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.