Let 'Em Drive

The Problem:

Find the x-intercepts of y = x² + 10x + 16.

Me:  How many different ways could we do this?

Class:  Quadratic Formula, completing the square and factoring.

Student 1:  Mr. Cox, why does y = 0?  Isn't this a function where y can equal a bunch of things?

Student 2:  Yeah, but we want y = 0 because that's when it crosses the x-axis.

Student 1: Ok, so we are just focusing on one possible value of y.

Student 3:  So then, if we let y = 0, then we are finding the x values that make y = 0, right?  But what if we want to know when y = 1?  Can we do that too?

Student 4:  I guess we should just let y = 1, but we'd have to subtract it from both sides so we get:
0 = x² + 10x + 15.  So now we are finding the x-intercept again.

Me: Is this an x-intercept?  What does y equal?

Student 4:  Oh, no, y = 1.

Student 5:  Mr. Cox, can we see what this looks like in GeoGebra or something?

We graphed it and then moved a point around on the parabola to verify our results and looked at how the points where y =1 maintained the same symmetry with the x-intercepts.  Then, a kid pipes up:

Are there graphs that have more than two x-intercepts?

Me:  Hmm.  Maybe.  I'll tell you what...you guys give me a function that has x intercepts at -4, 0 and 3.  You have 15 minutes.  Go!

About 5 minutes later, J.V. walks up with this:

y = (x + 4)(x - 0)(x - 3)

y = x³ + x² - 12x

And I ask him, "How did you come up with this?"

J.V.: Remember the other day when we had to come up with parabolas with x-intercepts?  I figured I could just work backwards like we did then.

I'm amazed at how often these kids will take the lesson into places I wouldn't have thought to go.  It's just a matter of letting go a little.  And the more I look for open ended opportunities, the return is exponential.

No Title Necessary

- Posted using BlogPress from my iPad

Some GeoGebra Love

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Inspired by Greg Hitt

I'm Holding Somebody Responsible For This!

Step 1: Choose values for 'a' and 'c'.
Step 2: Move the 'b' slider from -10 to 10.
Step 3: Select all the points step 2 puts in the spreadsheet. Right click and select "Create list of points."
Step 4: Click on the "Conic through 5 points" tool.
Step 5: Click on 5 of your points generated in step 3.
Step 6: Right click on the resulting function in the algebra window and change standard form
Step 7: Compare this function to the Starting Function
Step 8: Explain to me why I never knew this before.




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Quadratics Revisited: The Falling Object Model

I keep hoping that I can use this to help kids derive the falling object model. I'm getting close. I exported the video at 6fps so 1/6 second elapses between strobes. I'd like some feedback on this before I roll it out to my students. How intuitive is it to use both applets together?



This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com


Once you have plotted your points, use the FitPoly function. Simply enter "fitpoly[A,B,C,D,E,F,G,H,I,2]" to plot a quadratic function.



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If this proves to be useful, I'll dial it all in and post the still, video and applets for download.

Mixture Problems

You know this one too:

Given:

Solution A contains 20% acid

Solution B contains 60% acid

Find: The number of mL of each necessary to obtain 450mL a solution that contains 30% acid.

I usually teach this the same way I do work and rate problems. Find what we call a "guiding equation" that is based on the units available in the problem and set up some sort of table to organize the information. Depending on the given information, we may use two variables and solve a system of equations. Maybe not. But at the end of the day, this is going to be some complex variation of a rate problem.

I have a colleague who sees these problems in terms of proportions and our recent conversations have inspired this GeoGebra applet.



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There may be some rounding error, but I think this gives an interesting conceptual spin to a problem that can very easily become an algorithmic nightmare.

We can make some quick adjustments for the good ol' price per pound problems as well.




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Download % Mixture applet
Download Price per Pound applet

Work Problems

Y'all know the drill.

Person A does a job in A hours, Person B does the same job in B hours.  How long does it take them to do the job together?

Teaching the work problem can turn into a debacle fast.  Really fast.  Most teachers boil these down to an algorithm they themselves might not understand.

"Here's the formula, kids.  Learn it.  Know it.  Use it.  Now let's move on to something else."

That's never set well and a couple of years ago, it dawned on me that a work problem is nothing more than a complex rate problem.

Person A can do 1/A of the job per hour and person person B can do 1/B of the job per hour.  How much of the job do they complete per hour when working together?  

This has been a game changer.  Find the common denominator and add 'em up.  Now we have a common rate.  From there it's pretty easy to determine how long it takes to complete the job.  We give no formulas and kids understand how to find an answer and why it works.

We had kids derive two different algorithms for work problems.

x = (AB)/(A+B)

and

1/A + 1/B = 1/X

Sure, kids.  Use it.  G'ahead.

But then we added the dreaded twist.

Person A can do the job in A hours.  If person B helps, they can do the job in X hours.  How long would it take person B to do the job when working alone?

I give you, Clinton's Theorem:

I kid you not,  this kid came up with this all himself.  By "integer portion of X" he means, everything to the left of the decimal.  He's truncated the number to make things "simpler."  

As he was presenting this to us before we left for break, I was baffled.  It works, but why?  So today, I sat down next to him and had him explain the formula to me again, I took notes and then simplified the expression.  I won't deprive you the joy of doing so yourself.

Seriously, have at it.