What if We Start With the Axis of Symmetry?
Lesson 13 of 21
Objective: SWBAT to create "Five Point Graphs" of quadratic functions by considering the axis of symmetry and its relationship to important points on a parabola.
Today, students will have the opportunity to review what they know so far about quadratic functions: how to factor an expression to reveal the roots of the function it defines, how to determine the axis of symmetry and vertex of a function in standard form, and how to calculate the discriminant. They will also continue to practice graphing quadratic functions, as they learn a novel method that starts by considering the axis of symmetry.
For the opener, which is on the first slide of today's lesson notes, I ask students to find the roots and the axis of symmetry for three quadratic functions. The first two expressions can be factored into two binomials; the third cannot.
I give students space to get started, and I circulate to take a peek at what everyone is able to do. After a few minutes, I call everyone to attention to discuss the solutions. By this point, there's an energetic buzz that we can't get the roots for exercise (c) by factoring, and students are eager to share what they've figured out.
As I've done frequently over the last two weeks, I make a table on the side board to collect what we know about each of these functions so far. We're going to continue to use these functions in the next two sections of today's lesson, so I take a moment to ask if anyone has questions and to acknowledge some student ideas about the roots of the third function before moving on tackle just that challenge.
Immediately following the opener, we briefly review the formula for the axis of symmetry and how to calculate the discriminant of a quadratic function.
Slide #2 of the lesson notes is repeated from yesterday's lesson, when kids saw the axis of symmetry formula for the first time. When I post this slide, I ask students if they've already got this formula in their notes. I know they do, as I've just seen most students use the formula to complete that part of the opener. I write x = -b/2a on the board again, just to give everyone a chance to see it and feel confident that they're getting their own grasp of it.
With slide #3, I ask students if they can find the discriminant of each of the three functions on the opener. It's been a few days since we've used this tool, so I encourage students to find relevant notes. By encouraging group-mates to help each other out, I can count on all students being able to complete this task. As students finish up, I ask them what they notice about the discriminant values they've found for each function. I want them to notice that the first two functions have the same discriminant, 64, which is also a perfect square, but that for the third function the discriminant is 20, which is not a perfect square.
I hope to see that students can use this to explain why we could factor to find roots for the first two functions, but not for the third. I also hope to get students to wonder what else the first two functions have in common. If they have the same discriminant, then what else might they share?
Depending on how confident students are in finding the discriminant for each function, there are two versions of the assignment that follows. I will share the version that I use when most students can find and talk about the discriminant, and then I will share an adapted version of the lesson in the next section.
Five Point Graphs
Today, students will get started on a new graphing activity. Here is the Five Point Graphs handout, and a video in which I define what I mean by a "Five Point Graph". I'm going to share a lot of ideas in this section of the lesson, but you can the main gist of things by checking out that video.
The big idea here is that we can sketch a parabola by starting from the axis of symmetry. If we use the properties of symmetry and what the discriminant tells us about the relationship between the two roots, we can get all the points we need to make a complete sketch without factoring. The paths to sharing all of these ideas with students are as varied as their learning styles, and it's important for a teacher to be flexible here! I present all sorts of possibilities here, and by paying attention to my students, I'll lead them through as much as we can do today and tomorrow.
Let's use "Function A" as an example, although "Function B" works just as well. From the opener, we already know the axis of symmetry and the roots of this function, so now our first step is to briefly review how to find the vertex of this parabola. By evaluating the function at f(7), we can see that the vertex is (7, -16). I sketch a pair of axes on the board, draw a dotted line to represent that axis of symmetry, and I plot the vertex. After that, I plot the roots at (3,0) and (11,0). "What is the distance between these roots?" I ask. Students are quick to note that the distance is 8, and they recall from the last few lessons that the axis of symmetry is halfway in-between these two roots.
Now I challenge students to see what they notice. "So we know that the distance between the roots is 8," I say. "Is there anything else we know about this function that might have something to do with the number 8?" I'm fishing to see if anyone makes the connection that the square root of the discriminant is 8. If a student makes that observation, I congratulate them and say that they're onto something there. If one gets there quite yet, I'll return to this in a few minutes.
I point to the words "Five Point Graphs" at the top of the board, and I say that we've got three points so far, with two to go. "Can anyone tell me another key point on this graph?" I ask. If we stall here, I get more specific: "How about the y-intercept? What's that?" I ask. At that point, students are pretty quick to point out that the y-intercept is at (0,33), and most kids are satisfied that that point is simple enough to find. "Now we need one more point," I say. "Let's consider the axis of symmetry once again. You can see that the roots are each four spaces away from the axis of symmetry." I pause for a moment to allow students to think. "So how far is the y-intercept from the axis of symmetry?" I ask. We're all quick to agree that (0,33) is seven spaces to the left of the line x=7. "So what if move seven spaces to the right?" I ask, showing what I mean by counting those space on the board. I land on (14,33), and students are satisfied that this point appears to work out. If anyone is skeptical, I'm stoked: that's an opportunity to check whether or not f(14) = 33. If everyone is content taking my word for now, however, I don't press that.
So now we've got a Five Point Graph. I connect the dots with a rough sketch and encourage everyone to make sure they've got this example in their notes.
Finally, I revisit the discriminant from earlier in the lesson. "So we know that the discriminant of this function is 64," I say. "Note that the square root of the discriminant is 8. That's the distance between the two roots!" I say, again pausing to let students consider this connection. Someone may have noticed this idea when I prompted the class for it a few minutes before, but now I want to make it explicit. "Not only that, but each root is half that distance from the axis of symmetry. So if the discriminant is 64, and the square root of that number is eight, we can see that each root is four spaces away from the axis of symmetry."
It's a lot all at once, but I'm purposefully throwing everything on the table here. When they get to work, students will have a chance to make sense of what they've seen here.
I distribute the Five Point Graphs handout, review the instructions, and tell everyone to get started. Students should use what they've just seen to sketch a five point graph for each of these eight quadratic functions. I leave the initial example on the board as everyone gets started, and I circulate to answer clarifying questions about the assignment. Then, I await opportunities to share another example and a few more ideas.
What I really want students to understand today is that we can find the roots of a parabola without factoring if we use what we know about the axis of symmetry and the discriminant. The best way to help students understand this idea is to look for a chance to help them see this idea in the context of their own work.
With that in mind, I've prepared slides #6 through #9 of the Lesson Notes, and I plan at some point to use at least one function from the student handout as an example, but there's no prescription for exactly which one I'll use or when. This work will also continue into tomorrow's lesson, and if students are doing great work without my help today, then all of this can wait.
Here are a few scenarios for how I might share another example today:
- We might look at the other two functions from the opener. We'll see that although it was easy to find the roots of "Function B" by factoring, the same can't be said for "Function C". Extending from our observation that the square root of the discriminant gives us the distance between the two roots, we can use the square root of the discriminant for Function C to see that the distance between those roots is in-between 4 and 5, as is square root of 20. Half of that distance indicates that each root should be "a little more than 2" to the left and right of the axis of symmetry. Students will see that this method can help them create a Five Point Graph for any quadratic function, even those that are "un-factorable".
- If I start to see that students would welcome an alternative approach to creating a Five Point Graph for even those quadratic expressions that are factorable, I will use one of the problems from the Five Point Graphs handout as an example. In this case, we'll start with the axis of symmetry and the vertex, then use the discriminant to figure out how far each root should be to each side of the axis of symmetry. For many students it's very helpful to create a checklist that everyone can use to move through this activity. If this doesn't happen today, I use an example like this tomorrow, to ensure that kids get to see this idea.
- Students might cruise through the first six functions on the handout pretty quickly. If they do, the last two functions provide examples of equations that make it impossible to get a five point graph - after all, if there are no real roots, then we can't plot those points. Just like I note above, if this doesn't happen today, then I'll expect to address this problem tomorrow.
- Slides 7, 8, and 9 of the Lesson Notes present a few more opportunities. On slide #7, there's a function in which b = -5, which means that it won't have an integer value for its axis of symmetry. Slide #8 presents a function with just one root - and therefore a discriminant of 0 - which means that the vertex and the root are actually at the same point. Can we really make a "five point graph" in this case? Slide #9 presents another example of a function with no real roots.
A lot could have happened today, and as class comes to close, I tell students that as long as they worked hard today, whatever happened was exactly what was meant to happen. I encourage everyone to gather their work and take a good mental snapshot of where they're leaving off, because this work will continue tomorrow.
For teachers reading this, I hope that what I've shared here makes sense and equips you to give this activity a try on your own. I also hope that the sometimes rambling nature of what I've written makes it clear that I'm really trying to put students at the center of my classroom, and that although there's this rich, exciting idea that I'm trying to get across, the precise narrative of how everything develops depends on what my real students are actually doing.