What if we vary b

Playing a what if game with translation form

The main post where all my quadratic posts are listed can be found here.

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, \(f(x)=ax^2 +bx +c\) 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:

\[ f(x) = a \left( \left(x + \frac{b}{2a} \right)^2 - \left(\frac{b}{2a} \right)^2 + \frac{c}{a} \right) \]

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.

\[ k= - \left(\frac{b}{2a} \right)^2 + \frac{c}{a} \] So if we do a little algebra on this, we get:

\[ k=\left(c- \frac{b^2}{4a}\right) \] 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 \(x=\left(-\frac{b}{2a}\right)\) then substitute that into the right side of the k=, we get a new function \(v(x)=c-ax^2\)

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.

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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.