First off, I wanted to build a site that had totally separate parts. By this, I mean, I wanted a “Work” site that would allow me to open up links I used almost daily, without typing in the links, or keeping track of them. This is purely a selfish, ‘spend an hour building a site, so I can save seconds every day’ type of thing. But it works for me.

So my “Work” site needed a menu structure that was self contained. I wanted to be able to share the work site with colleagues and open up the work page in front of students, without giving away personal information like where I shop, or where I bank (that is a private site). I also wanted a public facing site for my blog and an about me page. Maybe there would be more there in the future, but for now, those are the most important things on that branch.

So that means I wanted a Public group of links, a Work set of links, and a Private set of links. It is all built on the same platform, but none of them are cross linked or interconnected.

The help above says it is a “Hybrid” navigation, but it was not clear. So, on to the learning!

Here is what I figured out.

In my _quarto.yml file, I have the following code:

  sidebar:
  - id: public
    style: "docked"
    contents: 
      - text: "Home"
        href: index.qmd
      - text: "About Me"
        href: about.qmd
        
  - id: worklinks
    style: "docked"
    contents: 
      - text: "Work Links"
        href: work.qmd

    - id: private
    style: "docked"
    contents: 
      - text: "Personal Links"

I have more than those few links, but the important chunk is the “id: public” or “id: worklinks” line. Those lines define all the links below as part of the menu I define.

Then, in the page I create, I have the command “sidebar: public” in the top Title Block. This is the actual sidebar used for all of the blog pages, and tells Quarto to add the particular menu structure I want to the particular page.

So how does it all work in reality? Pretty sweet.

This is the public menu, and the “Home” words will take a person always to this page. I circled the “Home” on the menu bar, because that takes me to my private menu structure (which I am not showing here because … well … private.) So, I can access any menu structure I want from any page, as long as I am logged into the computer and my settings are saved. If I am on a public computer, I am not using my private links anyway.

And from my private menu structure, I have a work link, which takes me to my work page, or I can just type in the link when I am in a classroom and need my work links.

You can literally have as many sidebars as you want for your site / pages. All you need to do is build them out in the _quarto.yml file, and give them a different “id: name”. Then use that name in the Title Block to tell Quarto to use sidebar menu structure A, instead of B.

Voila!

It took me quite some time to figure this out. It is really easy, and much easier than the Quarto help makes it seem.

I hope this can help someone else along the way. Drop me note at BlueSky if it does.

One of the ideas I like to play with often is the “What If-Not” approach to questions in mathematics. This formulation was created by Brown & Walter in their book “The Art of Problem Posing” now in its 3rd edition. It is a fabulous book that all teachers should read (in my opinion).

Anyway, applying this to the quadratic equation in standard form, we play this game often in class. What if we vary the ‘a’ value? What if we vary the ‘c’ value in the equation?

But we almost never ask, “what if the ‘b’ value is not constant?”

Have you ever done this? It has an extremely surprising outcome! Go over to Desmos and do it now. What do you see the vertex of a quadratic doing? Hmm, it is a little hard to see right off the bat, but let’s do some math and see if we can figure out what is going on.

If we have a quadratic function, then we can complete the square to find the vertex. I am going to skip over the large amount of typing again, but you can go to this post and see the steps

If we jump down in that link, we will see the vertex form of the function is:

This means the vertex of the function is:

$$ vertex = (-, f (-) )

$$ Easy enough to put into Desmos to graph, but let’s look at that “k” value for the vertex. We know that in translation form, the x value of the vertex is -h, and the rest of that function above is the k.

So if we do a little algebra on this, we get:

And look at that! The k value is a reflected quadratice in terms of a, b, and c! But that means that the vertex of the function that the vertex traces is also a quadratic!

And, in fact, if we let then substitute that into the right side of the k=, we get a new function

So let’s plot this in Desmos and see what happens!

Sure enough, when we vary the ‘b’ value in a quadratic in standard form, the vertex of the original function traces out a new quadratic function.

So of course, the next question is, “What happens if we do this with a cubic function or other polynomials?” My advice is to read this paper first: Algebraic Number Starscapes by Edmund Harriss, Katherine E. Stange, Steve Trettel

It is 59 pages long, with some amazing proofs, images, and discussion. What happens is we end up with some open questions that are not fully solvable! How awesome is that. The quadratic is easily worked out and clearly explained, but the cubic and higher polynomials end up with open questions.

This is the beauty of math! Love it!

This will also end (for now) this discussion of quadratic for me. This makes 7 different posts on the topic. I think I have gotten all my thoughts out for now.

The specific post I that inspired this extension on quadratics is his post on Pythagorean Parabolas. I am not going to duplicate what he said, but show some connections that I found truly interesting.

First off, let’s chat about the really simple way to build Pythagorean Triples he mentions. Honestly, I had not heard of this method before. (I also have to admit, I taught everything except geometry in HS, so that is probably why).

Take any odd number. Say, 3 for example. Square it, so you get 9, divide by 2, so you get 4.5. Now, do the floor and the ceiling on 4.5, and you get 4, and 5. The Pythagorean Triple is 3-4-5. There is an additional interesting thing with this series of triples, and that is because we do the floor and ceiling, the hypotenuse and the other long side will always have a difference of 1.

If we graph the two short sides, and ignore the hypotenuse, so in essence, the short side is the x value, and the long non-hypotenuse is the y value of a triangle, we get the following graph:

Looks a lot like a parabola! In fact, it is. It has an equation of:

Which makes sense, given the definition of a parabola using the focus and directrix. Pat puts it this way:

If we think about the definition of a parabola as the set of points equally distant from a focus and directrix, we realize the line of the directrix must be the line y = -1 so that, for instance the point (3,4) which is 5 units from the origin/focus will also be 5 units away from the directrix. 

So if we multiply all the original triples by 2 (so our first triple is 6-8-10) we get a triple that has a difference of 2 between the two longest sides, so the equation should be:

and sure enough!

But, there are other ways to create Pythagorean triples.

Multiplying the primitive Pythagorean Triples by 3, gives us an equation of

Does this always work for every Pythagorean Triple though? That is a very interesting question, because we used a very simple, primitive way to create the triples, but there are other ways.

For example, there is this way to create the triples. Take any two numbers, such that

.

So, for example, we can use consecutive Fibonacci numbers (for fun, why not), we get the following Pythagorean Triples.

3-4-5! Yay, that matches what we got before. But then it explodes.

5-12-13 , again, a simple triple, found above.

16-30-34, this is not a multiple of a primitive triple as above. but it is a multiple of 8-15-17. The 8-15 is on the ‘times 2’ parabola, while the 16-30 is on a ‘time 4’ parabola, .

39-80-89, This Triple is the first one that is different! It IS a primitive Triple. So, does it follow the pattern found above? Is it a “difference of 9” parabola? Yes! It is!

It seems the generalization of the equation is: where n is the difference between the hypotenuse and the next longest side, AS LONG AS IT IS A PRIMITIVE TRIPLE!

Let’s test it on this next triple using Fibonacci numbers.

105-208-233 ; This is a primitive Pythagorean Triple, as the 3 numbers are co-prime. The difference between the last two numbers is 25. So, the function that would create the parabola this triple should be:

Yes!

It works. Also, if you substitute 105 into the function, you get an output of 208.

Which begs the question of Why?

I don’t have that answer, nor do I have a proof. Right now, I have a conjecture that for every primitive Pythagorean Triple, such that , and where , if then the point will be on the parabola described by the function .

Edit: 24 October

Proof of my conjecture:

Shoutout to TrigOrTreat over on Mastodon who pointed out that if I take my statement that and reframe it as and insert it back into the Pythagorean Theorem and solve for b, the conjectured equation for the they parabola is an outcome.

Not a hard proof of the conjecture at all.

Pythagorean Triples are really interesting things, and the connections to quadratics is also fascinating!

Edit: 21 October

As I was exploring how to prove this, I came across this interesting video by 3Blue1Brown on calculating triples from complex numbers. Um, WOW! Every Pythagorean Triple can be thought of as the square of any complex number with integer components. (definitely watch the video. It is 14 min well spent)

So, for example, if you take the complex number and square it, the magnitude of this new complex number is ! And the Pythagorean Triple is (9,12,15).

When you do a whole bunch of these, and rotate the axes to show the new lattice points of the coordinate system, it looks like this:

This does not answer the question of proving the general equation above, but it does demonstrate the fact that the equations for these are quadratics. Still have a way to go in my proof, but this is a really interesting connection between complex numbers and Pythagorean Triples.

It also shows why one formula for finding Pythagorean Triples is the one given above. It is the way to find the new magnitude of the hypotenuse of the two vectors in imaginary space. The video does a wonderful job showing it and demonstrating the proof.

In talking to teachers, I have been told a variety of techniques to teach the transformations of different function families. The one thing that I realized is that there is no one way that teachers use to teach this idea. There is, however, and efficiency to be found in the way I approach this topic.

On day 1 of my classroom in Algebra 2, this is what the learners saw on the board, and we used this every day in class.

Sometime in around day 3 or 4 of the school year we would do a “what do you notice” and “what do you wonder” exercise with this, and we would build the ideas of similarity of a and (h,k) actions, and inverses of functions.

I leaned heavily into the similarity of how a, h, k, (and b, but that is typically outside of the Alg 2 curriculum and into the Trig/Precalc curriculum). I had a rule for myself in my teaching, which was: “Teach something once, use it every day.”

So for translations, I would teach it once. I used Desmos to assist with this, and we used a type of this calculator file. It has ALL the things. I would start it quite simply with this:

and work up to all the functions.

This is messy, ugly, and not clear from this image, but if you use the sliders, you will see that all the functions will move together. Rational functions, transcendental functions (sine, cosine, tangent), exponential functions, logarithms, as well as all the polynomial functions all work the same way.

Teach it once, use it every day.

This is essential to my teaching philosophy and approach, but it also saves so much time. We can talk about “what do you remember?” and challenge learners to dig into their notes and memory to reconnect knowledge. “Make a prediction as to what happens when I do this, based upon what you learned earlier,” and learners will be able to use what they learned, instead of learning something new.

The translation form of all the functions isn’t always useful, but it serves the purpose of tying all the different functions we talk about together, so that it is clear that as we are learning one idea or topic, we are not forgetting the other ideas we learned.

Everything is connected. We just have to spend a little bit of time thinking about how the topics are connected, and build the connections with the learners.

I am placing this post in the middle of my thoughts and writing on this topic, because this is more of a compilation of ideas from others.

I am doing this because there are so many different ways of solving and working quadratics, and it is so easy to lose these methods if not careful. I want to keep track of all the amazing ideas, and have a repository for myself that can store, but also show my additions to, these ideas.

So, get us started, I want to look at Pat Bellew’s blog, and a post he first made in 2007, and then reposted in Jan 2024. He also has a duplication of this post on Acadamia.edu. What is really interesting about his post, is that it starts with the 18 different methods written about in NCTM’s Mathematics Teacher in March, 1951, by Willian J. Hazard (pdf). Hazard wrote the list in 1924.

Pat, because he is amazing, goes through and gives an example of each one of the 18 methods in his post linked above. It is a great read. I suggest you spend some time doing so.

What are the 18 methods you ask?

1. By factoring by inspection.By
2. factoring after a substitution, z = ax, which leads to z^2 + bz + ac = 0.
3. By factoring in pairs by splitting bx into two terms.
4. By completing the square when a is 1 and b is even.
5. By completing the square as usual after dividing through by a.
6. By completing the square by the Hindu method (“the pulverizer”), i.e. by multiplying through by 4a and adding b^2 to both sides.
7. By completing square as given, adding b^2/4a.
8. By the formula.
9. By trigonometric methods (see Wentworth-Smith,‘Plane Trigonometry’).
10. By slide rule (see Joseph Lipka, ‘Graphical and Mechanical Computation’. John Wiley and Sons, Inc. [1918], p. 11 ff.
11. By graphing for real roots. (All modern textbooks.)
12. By graph, extended for complex roots. (See: Howard F. Fehr, “Graphical Representation of Complex Roots,” ‘Multi-Sensory Aids in the Teaching of Mathematics’, ‘Eighteenth Yearbook of the National Council of Teachers of mathematics’ [1945] pp. 130-138. George A. Yanosik, “Graphical Solutions for Complex Roots of Quadratics, Cubics, and Quartics,” ‘National Mathematics Magazine’, 17 [Jan. 1943], pp. 147-150.)
13. Real roots by Lill circle. (d’Ocagne, ‘Calcu graphique et nomographie’, from which L. E. Dickson got his reference to it in his ‘Elementary Theory of Equations’.) (Also see J. W. A. Young’s ‘Monographs on, Topics in Modern Mathematics’ “Constructions with Ruler and Compasses.”)
14. By extension of the Lill circle to include complex roots.
15. Using the graph of y = x^2 and y = -bx – c to find real roots. (Lipka, ‘op. cit.’ p. 26, modifies and extends this solution; Schultze, ‘Graphic Algebra’; Hamilton and Kettle, ‘Graphs and Imaginaries’.).
16. By extending (15) to include complex roots (Hamilton and Kettle, Schultze).
17. By use of a table of quarter squares. This is a practical method of handling an equation having large constants, as we already have the table in print (Jones’ ‘Mathematical Tables’).
18. By use of “Form Factors.”

Whew, that is a long list! What is awesome about this list is that there are still things unexplained on it. For example, number 18? No clue what that means. Neither does Pat. Professor Hazard just dropped that little tidbit there, and walked away.

Also, number 13, the Lill circle is what I was working with in my last post on Visualizing the imaginary roots! I didn’t know it, but that is what I was doing. Of course, there is a more general description than I did, and Pat does that on his blog as well in a followup post in 2024. In addition, John Golden created a really nice Geogebra display of it. In my post, I assumed that the center of the circle was on the x-axis, so the y-values were zero. That does not need to be assumed, and that is corrected with the Lill Circle ideas.

There are also books which give geometric solutions of quadratic equations. Chapter 19 of this book “Experience Geometry: Euclidean and Non-Euclidean with History” is entitled, “Geometric Solutions to Quadratic and Cubic Equations.” It gives a wonderful history of solving these equations, and explains why negative numbers have been so problematic in history.

I want to pick on number 8 though, “By the formula.”

This is a redo of a post from 27 July 2012. It was lost in the great blog destruction of 2024, but resurrected here.

Learners have a difficult 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 Descartes.”)

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. I will reconstruct that reasoning I saw here, and then prove it below.

Let’s begin with 2 equations and graphs that are simple, straight-forward and make sense.

Respectively, the equations are:

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 trouble maker.

Now they have to do the whole Quadratic formula 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:

To do this reflection, we first have to complete the square on the original equation to get

.

Now, with this equation, we can put a minus sign into the equation and get the reflection,

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.

Guess what the solution to the original quadratic equation is. If you guessed 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. That is easy to see and understand. On to the more abstract.

If we rotate the circle out of the page 90 degrees (or, if we rotate the circle orthogonal to the page) we will now have the circle in the IMAGINARY plane! And the radius of the circle is sqrt(2), which we add and subtract to the center for the imaginary roots (and we multiply by i, because we are in the imaginary plane, not the real plane).

[If we want to get sloppy for a second, we could say that the reflected parabola shows us where those two roots are. BUT, this is sloppy, and useful only to help learners understand that there is a connection between the original parabola and the roots we calculate using the Quadratic Function. ]

Okay, mind blown. Why? How could I prove this is always true in general?

Aha! now comes into play the hours I spend on a motorcycle every summer. How could I PROVE that this will always work? I have the proof.

Okay, on to the proof! (this was originally 2 different posts, but I am combining them since I can. And since I am rebuilding them anyway.)

The Goal:

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

We will begin with:

as our initial equation, with one requirement that the discriminate is negative;

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 introduce the minus sign to reflect.

Given the equation

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

This gives us a first coefficient of 1, which makes

Completing the square much, much easier. Next, we will complete the square by using they standard method, and square of the term. The square is added and subtracted from the same side of the equation. (It could be added to both sides, that is a personal preference.)

From here, we construct the perfect square trinomial with the extra two terms

And we now have the equation in its ‘vertex’ form, where the vertex is (h,k), and the h value is also the value of the axis of symmetry. It is also the x value of the center of the circle we are looking for, and since it is on the x axis, the center is:

We will need this later.

The reflected parabola is easy to find from equation 6, the vertex form. All we need to do is reflect the equation around the vertex.

Add & Subtract the constant terms from both sides, and multiply by -1:

Take the square root of both sides:

Finally subtract the constant term from both sides:

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

Distance formula comes in handy here:

Insert the values for x and y from our two points (thankfully the y’s are both zero), and using only the + value from the plus/minus:

This appears ugly, but the first two terms are zero, the third term is squared, leaving us with:

This must be positive because distances are always positive, so we know that this value is positive as well, so we do not have an imaginary number.

[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 distance and the regular discriminant are essentially equivalent.]

At this point, with some additional algebra, we can see the typical “right” part of the quadratic formula,

We are only missing the “left” part, the location of the center of the circle:

assembling the distance from the center, on either side of the center, of the circle, we get the both distances:

And voila! We have proven that the circle represents the roots of the quadratic! And along the way, have also derived the quadratic formula.

Yay!

And of course, I did not discover this, nor did I come up with this on my own. There was that website that gave me the initial idea, and there are other sources that explain this better.

For example, Francis Su does a clearer and better explanation on his site. This resembles what I described in 2012, but it was not this site or image. Clearly the image and idea has been around a long time. I spent hours trying to figure out how to prove this idea, and Francis gives a better method for proving it right there at the bottom!

There is a great video introducing complex numbers, found here.

And this amazing link pushes the idea into 3 dimensions. Love this!

Also, from Teaching Mathematics, 2nd edition by m. Sobel and E. Maletsky. This image gives a very clear, understandable way to connect this for learners. This book is now in the 3rd ed, and fairly inexpensive to purchase.

Also, in my recreating this post, I had the advantage of using RStudio and Quarto, which means I was able to write this in LaTeX. The first time, in 2012, I wrote all the equations in Word’s editor, and pasted them as images. So much easier just to write in LaTeX.

I want to start the exploration here with an introduction that maybe not every teacher or math person has thought about. We jump to quadratics, and sometimes don’t think about how we build a quadratic function from simpler, more basic functions. For example, look at the image:

I intentionally wrote the two lines in function notation. We have one line with a negative slope and x-intercept of (2,0) and another line with a positive slope and an x-intercept of (3,0). They have an intercept with each other at (2.5,-0.5).

But look what happens when we multiply the two functions.

We get the quadratic! In fact, we get the intercept form of the quadratic function! And, since the two lines intersect, we get the vertex as the x value of the line intersection, and the square of the y value, or (2.5,+0.25).

Of course, this doesn’t work if the lines are parallel or have different slopes. Then the vertex of the parabola is the average of the x-intercepts of the lines, and then the values multiplied.

In this case, the average is 0.5, so the y value of the vertex is found at f(0.5)*g(0.5), so the vertex is (0.5, -6.25).

Yeah, yeah, yeah, we have other tools to find these same values, but the lines literally give us the answers! A quadratic function is constructed from two lines multiplied. If we want to connect lines to the rest of the algebra curriculum, start here.

I also want to take a moment to reinforce some important language. I have tried to be careful in my explanations above, and use the appropriate language of roots, zeros, solutions, and intercepts. The language often gets muddled, and I know some teachers have even called these items “ZIRaS” which stands for “Zeros, Intercepts, Roots and Solutions” because they “mean the same thing.”

I want to argue they don’t. Why would we have four different words for the same thing? As math people, we would not. If we have four words, it is because they four things mean something different.

Sarah, at the blog “Math = Love” has a great post on this. She even has posters you can download.

The following is the correct usage of the terms:

Expressions have Roots: (x-3)(x+2)

Equations have Solutions: (x-3)(x+2) = 0

Functions have Zeros: f(x) = (x-3)(x+2)

Graphs have x-intercepts: see the graph above.

But wait, you say. The answers to each of the four different things above are the same values. Not quite, I say. The x-intercepts are points. (3,0) and (-2,0). That matters. A lot. We get them by looking at the graph, and seeing where the parabola crosses the x axis.We get that from the graph of the parabola (the actual shape of the quadratic function).

The zeros of the function are numbers that when substituted into the function produce a zero output, ie.where f(x)=0. In this case, the two values are x=3 and x=2. If we are not careful, we could assume that we just described two vertical lines, but we did not. We described two values to substitute into a function.

The solution is the values we get by using the zero product property, after setting the equation equal to zero. We set each of the two products equal to zero and solve, (x-3)=0 and (x+2)=0, which results in two answers, x=3, and x=-2. Again, if these were on a graph, they would be vertical lines. We are not graphing.

Finally, the roots. This one is more abstract to explain, because the roots are simply (x+2)(x-3). The roots are the factored form of the quadratic expression, and it isn’t equal to anything, so we can’t set it equal to zero. If it was equaling zero, it would be an equation!

This post sets the stage for the rest. I am going to be careful in language, but also push some boundaries into different aspects of quadratics. Let me know on Bluesky or Mastodon what you think. I haven’t figured out how to incorporate comments here yet.

Just look at it. It is so wonderful to view and behold. But the simplicity hides some really interesting features. One reason it is so interesting, is that it is one of the few functions that has all three forms for every function: an intercept form, a standard form, and translation form. Not every cubic function can have all three forms. See! Already some interesting features. With that in mind, onward!

In this post, I will give an overview of several different connections I have found over the last few years, and then dive into each one of them in a separate post. There are so many interesting things to explore. Below, I will list some interesting “discoveries” and connections. I put “discoveries” in quotes for a reason. These are all personal discoveries, not discoveries that had not been discovered before by much smarter people than I. This is just my path to rediscovering some really interesting things.

First, an introduction to quadratics. We often, as teachers, say that it is the squared term, but really, it is the fact that we multiplied two lines together that makes the squared term! See, THIS is why we need to really understand lines well. If we understand lines, then we build quadratics and parabolas out of lines! There is also a difference among the terms ‘zeros,’ ‘roots,’ ‘solutions,’ and ‘intercepts.’ I dive into those too in this post.

I recreate a post from 20212 where I uncovered that the imaginary roots of quadratics “are” the real roots of the same quadratic that was flipped over its own vertex. I put “are” in quotes, because it really requires rotating a circle into the imaginary plane from the real plane. There is my own proof of the idea in this post, but of course, there are other, easier proofs. I made it too complex.

Next, I follow up on a post written in 2007 by Pat Bellew. He writes about 18 different ways of analytically solving for the roots of quadratics. I didn’t see this in 2007, but instead read the 2024 version of this post. Amazing! A very interesting dive into the topic. I am expanding on a couple of things in this post and adding 2 more methods. One of the things I add to Pat’s amazing post, is a different way of teaching the quadratic formula. I worked with some international mathematics teachers who were here on the Fulbright Teaching Excellence program. They introduced me to a different (and I think better) way to teach and understand the formula. There are also two different methods introduced at the end.

Pat also did more with this topic. Also in 2024, He connected the idea of Pythagorean Triples with quadratics. I extend his idea here and generalize it for any Pythagorean Triple (including those found from Fibonacci numbers.) Did you know that every Pythagorean Triple is found as an integer point on a parabola? It’s true, and I prove it.

And there is more to consider. In the Alg curriculum, we dive into the translations of functions, and show that all of these translations are the same. Well, we should approach it that way. I know not all teachers do, but they miss out on making some fabulous connections if they don’t do them all together. These translations are connected to the first post.

Lastly, what if we vary the ‘b’ value of a quadratic in standard form? We know, and teach, what happens when the ‘a’ is varied, and what happens when the ‘c’ is varied. But what happens when the ‘b’ value is varied? Something interesting and exciting! I will not spoil it here, but I found it really interesting.

Ungrading as a teaching practice

I am not sure what the end result will be, but if the response from the first two days of classes is any indication, it will be positive. Ungrading is a practice where the instructor does not give grades to the assignments, papers, etc. In fact, you can go far enough that you can ungrade the entire course, and require the learners in the course to have a sit down discussion with the instructor on the final grade!

Guess what version of this I did? LOL.

Of course I went all the way. I explained where I got this idea from in my previous post. The Ungrading idea is really interesting. All too often, learners simply look at the final grade on an assignment, and call it good. Some will complain if they get a lower grade than they expected, but the majority of learners just take the grade they get and move on.

In fact, when I brought up the question of whether learners actually read the comments on assignments most of them said “no, they just look at the grade.” When I asked, “the why should the instructor give any comments if the majority of learners don’t read them?” they had no answer.

The end result then is to ONLY give comments, and not grades. Break the chain.

So far in the first week, almost all learners were receptive and positive. There were a couple of learners who were still freaked out, and I asked them to talk to me privately and we will address their concerns.

So far. So good. We will see though!

True story.

If you want some details on why and how to do it, see this book:

In addition, I am asking all the learners to read this article entitled, “Redefining Grades as Feedback, Not a Finish Line” by Donati et al. (2024). This will set the stage (hopefully) for a discussion on the purpose of grades, and what grades mean to the learner and do the instructor. Tomorrow is the first day of class, so we will see how it goes!

Thanks for coming along on this journey with me.

Starting over, but not from scratch

I had a huge problem with Wordpress and themes. A theme malfunctioned, corrupted my databases, and after spending a couple hundred dollars on Upwork, it continued to fail me.

It was not the problem of the people at Upwork. They were fabulous. They fixed the first problem I was having, and then several days later, a new problem cropped up. They fixed it, and then a couple of weeks later, and fatal problem occured.

At that point, I just gave up. Mainly because I learned about Quarto, and its integration with R. I use R for research, and knowing that I can write in R, using Quarto to publish the site and blog, means that I can control everything, write in HTML, CSS, and, most importantly YAML and Markdown.

A HUGE sigh of relief. Now I control everything, how it works, how it links, and all the things. Even more importantly, I understand how it works, where with Wordpress, I just trusted the software.

Huge sigh of relief.

Now, since my blog was essentially ruined, I have backups that I need to rebuild here. I can’t just copy and paste, because everything is in pieces. But I will start picking and choosing, and rebuild the important posts first.

Onward!