Ok, so I don't have another anecdote from my grandson today so I'll have to get back to mathematics -- problem-solving, teaching and learning.

The video below deals with an algebraic equation in 2 variables which should be straightforward for your stronger Algebra II or Precalculus student. But will it be? I invite you to predict how many in your classes will answer it correctly, then try it out. After all, it is multiple choice, so statistically some should get it right by some means or other!

Hopefully, the purpose of the problem and the video will become clear to all of you. If we want our students to demonstrate better reasoning and an understanding of important ideas in math, we need to feel comfortable in teaching for meaning and understanding. This doesn't mean we stop teaching algorithms and procedures, however. Exactly what all this means and how to do it is the reason for this blog. I certainly never claimed to know the answers or any other mystical secrets. I only know that I never gave up trying. Sometimes my efforts failed miserably, but I hopefully learned from these attempts.

It would mean a lot to me if you share your thoughts here or on my You Tube channel, MathNotationsVids, where you will find my other videos.

Note: Another subtle point I should have made in the video---

y(x-4) = 0 → y = 0 ** OR** x = 4

It is important for us to stress this point and distinguish it from "

**" logic. If the equation were in the form: y**

*AND*^{2}+ (x-4)

^{2}= 0, we would have (y = 0)

**(x = 4), whose graph would be the single point (4,0). Another instance where an exercise on the board can lead to a rich, fruiful and profound discussion. If all of this is seen as taking too much time away from content, remember this is precisely the kind of change in curriculum and instruction that Prof. Schmidt has been trying to tell us about for over 15 years! Well, I'm preaching to the converted, aren't I...**

*AND*-------------------------------------------------------------------

*"All Truth passes through Three Stages: First, it is Ridiculed...*

Second, it is Violently Opposed...

Third, it is Accepted as being Self-Evident."

- Arthur Schopenhauer (1778-1860)

You've got to be taught

To hate and fear,

You've got to be taught

From year to year,

It's got to be drummed

In your dear little ear

You've got to be carefully taught.

--from South Pacific

Second, it is Violently Opposed...

Third, it is Accepted as being Self-Evident."

- Arthur Schopenhauer (1778-1860)

You've got to be taught

To hate and fear,

You've got to be taught

From year to year,

It's got to be drummed

In your dear little ear

You've got to be carefully taught.

--from South Pacific

## 3 comments:

I'll try this, and get back to you. But in the meantime, why isn't "hyperbola" one of the choices?

Actually, I'll probably ask for the graph without offering choices.

And then, assuming that it goes relatively quickly --

(I'll get lots of single line kids, a fair number of stuck-from-the-start kids, a few oddball errors, but I predict I'll get 2 - 5 good graphs in each Algebra II class. Recall, I do this sort of thing from time to time) --

I'll ask them to help construct five multiple choices.

Let's see if I get to that.

Jonathan

Thanks, Jonathan.

I like the idea of having the students construct their own choices. Let me know what they come up with.

I didn't include 'hyperbola' as a choice because

(1) I was writing this question for many levels of students including Algebra 1

(2) The term 'hyperbola' could appear on the Math I/II subject tests but has never appeared in any SAT question.

(3) Technically, a pair of intersecting lines can be thought of as a special case of an hyperbola, so I didn't want to deal with that issue.

Interesting, if we consider any two non-vertical intersecting lines of the form y = ax+b and y = cx+d, where a ≠c, we produce the following quadratic:

(y-(ax+b)) (y-(cx+d)) = 0 →

acx^2 - (a+c)xy+ y^2 +... = 0

In this case, the discriminant, B^2-4AC =

(a+c)^2 - 4ac which has to be positive by the arithmetic-geometric mean inequality.

Therefore, it can be thought of as an hyperbola. If one of the lines is vertical, we can similarly show that the resulting quadratic falls into the same case!

Also recall that for a Monday challenge a few weeks ago I asked for the graphs of xy ≤ 0, (degenerate hyperbola) and |y| ≤ |x|, also a degenerate hyperbola.

But because of the shifted NY State exam, I flew through conics as a concept, with nothing to calculate beyond what they already knew about circles and parabolas, so they have not seen many xy terms at all.

Jonathan

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