Exam question – Precalc
It looks like most of my students did not reach the bonus question on the final exam. They had 2 hours and 15 minutes for 12 questions, and several were quite engaging. Deriving the quadratic formula is old hat, as is finding the locus of points equidistant from two given points, and pretty much as is finding all solutions (real and complex) to a cubic. However I thought this one clever, and am impressed by some of the solutions:
a) Consider a system of a cubic and a circle. How many solutions are possible?
b) Create such a system that has the maximum number of solutions.
(more commentary, and info about those graphs, below the fold –>)
I like that they had never encountered this before, and were forced to think “newly,” but at the same time that it was well within their grasp. I also appreciate the nice piece of thinking that part b) forces. Students used their knowledge of polynomials to create good guesses.
Today’s image is taken from: A Dozen Low-dimensional Problem, compiled and typeset by S. V. Duzhin, October 5, 2001. The particular graphs accompany a discussion of a case of Hilbert’s Sixteenth Problem. The website is ras.ru, which I assume is Russian Academy of Sciences, (Akademiya Nauk). Click the graph to take a look – it’s just a brief discussion at the bottom of the page.
JD2718 
well, of course the maximum number of possible real solutions is six, since this is the product of the degrees of the curves.
So, let us consider the circle x2 + y2 = 4 and the cubic y = ax(x-1)(x+1). If a is large enough, you’ll get all six solutions. A quick way to find a suitable value of a is to notice that the cubic has a local maximum at x=(-1/sqrt(3)) whose value is 4a*sqrt(3)/3; if you choose a so the value is >=2, you’re done.
of course, of course, but these were younger students with no calculus – so this was a challenge. At our level it is sort of routine. But I made the little kids think.
but I think that even without calculus they knew how to draw a cubic which has three real roots, right? This, coupled to the idea that you may stretch the curve in one direction, should be sufficient to find that there may be at least six solutions. I admit that I solved the problem by drawing a sketch and then translating it to formulae :-)
Yes, yes, that’s exactly what I wanted them to do. The nice thing here, they had never seen this question before. The trick then is not “give them a very hard question,” the trick is “make them think about what they already know, but in a new way”