Nov 062014
 

I haven’t posted in a while, mainly because I am just so happy with how my classes are going. I will focus on Alg 2 here, because these awesome learners are just knocking my socks off.

I am in the polynomial unit, knee deep in graphing, and increasing, decreasing, relative mins, relative max’s, absolute mins, etc. This is the problem set we were working on today in class:

problemset

Here are the questions I ask (docx format) for every single graph, from lines all the way through sin & cos at the end of the year.

Yes, some of these are going to be Does Not Exist. That is okay. Just because we don’t need to think about asymptotes with cubics does not mean we shouldn’t ask about them.

A little back story before I say something about my learners. I used to teach the textbook. I admit it. I sucked, horribly. My learners did not connect anything with anything and they did not see how to connect topic from one unit to the next. I was frustrated. So I first came up with my list of functions in (h,k) form, wrote it on my board and changed how I approached algebra.

2014-08-10 15.43.46

That was a win. But, then I was frustrated because every time I changed the graph, added an exponent, I had to teach a new set of vocab, but everything was the same; so why was I teaching new stuff? Why couldn’t I teach all the vocab up front, and then just explore the heck out of each function family?

Short answer was, I could. So, I did. That is where the form above came from. I introduced it last last year, and used it and modified it and tweaked it and the learners responded.

Enter this year, this class. I have everything set on day 1. We entered the year thinking about connections and planning our math and discussing end behaviors of lines (wow, that was easy, hey, they are always the same!, etc). Then quadratics, and we completed the square to get vertex forms, and we factored, and saw how intercept, standard and vertex forms were all the same function, and and and.

Enter polynomials.

We have done them from standard form, and done the division to get intercept form, we have broken these guys down every which way. I have tossed them fifth degree and fourth degree polynomials, they didn’t even blink. “Oh, so this just adds a hump to it.” I have explored more in polynomials this year than ever before.

And, since it is a constant review of prior material (“If this works with quartics, will it work with quadratics too? Yes”) I am constantly cycling and eliminating the mistakes my learners made in previous sections and on previous exams.

Which brings us to the problem set above. That is a killer set. The 4th and 5th are tricky, and they struggled. Until one of the class members said, “Don’t all we have to do is distribute them and so it is just a bigger distribution problem?”

Done. And. Done.

Now, of course there is a nicer way to do it. Substitute “u” or some other variable in for (x-3) in the fourth problem so you are multiplying binomials first. It saves time. BUT, it was not necessary to show it. They know distributing, so distributing is what works and they rocked the socks of of it.

So, why have I not been posting much? Because I have been enjoying the heck out of teaching. These learners are taking these ideas and running with them.  And I love it and them.

Jul 292012
 

Many people ask me why I ride my motorcycle long distances in the summer. This summer I traveled from Reno, NV to St. Louis, MO. It was around 4000 miles, round trip, and brutally hot for a couple of states worth of riding.

But, that traveling allows me one single thing I rarely get. Time away from all distractions. It worked. I thought long and hard about the problem I talked about last post; A visual representation to imaginary solutions of quadratics. Somewhere in Wyoming I had the idea on how to prove it. By the time I hit Utah, I had the solution worked out in my head, and I needed to jot some notes. It honestly took me several hours to type up the solution, and without further ado, here it is.

graph5

The Goal:

To prove that in a general case, the circle that is created by reflecting a parabola with imaginary roots (the orange one) about its vertex (the black one) will have as its radius the value of the imaginary roots of the original.

We will begin with clip_image002[12] as our initial equation, with one requirement that the discriminant is negative;  clip_image004[10]. This will ensure that our initial quadratic equation has imaginary roots and the parabola exists above the x axis as shown.

 

Now, we need to reflect this equation around the vertex, but just adding a negative sign in front of the “a” will not do it. If we add that sign in and make it “-a” it will reflect around the x-axis, not the vertex. Therefore, we are going to need to complete the square, get the original equation in vertex form and then add the minus sign to reflect.

 

Given equation                                                                       clip_image002[13]

 

First, divide all terms by “a” and set the y = 0                              clip_image006 

This gives us a first coefficient of 1, which makes

Completing the square possible. Next, we will complete

the square by using clip_image008 and its square.                        clip_image010

 

now that the perfect square trinomial has been constructed       clip_image012

we can factor the trinomial into vertex form.

 

The center of the circle above can be clearly seen in this form, and is: clip_image014 We will need this later.

 

Now we need to solve the reflected parabola for x.               clip_image016

 

Add & Subtract the constant terms from both sides to get:          clip_image018

 

Move the negative sign from the right to the left side:                 clip_image020

 

Take the square root of both sides:                                       clip_image022

 

Finally subtract the constant term from both sides:                  clip_image024


Notice that we have essentially derived a version of the quadratic formula. It doesn’t look exactly like the standard version we all memorize, but it is the same, with one important difference. There is a sign change to the terms inside the radical sign! That will be very important.

 

This formula gives us where the reflected parabola crosses the x-axis, so we now have 2 points on the circle, the plus and minus, and the center of the circle.

 

The final step of the proof is to show that the radius of the circle, or to put it in another way, the distance from the center of the circle to one of the roots of the reflected parabola, is identical to the imaginary part of the solution / roots of the non-inverted parabola. So, onward to the distance formula.

 

We need to find the distance from clip_image014[1] to clip_image026.

 

Distance formula:                                            clip_image028

Insert the point values for x and y       clip_image030


Using just 1 of the 2 values for the + or -.

 

Simplify the subtractions:                                               clip_image032

 

Finally, square the inside term leaving the following:             clip_image034

 

This leaves us with a pseudo-determinant of:                     clip_image036

 

 

However, in setting up the problem initially, we stipulated that the determinant clip_image038 would be negative. If that is true, then the value of inside the radical sign in our last step must be positive!

 

[And yes, I am cheating. I am leaving it to the reader to show that the way it is written above in the last step as the “pseudo-determinant” and the regular determinant are essentially equivalent.]

 

Not only that, but the value of clip_image040 which is from our inverted quadratic, is the same value but opposite sign of the more familiar clip_image042 from the quadratic equation.  If clip_image038[1] is negative, our inverted quadratic will be positive with the same value (oh, and it works in reverse too!)

 

There, I now proved that the reflecting a parabola with imaginary roots around its vertex will allow you to calculate the imaginary part of the complex answer as the radius of a circle created by the reflection.

 

QED.

Jul 272012
 

For today’s #myfavfriday I am presenting an idea that has been percolating in my head for a while. If you want to know what a #myfavfriday is, then see Druinok’s blog here.

MyFavFri-t

Learners have a devil of a time with quadratics. Afterall, there can be 2 solutions, 1 solution, or no solutions in Algebra 1, and then in Algebra 2, we come at them with the fact that those equations with no solution really do have 2 solutions after all, they are just “imaginary” (could there be a worse name for them, really? Thanks a lot Descarte.”)

But I came across a picture on some site one day, and it has stuck with me. I never bookmarked it, or wrote down the site, so it is lost to me (and I have searched hard for it) but the work blew my mind, and as I have shown it to learners, they have at least gotten a sense that the “imaginary” really does have meaning.

Let’s begin with 2 equations and graphs that are simple, straight-forward and make sense. [all images are clickable to see full size]

graph1and graph2

The equations are y = x^2 – 4x + 3 and y = x^2 – 4x + 4.

A simple change of one number changes the number of solutions from 2 distinct to 2 repeating solutions, and learners don’t have a problem with that idea, generally. Then comes this bad boy.

graph3 y = x^2 –4x + 6

Now they have to do the whole Quadratic formula on it to get the solution, and the solution has those i thingies in it, which makes them all confused and irritated until they wrap their heads around it. And why does it still have 2 solutions? It doesn’t touch at all!

But wait! We can play a game with this quadratic function. What if we reflect the parabola around the vertex in the downward direction? Then we end up with something that looks like this:

graph4a To do this reflection, we first had to complete the square on the original equation to get y = (x-2)^2 + 2. Now, with this equation, we can put the – sign into the equation and get the reflection, y = -(x-2)^2 + 2.

But hold on, see those 2 points where it crosses the X-axis? And see the Axis of Symmetry that goes through both equations? If we use those three points as definitions for a circle, we get the following graph and equation.

graph5 (x-2)^2 + y^2 = 2

Guess what the solution to the quadratic equation y = x^2 –4x + 6 is. If you guessed 2 + root(2)i and 2 – root(2)i  then you are absolutely correct.

The real number part of the complex solution of a quadratic with two imaginary roots is the X value of the Axis of Symmetry, and the imaginary part of the solution is the radius of the circle created by the center and endpoints created when the inverted parabola crosses the X-Axis!

Okay, mind blown. Why? How could I prove this?

Aha! now come into play the hours I spend on a motorcycle every summer. How could I PROVE that this will always work? I have the proof. I am working it up, but it is a pain to type. That, I think, will be the focus of a future, #myfavfriday!

[And I really need to look in to a LaTex module for my blog if I am going to do math. The equations look horrible.]

Edit: 29 July 2012: I proved this assertion, at least to my satisfaction in a followup post: http://blog.mrwaddell.net/archives/348

Edit: 4 August 2012: I found, stuffed in the bottom of my backpack, a rumpled piece of paper with this link on it. I think I did this page justice with my treatment. I wish I had found the page before I spent hours thinking about how to prove it, it gives the suggestion right there at the bottom!

Edit: 18 Dec 2012: @Mythagon posted this picture on Twitter. It is a great visual of what is discussed above, and clearly shows why the rotation is so important.

From: Teaching Mathematics, 2nd edition by m. Sobel and E. Maletsky

Edit: 27 Sep 2015: Wow, a long time since the original post, however I still come back to this every year. Love it. Now, Luke Walsh, aka @LukeSelfwalker added this to the mix. Love it. Click it for the live Desmos file.