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.