 20-30 problems they work through; if they finish one page, I give them another with more complex problems. Anyone who can do the work peels off from the class discussion and works independently from the beginning, the rest are “released” after the class discussion. I put worked examples all over the whiteboards to give them models to follow. Many of my struggling students don’t move past a=1. Some of the weakest will only be reasonable competent at c>0 in the first go-round and struggle with finding the difference of two terms for a while. So over the next two-three days, the kids work on factoring at their own pace. The strongest kids are working a>1 by the last day (and their third page of problems), and working problems like x2 – 9x = 10, learning to set it equal to 0 and factor.

Here’s an example with c<0:

Here’s a>1—and this, by the way, is where anyone can benefit from the generic rectangle. Any other method of factoring a>1 trinomials is a pain in comparison:

I return to factoring throughout the year. Every so often I’ll declare it time to build on existing skills, so kids who had just gotten competent at c>0 can get more practice time on c1, and then the strongest kids start to identify patterns—identifying perfect squares, difference of two squares, and so on.

As time goes on, I give fewer worked examples and just the general outline below, to see how they do at moving from general to specific:

Next Steps

I have traditionally gone from this to completing the square and quadratic formula, then onto graphing parabolas. I am going to reverse these two topics this year. Teach factoring trinomials, then graph parabolas. Get that going well, and then move onto completing the square, quadratic formula, and then graphing those cases. See how that goes.

Finally, I can’t stress this enough: a quarter or more of my algebra classes are low ability kids, so if you’re thinking Jesus, two weeks or more for multiplying and factoring quadratics? then you aren’t teaching low ability kids or you’re just ignoring the fact that they’re flunking your class. My top kids are doing in depth work on the topic or, in some cases, moving onto another topic entirely.

I’ve been getting some people lately asking, or complaining, that “low ability” is vague. I’m sorry, but it’s not. Potter Stewart was right: you know it when you see it. If you want a specific metric, it’s a kid with cognitive abilities measured at the 50th percentile (say, IQ from 95-105, but that’s a guess). In other words, kids that are perfectly functional in the real world, but simply don’t have the interest or ability for advanced math. Kids with cognitive abilities any lower than that aren’t, as a rule, going to be able to even fake it in algebra, much less anything past that. There are always exceptions.

It’s the delusion of eduformers and progressives, one and all, that if teachers find the right approach, a low ability kid is transformed into a competent high ability kid. In reality, success in teaching low ability kids comes when they start to feel a sense of competence at some level of math. When a kid goes from staring blankly at a trinomial to thinking “Oh, yeah” when I draw the rectangle, that’s a big goddamn win. I believe a lot of kids in this category could learn specific high level math in the context of a concrete task, although I have no evidence of this. But we’d have to sort kids into different groups and sorting’s just one big no-no.

However, this method is helpful for kids of all abilities. High ability kids get a real kick out of seeing the link between the area model and the algebra, and I’ve rarely met a kid who didn’t appreciate the utility of this method for a>1.

I don’t have a handout per se for this whole method; what I’ve just laid out is 8-10 days of practice, followed by days interspersed here and there throughout the year. However, if I get a kid who comes in late, or who wants a specific tutorial, I have a document that I really need to rework, which is why I spent some time creating images for this writeup. But remember, all of this is religion and on factoring, I’m in a state of epistemic closure. Convert or live life as a heretic. I was going to say “Die, infidel”, but really, the current insanity in the mideast takes all the hyperbole out of that statement and thus all the fun.

*If you are a mathematician who is also a teacher, stop hyperventilating. It’s true. You know it is. Embrace the reality we live.