Tag Archives: degenerate

Cubics and CAS

Here’s a question I posed to one of my precalculus classes for homework at the end of last week along with three solutions we developed.

Suppose the graph of a cubic function has an inflection point at (1,3) and passes through (0,-4).

  1. Name one other point that MUST be on the curve, and
  2. give TWO different cubic equations that would pass through the three points.

SOLUTION ALERT!  Don’t read any further if you want to solve this problem for yourself.

The first question relies on the fact that every cubic function has 180 degree rotational symmetry about its inflection point.  This is equivalent to saying that the graph of a cubic function is its own image when the function’s graph is reflected through its inflection point.

That means the third point is the image of (0,-4) when point-reflected through the inflection point (1,3), which is the point (2,10) as shown graphically below.

From here, my students came up with 2 different solutions to the second question and upon prodding, we created a third.

SOLUTION 1:  Virtually every student assumed y=a\cdot x^3 was the parent function of a cubic with unknown leading coefficient whose “center” (inflection point) had been slid to (1,3).  Plugging in the given (0,-4) to (y-3)=a\cdot (x-1)^3 gives a=7.  Here’s their graph.

SOLUTION 2:  Many students had difficulty coming up with another equation.  A few could sketch in other cubic graphs (curiously, all had positive lead coefficients) that contained the 3 points above, but didn’t know how to find equations.  That’s when Sara pointed out that if the generic expanded form of a cubic was a\cdot x^3+b\cdot x^2 +c\cdot x+d , then any 4 ordered pairs with unique x-coordinates should define a unique cubic.  That is, if we picked any 4th point with x not 0, 1, or 2, then we should get an equation.  That this would create a 4×4 system of equations didn’t bother her at all.  She knew in theory how to solve such a thing, but she was thinking on a much higher plane.  Her CAS technology expeditiously did the grunt work, allowing her brain to keep moving.

A doubtful classmate called out, “OK.  Make it go through (100,100).”  Following is a CAS screen roughly duplicating Sara’s approach and a graph confirming the fit.  The equation was onerous, but with a quick copy-paste, Sara had moved from  idea to computation to ugly equation and graph in just a couple minutes.  The doubter was convinced and the “wow”s from throughout the room conveyed the respect for the power of a properly wielded CAS.

In particular, notice how the TI-Nspire CAS syntax in lines 1 and 3 keep the user’s focus on the type of equation being solved and eliminates the need to actually enter 4 separate equations.  It doesn’t always work, but it’s a particularly lovely piece of scaffolding when it does.

SOLUTION 3:  One of my goals in Algebra II and Precalculus courses is to teach my students that they don’t need to always accept problems as stated.  Sometimes they can change initial conditions to create a much cleaner work environment so long as they transform their “clean” solution back to the state of the initial conditions.

In this case, I asked what would happen if they translated the inflection point using T_{-1,-3} to the origin, making the other given point (-1,-7).  Several immediately called the 3rd point to be (1,7) which “untranslating” — T_{1,3}(1,7)=(2,10) — confirmed to be the earlier finding.

For cases where the cubic had another real root at x=r, then symmetry immediately made x=-r another root, and a factored form of the equation becomes y=a\cdot (x)(x-r)(x+r) for some value of a.  Plugging in (-1,-7) gives a in terms of r.

The last line slid the initially translated equation using T_{1,3} to re-position the previous line according to the initial conditions.  While unasked for, notice how the CAS performed some polynomial division on the right-side expression.

I created a GeoGebra document with a slider for the root using the equation from the last line of the CAS image above.  The image below shows one possible position of the retranslated root.  If you want to play with a live version of this, you will need a free copy of GeoGebra to run it, but the file is here.


What’s nice here is how the problem became one of simple factors once the inflection point was translated to the origin.  Notice also that the CAS version of the equation concludes with +7x-4, the line containing the original three points.  This is nice for two reasons.  The numerator of the rational term is -7x(x-2)(x-1) which zeros out the fraction at x=0, 1, or 2, putting the cubic exactly on the line y=7x-4 at those points.

The only r-values are in the denominator, so as r\rightarrow\infty, the rational term also becomes zero.  Graphically, you can see this happen as the cubic “unrolls” onto y=7x-4 as you drag |x|\rightarrow\infty.  Essentially, this shows both graphically and algebraically that y=7x-4 is the limiting degenerate curve the cubic function approaches as two of its transformed real roots grow without bound.


Quadratic Surprise

Summer is giving me some time to tie up loose ends that inevitably get dropped during the busy-ness of the school year. Here’s one of those.

I hinted in a post from several months ago about a cool underlying pattern in the quadratic function family.  Most algebra students know how the coefficients a and c control the graph of y=a\cdot x^2+b\cdot x+c, but what does b do?

I wrote a Geogebra worksheet to allow exploration of a, b, and c.  On this page, there are sliders on the right side that allow users to vary the values of these three coefficients while the graph of y=a\cdot x^2+b\cdot x+c changes live to reflect those values.

Over the past decade, I’ve become increasingly enamored with this approach to exploring the behavior of function families in advance of more formal analyses.  “Seeing” the effects of parameter can inform and guide the work you do later.  Students quickly recognize that c vertically changes the parabola’s position; closer inspection notes that c is the y-intercept.

Most also note that a changes the “width” of the parabola.  This is true enough, but (in my opinion) a clearer description is that a changes the quadratic’s height.  For any value of x, the y-values of y=2x^2 (a=2) are twice the y-values of y=x^2.  If you attempt to quantify the width, then a=2 means the corresponding points are \displaystyle \frac{1}{\sqrt{2}} “wider”.  That just isn’t intuitive to anyone I know, and describing lead coefficients as vertical scale changes is an idea that applies to all functions. I eventually refocus these descriptions to vertical scale changes, but that’s not the point right now.

So what happens when you change b?  If you don’t already know the answer, I encourage you to explore the Geogebra worksheet before reading further.  Try to be precise.

SOLUTION ALERT!  Don’t read further if you want to solve this first.

Like c, varying b changes the position of the parabola, but not its shape.  The difference is that b moves the parabola both horizontally and vertically.   Closer observation suggests that the motion might be along a parabolic path.  Using a new Geogebra worksheet, I placed a trace on the vertex to record the “footprints” of the vertex as b changed.

That’s pretty compelling evidence.  The challenge students face at this point is defining an equation for the suspected parabola. Following the vertex of a parabola is a good proxy for following the entire parabola.

FAST FORWARD:  Through lots of trial-and-error, students eventually propose y=-a\cdot x^2+c.  That’s nice, but writing an equation isn’t a proof.  One of the most elegant proofs I’ve seen solves the system of equations defined by the original generic quadratic family and the proposed path of the vertex. A CAS is obviously an appropriate tool in this situation.

There are two solutions:  (\frac{-b}{2\cdot a},\frac{4\cdot a\cdot c-b^2}{4\cdot a}) and (0,c) implying two graphical intersections, a fact verified by the vertex trace image above.  The proof lies in the first ordered pair–the generic form of the coordinates of the vertex of y=a\cdot x^2+b\cdot x+c–clearly establishing that the generic vertex always travels on the proposed path.  Nice.

What amazed me most about this problem is that I had been teaching quadratic equations for years and remembered from my time as a student what a and c did to the graph.  How is it that I had never explored b ?  How could such a pretty result have been overlooked?  No longer.  This is a project every time I teach an algebra class.