## Best Algebra 2 Lab Ever

This post shares what I think is one of the best, inclusive, data-oriented labs for a second year algebra class.  This single experiment produces linear, quadratic, and exponential (and logarithmic) data from a lab my Algebra 2 students completed this past summer.  In that class, I assigned frequent labs where students gathered real data, determined models to fit that data, and analyzed goodness of the models’ fit to the data.   I believe in the importance of doing so much more than just writing an equation and moving on.

For kicks, I’ll derive an approximation for the coefficient of gravity at the end.

THE LAB:

On the way to school one morning last summer, I grabbed one of my daughters’ “almost fully inflated” kickballs and attached a TI CBR2 to my laptop and gathered (distance, time) data from bouncing the ball under the Motion Sensor.  NOTE:  TI’s CBR2 can connect directly to their Nspire and TI84 families of graphing calculators.  I typically use computer-based Nspire CAS software, so I connected the CBR via my laptop’s USB port.  It’s crazy easy to use.

One student held the CBR2 about 1.5-2 meters above the ground while another held the ball steady about 20 cm below the CBR2 sensor.  When the second student released the ball, a third clicked a button on my laptop to gather the data:  time every 0.05 seconds and height from the ground.  The graphed data is shown below.  In case you don’t have access to a CBR or other data gathering devices, I’ve uploaded my students’ data in this Excel file.

Remember, this is data was collected under far-from-ideal conditions.  I picked up a kickball my kids left outside on my way to class.  The sensor was handheld and likely wobbled some, and the ball was dropped on the well-worn carpet of our classroom floor.  It is also likely the ball did not remain perfectly under the sensor the entire time.  Even so, my students created a very pretty graph on their first try.

For further context, we did this lab in the middle of our quadratics unit that was preceded by a unit on linear functions and another on exponential and logarithmic functions.  So what can we learn from the bouncing ball data?

LINEAR 1:

While it is very unlikely that any of the recorded data points were precisely at maximums, they are close enough to create a nice linear pattern.

As the height of a ball above the ground helps determine the height of its next bounce (height before –> energy on impact –> height after), the eight ordered pairs (max height #n, max height #(n+1) ) from my students’ data are shown below

This looks very linear.  Fitting a linear regression and analyzing the residuals gives the following.

The data seems to be close to the line, and the residuals are relatively small, about evenly distributed above and below the line, and there is no apparent pattern to their distribution.  This confirms that the regression equation, $y=0.673x+0.000233$, is a good fit for the = height before bounce and = height after bounce data.

NOTE:  You could reasonably easily gather this data sans any technology.  Have teams of students release a ball from different measured heights while others carefully identify the rebound heights.

The coefficients also have meaning.  The 0.673 suggests that after each bounce, the ball rebounded to 67.3%, or 2/3, of its previous height–not bad for a ball plucked from a driveway that morning.  Also, the y-intercept, 0.000233, is essentially zero, suggesting that a ball released 0 meters from the ground would rebound to basically 0 meters above the ground.  That this isn’t exactly zero is a small measure of error in the experiment.

EXPONENTIAL:

Using the same idea, consider data of the form (x,y) = (bounce number, bounce height). the graph of the nine points from my students’ data is:

This could be power or exponential data–something you should confirm for yourself–but an exponential regression and its residuals show

While something of a pattern seems to exist, the other residual criteria are met, making the exponential regression a reasonably good model: $y = 0.972 \cdot (0.676)^x$.  That means bounce number 0, the initial release height from which the downward movement on the far left of the initial scatterplot can be seen, is 0.972 meters, and the constant multiplier is about 0.676.  This second number represents the percentage of height maintained from each previous bounce, and is therefore the percentage rebound.  Also note that this is essentially the same value as the slope from the previous linear example, confirming that the ball we used basically maintained slightly more than 2/3 of its height from one bounce to the next.

And you can get logarithms from these data if you use the equation to determine, for example, which bounces exceed 0.2 meters.

So, bounces 1-4 satisfy the requirement for exceeding 0.20 meters, as confirmed by the data.

A second way to invoke logarithms is to reverse the data.  Graphing x=height and y=bounce number will also produce the desired effect.

Each individual bounce looks like an inverted parabola.  If you remember a little physics, the moment after the ball leaves the ground after each bounce, it is essentially in free-fall, a situation defined by quadratic movement if you ignore air resistance–something we can safely assume given the very short duration of each bounce.

I had eight complete bounces I could use, but chose the first to have as many data points as possible to model.  As it was impossible to know whether the lowest point on each end of any data set came from the ball moving up or down, I omitted the first and last point in each set.  Using (x,y) = (time, height of first bounce) data, my students got:

What a pretty parabola.  Fitting a quadratic regression (or manually fitting one, if that’s more appropriate for your classes), I get:

Again, there’s maybe a slight pattern, but all but two points are will withing  0.1 of 1% of the model and are 1/2 above and 1/2 below.  The model, $y=-4.84x^2+4.60x-4.24$, could be interpreted in terms of the physics formula for an object in free fall, but I’ll postpone that for a moment.

LINEAR 2:

If your second year algebra class has explored common differences, your students could explore second common differences to confirm the quadratic nature of the data.  Other than the first two differences (far right column below), the second common difference of all data points is roughly 0.024.  This raises suspicions that my student’s hand holding the CBR2 may have wiggled during the data collection.

Since the second common differences are roughly constant, the original data must have been quadratic, and the first common differences linear. As a small variation for each consecutive pair of (time, height) points, I had my students graph (x,y) = (x midpoint, slope between two points):

If you get the common difference discussion, the linearity of this graph is not surprising.  Despite those conversations, most of my students seem completely surprised by this pattern emerging from the quadratic data.  I guess they didn’t really “get” what common differences–or the closely related slope–meant until this point.

Other than the first three points, the model seems very strong.  The coefficients tell an even more interesting story.

GRAVITY:

The equation from the last linear regression is $y=4.55-9.61x$.  Since the data came from slope, the y-intercept, 4.55, is measured in m/sec.  That makes it the velocity of the ball at the moment (t=0) the ball left the ground.  Nice.

The slope of this line is -9.61.  As this is a slope, its units are the y-units over the x-units, or (m/sec)/(sec).  That is, meters per squared second.  And those are the units for gravity!  That means my students measured, hidden within their data, an approximation for coefficient of gravity by bouncing an outdoor ball on a well-worn carpet with a mildly wobbly hand holding a CBR2.  The gravitational constant at sea-level on Earth is about -9.807 m/sec^2.  That means, my students measurement error was about $\frac{9.807-9.610}{9.807}=2.801%$.  And 2.8% is not a bad measurement for a very unscientific setting!

CONCLUSION:

Whenever I teach second year algebra classes, I find it extremely valuable to have students gather real data whenever possible and with every new function, determine models to fit their data, and analyze the goodness of the model’s fit to the data.  In addition to these activities just being good mathematics explorations, I believe they do an excellent job exposing students to a few topics often underrepresented in many secondary math classes:  numerical representations and methods, experimentation, and introduction to statistics.  Hopefully some of the ideas shared here will inspire you to help your students experience more.

## CAS Presentations at USACAS-9

I had two presentations at last Saturday’s USACAS-9 conference at Hawken School in Cleveland, OH.  Following are outline descriptions of the two sessions with links to the PowerPoint, pdf, and .tns files I used.  I’m also adding all of this information to the Conference Presentations tab of this ‘blog.

Powerful Student Proofs

This session started with a brief introduction to a lab that first caught my eye at the first USACAS conference years ago.

You know how the graph of $y=ax^2+bx+c$ behaves when you vary a and c, but what happens when you change b?

I ‘blogged on this problem here and here.  In the session, we used TI-Nspire file QuadExplore.

Next, we explored briefly the same review of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.  In the session, we used TI-Nspire file Intro Polar.

Having a complete grasp of polar graphs of limacons, cardioids, rose curves, and hybrids of these, I investigated what would happen for curves of the family $r=cos \left( \frac{\theta}{k} \right)$.  Curiously, for $k=3$, I encountered a curve that looked like a horizontal translation of limacons–something that just shouldn’t happen within polar coordinates.

One of my former students, Sara, used a CAS to convert a polar curve to Cartesian, translate the curve, and convert back to polar.  She then identified and solved a trig identity to confirm what the graph suggested.  A complete description of Sara’s proof is below.  I originally ‘blogged on Sara’s work here which was a much more elegant solution to the problem than my initial attempt.  It’s always cool when a student’s work is better than her teacher’s!  I used TI-Nspire file Polar Fractions in Saturday’s session.

The last example presented itself when I created a document to model the family of conic curves resulting from manipulating the coefficients of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.  After I created  dynamic points for the foci of the conics, something unusual happened when the E parameter for horizontal ellipses and hyperbolas varied.

The foci for hyperbolas followed an ellipse, and the locus of elliptical foci appeared to be a hyperbola.  Another former student, Lilly, proved this property to be true.  A detailed explanation of Lilly’s proof is below.  We were fortunate to have Lilly’s work published in the Mathematics Teacher in May, 2014.

To demonstrate this final part of the session, I used TI-Nspire file Hidden Conic Behavior.

Here is my PowerPoint file for Powerful Student Proofs.  A more detailed sketch of the session and the student proofs is below.

Bending Asymptotes, Bouncing Off Infinity, and Going Beyond

The basic proposal was that adding the Reciprocal transformation to the palette of constant dilations and translations dramatically simplified understanding of the behavior of rational functions around even and odd vertical asymptotes (bouncing off and passing through infinity).  Just like lead coefficients of polynomials determine their end behavior, so, too, do the lead coefficients of proper rational expressions define the end behavior of rational functions.

Extending the idea of reciprocating and transforming functions, you can quickly explain exponential decay from exponential growth, derive the graphs of $y=\frac{1}{x}$ and $y=\frac{1}{x^2}$, and completely explain why logistic functions behave the way they do.

We finished with a quick exploration of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.

I used TI-Nspire Bending and Intro Polar files in the demonstration.  Here is my outline PowerPoint file for Bending Asymptotes.

I’m teaching Algebra 2 this summer for my school.  In a recent test on quadratic functions, I gave a question I thought would be a little different, but still reachable for those willing to make connections or exert a little creativity.

Write a system of quadratic functions that has exactly one solution:  (1,1).

Their handheld graphing calculators were allowed.  Some students definitely had difficulty with the challenge, some gave a version of the answer I expected, and one adopted a form I knew was possible, but doubted anyone would actually find during a test situation.

I show my students’ solutions below.  But before you read on, can you give your own solution?

WHAT I EXPECTED

We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, algebraically, and verbally.  When you get stumped in one representation, being able to shift to a different form is often helpful.  That could mean a different algebraic representation, or a different Rule of 4 representation altogether.

The question is phrased verbally asking for an algebraic answer.  But it asks about a solution to a system of equations.  I hoped my students would recall that the graphical version of a system solution is equivalent to the point(s) where the graphs of the equations intersected.  In my mind, the easiest way to do this is to write quadratic functions with coincident vertices.  And this is most easily done in vertex form.  The cleanest answer I ever got to this question was

A graphical representation verifies the solution.

Another student recognized that if two parabolas shared a vertex, but had different “slopes”, their only possible point of intersection was exactly the one the question required.  Here’s a graphical version of her answer.

From these two, you can see that there is actually an infinite number of correct solutions.  And I was asking them for just one of these!  🙂

WHAT I KNEW, BUT DIDN’T EXPECT

Another way to solve this question makes use of the geometry of quadratic graphs.  If two quadratics have the same leading coefficients, they are the same graph, intersect exactly once, or never intersect.  This is a very non-trivial idea for most students.  While I’m not convinced the author of the following solution had this in mind when he answered the question, his solution works because of that fact.  Here’s what J wrote on last week’s test and its graph.

J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.

This is a very different (and super cool) answer than what I expected my students to produce.  Lesson re-learned:  Challenge your students, give them room to express creativity and individuality, and be prepared to be amazed by them.

NEXT STEPS

J’s answer actually opens the door to other avenues of exploration.

1. Can you generalize the form of all of J’s equations, essentially defining a family of quadratics?  Can you prove that all members of your generalization satisfy the question posed and that no other answers are possible?
2. Can you find forms of other generalized families of quadratic functions whose only solution is (1,1)?
3. Notice that there were two types of solutions above:  A) those with coincident vertices and different lead coefficients and B) those with identical lead coefficients and different vertices.  Are these the only types of quadratics that can answer this question?  That is, is there a system of quadratics with (1,1) as the only solution that have identical vertices and lead coefficients?  Could both be different and (1,1) be the only solution?
4. If I relax the requirement that the quadratics be functions, what other types of quadratics are possible?  [This could be a very nice calculus question!]

For my part, I’m returning to some of these questions this week to stretch and explore my student’s creativity and problem-solving.

I’d love to hear what you or your students discover.

## Traveling Dots, Parabolas, and Elegant Math

Toward the end of last week, I read a description a variation on a paper-folding strategy to create parabolas.  Paraphrased, it said:

1. On a piece of wax paper, use a pen to draw a line near one edge.  (I used a Sharpie on regular copy paper and got enough ink bleed that I’m convinced any standard copy or notebook paper will do.  I don’t think the expense of wax paper is required!)
2. All along the line, place additional dots 0.5 to 1 inch apart.
3. Finally, draw a point F between 0.5 and 2 inches from the line roughly along the midline of the paper toward the center of the paper.
4. Fold the paper over so one of the dots on line is on tope of point F.  Crease the paper along the fold and open the paper back up.
5. Repeat step 4 for every dot you drew in step 2.
6. All of the creases from steps 4 & 5 outline a curve.  Trace that curve to see a parabola.

I’d seen and done this before, I had too passively trusted that the procedure must have been true just because the resulting curve “looked like a parabola.”  I read the proof some time ago, but I consumed it too quickly and didn’t remember it when I was read the above procedure.  I shamefully admitted to myself that I was doing exactly what we insist our students NEVER do–blindly accepting a “truth” based on its appearance.  So I spent part of that afternoon thinking about how to understand completely what was going on here.

What follows is the chronological redevelopment of my chain of reasoning for this activity, hopefully showing others that the prettiest explanations rarely occur without effort, time, and refinement.  At the end of this post, I offer what I think is an even smoother version of the activity, freed from some of what I consider overly structured instructions above.

CONIC DEFINITION AND WHAT WASN’T OBVIOUS TO ME

A parabola is the locus of points equidistant from a given  point (focus) and line (directrix).

What makes the parabola interesting, in my opinion, is the interplay between the distance from a line (always perpendicular to some point C on the directrix) and the focus point (theoretically could point in any direction like a radius from a circle center).

What initially bothered me about the paper folding approach last week was that it focused entirely on perpendicular bisectors of the Focus-to-C segment (using the image above).  It was not immediately obvious to me at all that perpendicular bisectors of the Focus-to-C segment were 100% logically equivalent to the parabola’s definition.

1. I knew without a doubt that all parabolas are similar (there is a one-to-one mapping between every single point on any parabola and every single point on any other parabola), so I didn’t need to prove lots of cases.  Instead, I focused on the simplest version of a parabola (from my perspective), knowing that whatever I proved from that example was true for all parabolas.
2. I am quite comfortable with my algebra, geometry, and technology skills.  Being able to wield a wide range of powerful exploration tools means I’m rarely intimidated by problems–even those I don’t initially understand.  I have the patience to persevere through lots of data and explorations until I find patterns and eventually solutions.

I love to understand ideas from multiple perspectives, so I rarely quit with my initial solution.  Perseverance helps me re-phrase ideas and exploring them from alternative perspectives until I find prettier ways of understanding.

In my opinion, it is precisely this willingness to play, persevere, and explore that formalized education is broadly failing to instill in students and teachers.  “What if?” is the most brilliant question, and the one we sadly forget to ask often enough.

ALGEBRAIC PROOF

While I’m comfortable handling math in almost any representation, my mind most often jumps to algebraic perspectives first.  My first inclination was a coordinate proof.

PROOF 1:  As all parabolas are similar, it was enough to use a single, upward facing parabola with its vertex at the origin.  I placed the focus at $(0,f)$, making the directrix the line $y=-f$.  If any point on the parabola was $(x_0,y_0)$, then a point C on the directrix was at $(x_0,-f)$.

From the parabola’s definition, the distance from the focus to P was identical to the length of CP:

$\sqrt{(x_0-0)^2-(y_0-f)^2}=y_0+f$

Squaring and combining common terms gives

$x_0 ^2+y_0 ^2-2y_0f+f^2=y_0 ^2+2y_0f+f^2$
$x_0 ^2=4fy$

But the construction above made lines (creases) on the perpendicular bisector of the focus-to-C segment.  This segment has midpoint $\displaystyle \left( \frac{x_0}{2},0 \right)$ and slope $\displaystyle -\frac{2f}{x_0}$, so an equation for its perpendicular bisector is $\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)$.

Finding the point of intersection of the perpendicular bisector with the parabola involves solving a system of equations.

$\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)=\frac{x^2}{4f}$
$\displaystyle \frac{1}{4f} \left( x^2-2x_0x+x_0 ^2 \right) =0$
$\displaystyle \frac{1}{4f} \left( x-x_0 \right) ^2 =0$

So the only point where the line and parabola meet is at $\displaystyle x=x_0$–the very same point named by the parabola’s definition.  QED

Proof 2:  All of this could have been brilliantly handled on a CAS to save time and avoid the manipulations.

Notice that the y-coordinate of the final solution line is the same $y_0$ from above.

MORE ELEGANT GEOMETRIC PROOFS

I had a proof, but the algebra seemed more than necessary.  Surely there was a cleaner approach.

In the image above, F is the focus, and I is a point on the parabola.  If D is the midpoint of $\overline{FC}$, can I conclude $\overline{ID} \perp \overline{FC}$, proving that the perpendicular bisector of $\overline{FC}$ always intersects the parabola?

PROOF 3:  The definition of the parabola gives $\overline{FI} \cong \overline{IC}$, and the midpoint gives $\overline{FD} \cong \overline{DC}$.  Because $\overline{ID}$ is self-congruent, $\Delta IDF \cong \Delta IDC$ by SSS, and corresponding parts make the supplementary $\angle IDF \cong \angle IDC$, so both must be right angles.  QED

PROOF 4:  Nice enough, but it still felt a little complicated.  I put the problem away to have dinner with my daughters and when I came back, I was able to see the construction not as two congruent triangles, but as the single isosceles $\Delta FIC$ with base $\overline{FC}$.  In isosceles triangles, altitudes and medians coincide, automatically making $\overline{ID}$ the perpendicular bisector of $\overline{FC}$.  QED

Admittedly, Proof 4 ultimately relies on the results of Proof 3, but the higher-level isosceles connection felt much more elegant.  I was satisfied.

TWO DYNAMIC GEOMETRY SOFTWARE VARIATIONS

Thinking how I could prompt students along this path, I first considered a trace on the perpendicular lines from the initial procedure above (actually tangent lines to the parabola) using to trace the parabolas.  A video is below, and the Geogebra file is here.

http://vimeo.com/89759785

It is a lovely approach, and I particularly love the way the parabola appears as a digital form of “string art.”  Still, I think it requires some additional thinking for users to believe the approach really does adhere to the parabola’s definition.

I created a second version allowing users to set the location of the focus on the positive y-axis and using  a slider to determine the distances and constructs the parabola through the definition of the parabola.  [In the GeoGebra worksheet (here), you can turn on the hidden circle and lines to see how I constructed it.]  A video shows the symmetric points traced out as you drag the distance slider.

A SIMPLIFIED PAPER PROCEDURE

Throughout this process, I realized that the location and spacing of the initial points on the directrix was irrelevant.  Creating the software versions of the problem helped me realize that if I could fold a point on the directrix to the focus, why not reverse the process and fold F to the directrix?  In fact, I could fold the paper so that F touched anywhere on the directrix and it would work.  So, here is the simplest version I could develop for the paper version.

1. Use a straightedge and a Sharpie or thin marker to draw a line near the edge of a piece of paper.
2. Place a point F roughly above the middle of the line toward the center of the paper.
3. Fold the paper over so point F is on the line from step 1 and crease the paper along the fold.
4. Open the paper back up and repeat step 3 several more times with F touching other parts of the step 1 line.
5. All of the creases from steps 3 & 4 outline a curve.  Trace that curve to see a parabola.

This procedure works because you can fold the focus onto the directrix anywhere you like and the resulting crease will be tangent to the parabola defined by the directrix and focus.  By allowing the focus to “Travel along the Directrix”, you create the parabola’s locus.  Quite elegant, I thought.

As I was playing with the different ways to create the parabola and thinking about the interplay between the two distances in the parabola’s definition, I wondered about the potential positions of the distance segments.

1. What is the shortest length of segment CP and where could it be located at that length?  What is the longest length of segment CP and where could it be located at that length?
2. Obviously, point C can be anywhere along the directrix.  While the focus-to-P segment is theoretically free to rotate in any direction, the parabola definition makes that seem not practically possible.  So, through what size angle is the focus-to-P segment practically able to rotate?
3. Assuming a horizontal directrix, what is the maximum slope the focus-to-P segment can achieve?
4. Can you develop a single solution to questions 2 and 3 that doesn’t require any computations or constructions?

CONCLUSIONS

I fully realize that none of this is new mathematics, but I enjoyed the walk through pure mathematics and the enjoyment of developing ever simpler and more elegant solutions to the problem.  In the end, I now have a deeper and richer understanding of parabolas, and that was certainly worth the journey.

## Fun with Series

Two days ago, one of my students (P) wandered into my room after school to share a problem he had encountered at the 2013 Walton MathFest, but didn’t know how to crack.  We found one solution.  I’d love to hear if anyone discovers a different approach.  Here’s our answer.

PROBLEM:  What is the sum of $\displaystyle \sum_{n=1}^{\infty} \left( \frac{n^2}{2^n} \right) = \frac{1^2}{2^1} + \frac{2^2}{2^2} + \frac{3^2}{3^3} + ...$ ?

Without the $n^2$, this would be a simple geometric series, but the quadratic and exponential terms can’t be combined in any way we knew, so the solution must require rewriting.  After some thought, we remembered that perfect squares can be found by adding odd integers.  I suggested rewriting the series as

where each column adds to the one of the terms in the original series.  Each row was now a geometric series which we knew how to sum.  That  meant we could rewrite the original series as

We had lost the quadratic term, but we still couldn’t sum the series with both a linear and an exponential term.  At this point, P asked if we could use the same approach to rewrite the series again.  Because the numerators were all odd numbers and each could be written as a sum of 1 and some number of 2s, we got

where each column now added to the one of the terms in our secondary series.  Each row was again a geometric series, allowing us to rewrite the secondary series as

Ignoring the first term, this was finally a single geometric series, and we had found the sum.

Does anyone have another way?

That was fun.  Thanks, P.

## Quadratics, Statistics, Symmetry, and Tranformations

A problem I assigned my precalculus class this past Thursday ended up with multiple solutions by the time we finished.  Huzzah for student creativity!

The question:

Find equations for all polynomial functions, $y=f(x)$, of degree $\le 2$ for which $f(0)=f(1)=2$ and $f(3)=0$.

After they had worked on this (along with several variations on the theme), four very different ways of thinking about this problem emerged.  All were valid and even led to a lesson I hadn’t planned–proving that, even though they looked different algebraically, all were equivalent.  I present their approaches (and a few extras) in the order they were offered in our post-solving debriefing.

The commonality among the approaches was their recognition that 3 non-collinear points uniquely define a vertical parabola, so they didn’t need to worry about polynomials of degree 0 or 1.  (They haven’t yet heard about rotated curves that led to my earlier post on rotated quadratics.)

Solution 1–Regression:  Because only 3 points were given, a quadratic regression would derive a perfectly fitting quadratic equation.  Using their TI-Nspire CASs, they started by entering the 3 ordered pairs in a Lists&Spreadsheets window.  Most then went to a Calculator window to compute a quadratic regression.  Below, I show the same approach using a Data&Statistics window instead so I could see simultaneously the curve fit and the given points.

The decimals were easy enough to interpret, so even though they were presented in decimal form, these students reported $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$.

For a couple seconds after this was presented, I honestly felt a little cheated.  I was hoping they would tap the geometric or algebraic properties of quadratics to get their equations.  But I then I remembered that I clearly hadn’t make that part of my instructions.  After my initial knee-jerk reaction, I realized this group of students had actually done exactly what I explicitly have been encouraging them to do: think freely and take advantage of every tool they have to find solutions.  Nothing in the problem statement suggested technology or regressions, so while I had intended a more geometric approach, I realized I actually owed these students some kudos for a very creative, insightful, and technology-based solution.  This and Solution 2 were the most frequently chosen approaches.

Solution 2–Systems:  Equations of quadratic functions are typically presented in standard, factored, or vertex form.  Since neither two zeros nor the vertex were explicitly given, the largest portion of the students used the standard form, $y=a\cdot x^2+b\cdot x+c$ to create a 3×3 system of equations.  Some solved this by hand, but most invoked a CAS solution.  Notice the elegance of the solve command they used, working from the generic polynomial equation that kept them from having to write all three equations, keeping their focus on the form of the equation they sought.

This created the same result as Solution 1, $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$.

CAS Aside: No students offered these next two solutions, but I believe when using a CAS, it is important for users to remember that the machine typically does not care what output form you want.  The standard form is the only “algebraically simple” approach when setting up a solution by hand, but the availability of technology makes solving for any form equally accessible.

The next screen shows that the vertex and factored forms are just as easily derived as the standard form my students found in Solution 2.

I was surprised when the last line’s output wasn’t in vertex form, $y=-\frac{1}{3}\cdot \left ( x-\frac{1}{2} \right )^2+\frac{25}{12}$, but the coefficients in its expanded form clearly show the equivalence between this form and the standard forms derived in Solutions 1 and 2–a valuable connection.

Solution 3–Symmetry:  Two students said they noticed that $f(0)=f(1)=2$ guaranteed the vertex of the parabola occurred at $x=\frac{1}{2}$.  Because $f(3)=0$ defined one real root of the unknown quadratic, the parabola’s symmetry guaranteed another at $x=-2$, giving potential equation $y=a\cdot (x-3)(x+2)$.  They substituted the given (0,2) to solve for a, giving final equation $y=-\frac{1}{3}\cdot (x-3)(x+2)$ as confirmed by the CAS approach above.

Solution 4–Transformations:  One of the big lessons I repeat in every class I teach is this:

If you don’t like how a question is posed.  Change it!

Notice that two of the given points have the same y-coordinate.  If that y-coordinate had been 0 (instead of its given value, 2), a factored form would be simple.  Well, why not force them to be x-intercepts by translating all of the given points down 2 units?

The transformed data show x-intercepts at 0 and 1 with another ordered pair at $(3,-2)$.  From here, the factored form is easy:  $y=a\cdot (x-0)(x-1)$.  Substituting $(3,-2)$ gives $a=-\frac{1}{3}$ and the final equation is $y=-\frac{1}{3}\cdot (x-0)(x-1)$ .

Of course, this is an equation for the transformed points.  Sliding the result back up two units, $y=-\frac{1}{3}\cdot (x-0)(x-1)+2$, gives an equation for the given points.  Aside from its lead coefficient, this last equation looked very different from the other forms, but some quick expansion proved its equivalence.

Conclusion:  It would have been nice if someone had used the symmetry noted in Solution 3 to attempt a vertex-form answer via systems.  Given the vertex at $x=\frac{1}{2}$ with an unknown y-coordinate, a potential equation is $y=a\cdot \left ( x-\frac{1}{2} \right )^2+k$.  Substituting $(3,0)$ and either $(0,2)\text{ or }(1,2)$ creates a 2×2 system of linear equations, $\left\{\begin{matrix} 0=a\cdot \left ( 3-\frac{1}{2} \right )^2+k \\ 2=a\cdot \left ( 0-\frac{1}{2} \right )^2+k \end{matrix}\right.$.  From there, a by-hand or CAS solution would have been equally acceptable to me.

That the few alternative approaches I offered above weren’t used didn’t matter in the end.  My students were creative, followed their own instincts to find solutions that aligned with their thinking, and clearly appreciated the alternative ways their classmates used to find answers.  Creativity and individual expression reigned, while everyone broadened their understanding that there’s not just one way to do math.

It was a good day.

## Rotating Parabolas

Here’s a fun problem that continues to grow that Nurfatimah Merchant and I included in our textbook .

How many uniquely defined curves can you find whose graphs contain the points (1,1), (6,-3), and (7,3)?

NOTE:  Some of the algebra below is very intimidating to those who aren’t pretty good friends with mathematical symbol-pushing.  If you like, you can skip to the brief video clip at the end of this post showing all solutions.

SOLUTION ALERT — Don’t read any further if you want to play with this problem yourself.

Many students instantly think of vertical parabolas of the form $y=a\cdot x^2+b\cdot x+c$, but when I presented this at the MMC meeting in Chicago tonight, two stellar Geometry teachers on the front row suggested circles.  Others correctly noted that there are infinitely many curves defined solely by those three points, but I’ll talk about that in a second.  Most students first think about vertical parabolas and circles.

From the generic equation for a vertical parabola, one could plug in the three given ordered pairs and create a 3×3 system of linear equations.  I argue that a CAS is a good call here, especially when it can give the quadratic equation in multiple forms with equal ease.

That’s lots of algebra, but it was all completed very quickly using my CAS, proving another technology advantage is that all forms of an equation are equally easy to compute on a CAS.  When people try to solve this problem by hand, they invariably use the standard form only because the algebraic manipulations in the other cases are unwieldy, at best.  Here’s an image of the curve using the factored form of the equation.

The circle was also easy to get using a CAS to drive the algebra.

This graph shows the circle and the vertical parabola together.

A horizontal parabola can be obtained just as easily, but I’ll leave that one for you to discover.

This past summer, I was prodded by a teacher at the institute Nurfatimah and I ran for Westminster’s Center for Teaching to find all rotated parabolas which contained those points.  To do so, I thought that if a parabola rotated some $\theta$ units contained those points, then if I could rotate the points back $\theta$ units, the corresponding parabola would be vertical, and I could solve it using my CAS as above.

Using a transformation matrix on my original 3 points, I found their rotation images and substituted these into the standard form of a vertical parabola, $a\cdot x^2+b\cdot x+c$ as shown below.

Those bottom three lines definitely look intimidating, but for any constant value of $\theta$, each is just a linear equation in a, b, and c.  While my Nspire CAS couldn’t solve that system directly and WolframAlpha timed out, I could accomplish the algebra step-by step on the Nspire.  (Click here if you have TI-Nspire CAS software and want to see my algebra.)

Geogebra 4.2 Beta did accomplish the solution in a single line.  Its solution to each coefficient is shown here:

Substituting the expressions in terms of $\theta$ for a, b, and c back into the standard quadratic, $y=a\cdot x^2+b\cdot x+c$ and rotating the equation back into place created a generic equation for a rotatable parabola through the given three points.  I used this equation to define my rotatable parabola through the 3 points with a slider for $\theta$ here on GeoGebraTube (here is the original GeoGebra document).  You don’t get to manipulate it, but the following vimeo clip shows all of the rotated parabolas for $0\leq\theta\leq 2\pi$.

As a final bang for the presentation, one teacher in the audience wondered what it would look like if a trace was placed on the rotating parabolas.  Easily done.  Whether doing this in an Nspire CAS or on GeoGebra, right click the curve and select trace on.  Dragging the slider for the angle through all of its possible values creates the following graph.

Now that’s just pretty!  The enveloped triangle in the center is precisely the triangle circumscribed by the circle found above.  It’s straight edges are defined by the degenerate cases of the parabolas at the instances when the vertices were stretched to infinity.

As some in tonight’s audience pointed out, there are also infinitely many rotated circles and ellipses defined by these points.  Another wondered what would happen if a trace was placed on the rotating vertex instead of the entire parabola. On that last question, we’re pretty certain that the curve is some sort of “tri-perbola”–a ‘hyperbola’ with three branches–whose vertices are the three given points.  We don’t know an equation for it (yet), so that and other great extensions of this are now problems for another day.  I’d love to hear what others think or find in this problem.