# In Plain Sight, but Unseen

Cross-posted from CAS Musings:

Thanks to a comment from Doug Kuhlmann on my last post, I’ve got a few new cool connections on the transformational effects of the parameters in $y=a\cdot x^2+b\cdot x+c$ on its graph.  This is exactly why I share.  Thanks, Doug!!

THE NEW PATTERN:  Use this GeoGebraTube Web document to model the problem.  Set the value of b to any non-zero value and vary a.  The parabola’s vertex moves along a line, as shown below.

As with the changes in the b parameter, define that line and prove your claims.

[The GeoGebra link above does not produce the vertex geometry trace footprints shown in the image.  If you want to create these, download GeoGebra and create this simple document for yourself.  It is FREE.  If anyone wants explicit instructions for how to do this, email me and I’ll post instructions on the ‘blog.]

Another option to see the line is to use Geogebra’s locus tool.  It requires two inputs:  which object is the locus following, and which variable driving the variation.  After selecting the locus tool, click on the vertex and then the slider for a.  You get the next image.

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

I knew the line contained the vertex and noticed that it seemed to pass through y-intercept.  Predicting the y-intercept was c, all I needed was the slope.  With my prediction of the two generic points, I could compute that, too.  I enjoy symbol manipulation for the mental exercise.  The symbols (to me) weren’t all that complicated, so I took a brief moment of fun solving that by hand.  But this is another of those situations where the symbol manipulation isn’t the point, so using my CAS is 100% legitimate.  It is also a great leveler of ability for those intimidated for any reason by algebraic manipulations.

The next image is a great use of CAS commands to find the line’s slope.  In particular, notice the use of a function definition to minimize the algebraic clutter through function notation.

Lovely and surprisingly simple.  That means the line the parabola’s vertex follows when a varies for non-zero b is $y=\frac{b}{2}\cdot x+c$.

Students often overlook the domain warning.  It doesn’t matter for the creation of the line, but ultimately lies at the heart of the unequal spacing of the vertex footprints in the first image and explains the unique behavior of the parabola’s movement.

If a student didn’t use the vertex and y-intercept to derive the linear equation, a CAS solve command could legitimately be used to show that those two generic points were always on the line.

MOTION ALONG THE LINE:  One of the interesting parts of this problem is how the parabola moves along $y=\frac{b}{2}\cdot x+c$.  After some play with the GeoGebra document, you can see that as $|a|\rightarrow 0$ the parabolas’ vertices move infinitely far away from the y-axis, and as $|a|\rightarrow\infty$ the vertices approach the y-axis. This can also be seen numerically from the generic x-coordinate of the vertex, $-\frac{b}{2a}$.  For a fixed, non-zero value of b, the fraction representing the x-coordinate of the vertex increases in magnitude as $|a|\rightarrow 0$ and decreases in magnitude toward 0 as $|a|\rightarrow\infty$.

The vertex trace points in the first image above are separated by $\Delta a=.01$.  The reason for the differences in distances between the points noted above is because $-\frac{b}{2a}$ does not change linearly when a changes linearly.  As $|a|\rightarrow\infty$, $-\frac{b}{2a}\rightarrow 0$ slower and slower, explaining the increasing density of the vertex trace points near the y-axis.

When $a=0$, the x-coordinate of the vertex is undefined.  At that moment, the generic quadratic, $y=a\cdot x^2+b\cdot x+c$, becomes the degenerate $y=b\cdot x+c$, a line.  Graphing that line against a trace of all possible parabolas as a varies (the red dashed line below), the degenerate parabola resulting when $a=0$ is precisely the tangent line to all of these parabolas at their y-intercept, $(0,c)$, a pretty extension on a connection suggested by Dave Radcliffe on my previous post.  Nice.

A FINAL NOTE:  My memory suggests that I’ve seen this pattern before in some of the numerous times I’ve presented the b-variation of this problem in conferences and assigned it in classes.  Despite all the times I must have seen it, the pattern never rose to my active conscience.  Serendipitously, I’m currently reading Tina Seelig’s inGenius:  A Crash Course on Creativity. I offer two quotes from her Are You Paying Attention? chapter:

• We think we understand the world and look for the patterns that we already recognize. (p. 71)
• We focus predominantly on things that are at our eye level rather than looking around more broadly.  In addition, we pay attention to objects that we expect to find and ignore those things that don’t fit. (p. 71)

The moral:  Even after all of my attempts and success at finding unique patterns, I missed this one until Doug pointed it out to me.  I suspect my focus on what I knew about b‘s effect blinded me to the a effect.  This is a great reminder to me to always hold myself ready to see beauty and pattern in unexpected places.

This is cross-posted on CAS Musings.

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:  $\displaystyle (\frac{-b}{2\cdot a},\frac{4\cdot a\cdot c-b^2}{4\cdot a})$ 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.

# More on Transformations

If anyone is interested in learning more about how transformations like those described in our Transforming Polar Functions post can dramatically simplify existing secondary mathematics while opening doors to other, richer ideas and connections, consider attending our 3-day PreCalculus Transformed summer institute (July 9-11, 2012) in Atlanta, GA.

By its title, the ideas are obviously primarily pre-calculus, but they easily extend before and after that point in the mathematics curriculum along with a solid look at how technology can be used to enhance student learning and exploration.

# Transforming Polar Functions

This is cross-posted on CAS Musings.

Nurfatimah and I were playing around with polar graphs, trying to find something that would stretch students beyond simple circles and types of limacons while still being within the conceptual reach of those who had just been introduced to polar coordinates roughly two weeks earlier.

We remembered that Cartesian graphs of trigonometric functions are much more “interesting” with different center lines.  That is, the graph of $y=cos(x)+3$ is nothing more than a standard cosine graph oscillating around $y=3$.

Likewise, the graph of $y=cos(x)+0.5x$ is a standard cosine graph oscillating around $y=0.5x$.

We teach polar graphing the same way.  To graph $r=3+cos(2\theta )$, we encourage our students to “read” the function as a cosine curve of period $\pi$ oscillating around the polar function $r=3$.  Because of its period, this curve will complete a cycle in $0\le\theta\le\pi$.  The graph begins this interval at $\theta =0$ (the positive x-axis) with a cosine graph 1 unit “above” $r=3$, moving to 1 unit “below” the “center line” at $\theta =\frac{\pi}{2}$, and returning to 1 unit above the center line at $\theta =\pi$.  This process repeats for $\pi\le\theta\le 2\pi$.

Our students graph polar curves far more confidently since we began using this approach (and a couple extensions on it) than those we taught earlier in our careers.  It has become a matter of understanding what functions do and how they interact with each other and almost nothing to do with memorizing particular curve types.

So, now that our students are confidently able to graph polar curves like $r=3+cos(2\theta )$, we wondered how we could challenge them a bit more.  Remembering variable center lines like the Cartesian $y=cos(x)+0.5x$, we wondered what a polar curve with a variable center line would look like.  Not knowing where to start, I proposed $r=2+cos(\theta )+sin(\theta)$, thinking I could graph a period $2\pi$ sine curve around the limacon $r=2+cos(\theta )$.

There’s a lot going on here, but in its most simplified version, we thought we would get a curve on the center line at $\theta =0$, 1 unit above at $\theta =\frac{\pi}{2}$, on at $\theta =\pi$, 1 unit below at $\theta =\frac{3\pi}{2}$, and returning to its starting point at $\theta =2\pi$.  We had a very rough “by hand” sketch, and were quite surprised by the image we got when we turned to our grapher for confirmation.  The oscillation behavior we predicted was certainly there, but there was more!  What do you see in the graph of $r=2+cos(\theta )+sin(\theta)$ below?

This looked to us like some version of a cardioid.  Given the symmetry of the axis intercepts, we suspected it was rotated $\frac{\pi}{4}$ from the x-axis.  An initially x-axis symmetric polar curve rotated $\frac{\pi}{4}$ would contain the term $cos(\theta-\frac{\pi}{4})$ which expands using a trig identity.

$\begin{array}{ccc} cos(\theta-\frac{\pi}{4})&=&cos(\theta )cos(\frac{\pi}{4})+cos(\theta )cos(\frac{\pi}{4}) \\ &=&\frac{1}{\sqrt{2}}(cos(\theta )+sin(\theta )) \end{array}$

Eureka!  This identity let us rewrite the original polar equation.

$\begin{array}{ccc} r=2+cos(\theta )+sin(\theta )&=&2+\sqrt{2}\cdot\frac{1}{\sqrt{2}} (cos(\theta )+sin(\theta )) \\ &=&2+\sqrt{2}\cdot cos(\theta -\frac{\pi}{4}) \end{array}$

And this last form says our original polar function is equivalent to $r=2+\sqrt{2}\cdot cos(\theta -\frac{\pi}{4})$, or a $\frac{\pi}{4}$ rotated cosine curve of amplitude $\sqrt{2}$ and period $2\pi$ oscillating around center line $r=2$.

This last image shows a cosine curve starting at $\theta=\frac{\pi}{4}$ beginning $\sqrt{2}$ above the center circle $r=2$, crossing the center circle $\frac{\pi}{2}$ later at $\theta=\frac{3\pi}{4}$, dropping to $\sqrt{2}$ below the center circle at $\theta=\frac{5\pi}{4}$, back to the center circle at $\theta=\frac{7\pi}{4}$ before finally returning to the starting point at $\theta=\frac{9\pi}{4}$.  Because the radius is always positive, this also convinced us that this curve is actually a rotated limacon without a loop and not the cardioid that drove our initial investigation.

So, we thought we were departing into some new territory and found ourselves looking back at earlier work from a different angle.  What a nice surprise!

One more added observation:  We got a little lucky in guessing the angle of rotation, but even if it wasn’t known, it is always possible to compute an angle of rotation (or translation in Cartesian) for a sum of two sinusoids with identical periods.  This particular topic is covered in some texts, including Precalculus Transformed.

# PreCalculus Summer Institute 2012

If you are in Atlanta, GA July 9-11, 2012, you might be interested in attending a workshop my co-author and I are offering at Westminster through the Center for Teaching.  A description of the workshop follows.  We hope to see some of you!

Title of Workshop:  PreCalculus Transformed
(Registration information will be available soon here.)

PresentersChris Harrow & Nurfatimah Merchant from The Westminster Schools

Workshop Dates:  July 9-11, 2012 (Monday-Wednesday)

Workshop Description: PreCalculus Transformed highlights the under-explored role of non-standard transformations and function composition in learning algebra and precalculus concepts. Families of functions are identified first by immutable distinguishing characteristics and then modified through multiple representations & transformations. Participants will discover that many historically complicated precalculus problems and concepts are both richer and greatly simplified in this process. The course integrates computer algebra system (CAS) technology, but it is certainly possible to use and grasp its concepts without this technology.  Potential topics include expanded transformations, polynomials, rational functions, exponentials, logistics, and trigonometric functions.  Additional topics may be explored depending on time and participant needs or experience.  Textbook is included in the workshop price.

Target Audience:  Algebra II, Precalculus, and Calculus teachers at the high school or junior college level

Workshop Location:
The Center for Teaching at The Westminster Schools
1424 West Paces Ferry Road NW
Atlanta, GA 30327

Contact e-mail: chrish@westminster.net

# When will it end?

The following expressions are written so that they require an infinite number of terms.  Can you rewrite either of them in a closed form?

1.   $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}$

2.   $1+\displaystyle\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}$

# How many zeros — student responses

In the earlier post, students were asked to determine the length of the expansion of 300!, the number’s last digit, and the number of that digit at the end. My students had powerful TI-Nspires available, so I deliberately asked about a number that would be difficult to deal with by hand, encouraging them to let computation tools do accurate work while they did the thinking.

Because 10 is one of the factors of 300!, the last digit was rapidly declared to be 0.  The two length questions took longer.

Most computed 300! on their Nspires and were amazed by the machines’ exact answers for such a large number even though they have difficulty displaying the result.  The following image is from Wolfram Alpha.  It shows the entire number, but also some additional information I wanted my students to reason for themselves.  Unfortunately, the Wolfram Alpha result also states the length of the number, so I was happier that they were “limited” to their Nspires.

Following is a summary of the track of their reasoning.

• One student unintentionally pressed ctrl-enter when computing 300! and got $300!=3.06...*10^{614}$, not the exact answers his classmates discovered.  After some thought about the length of numbers and their scientific notation equivalents, he declared the number to be 615 digits long.
• Another student declared the same answer after computing $log(300!)=614.486...$.  The class knew this answer was correct, but didn’t initially understand why until one suggested that $log(300!)=log(3.06...*10^{614})=log(3.06...)+614$, convincing them that both the scientific notation and the logarithm were giving the same results for this problem, but in different forms.

Then attention turned to the number of zeros, a more challenging question if judged by how long it took them to decide on an answer.

• A couple students opted for a careful count, arriving at 74 zeros at the end of 300!.  But most were unwilling to do that count on their Nspires because of the challenge of the display–a fact we well understood when we created the problem.
• Most of the others realized quickly that the problem hinges on the number of 10s used to create 300!.  Some used this to get a first guess of 30 zeros because 300! includes factors of 10, 20, 30, …, 290, 300.  This was upgraded to 33 to account for the double factors of 10 in 100, 200, and 300.
• Another spoke up to say that factors of 5, 15, 25, etc. could create additional final 0s when multiplied by other factors containing additional 2s.  Because these factors where spaced half way between the earlier multiples of 10, the class estimate jumped another 30 to 63.
• Many realized that this needed to be increased further because every multiple of 25 contained two 5s.  There are 12 of those (25, 50, …, 300) increasing their count to 75.  The numbers 125 & 250 added two more to the count because each had three 5s in their factorizations giving a final guess of 77.
• By this point, one questioned why we were going through all this trouble to account for the locations of both the 5s and 10s.  Because every 10 contained a factor of 5, this really just depended on the number of 5s.  Ultimately, she said, it also didn’t matter where the 5s happened.  We just needed to know how many 5s there were in the factorization of 300!.  Her Nspire had a factor command, so she guessed correctly that it could handle the following which suggested the correct answer of 74 zeros.  I asked if she was worried about having enough 2s, and she noted that there were 296 of those, so like limiting factors in her chemistry class last year, the zeros at the end of 300! were limited by the number of 5s.

• It took some guidance, but they eventually discovered that the difference between their initial 77 and their final 74 could be accounted for by the unintended over-counting of 100, 200, and 300 in their shift from counting 10s to counting 5s.

A secondary realization from this great conversation was that the factorization of a factorial number created a list of all of the prime numbers less than or equal to the integer in the factorial if you ignored all the exponents.  This Wolfram Alpha page shows this result nicely.