November 5, 2012

Memorizing Rules

I don't like the idea of memorizing rules but Glenn Stevens of the PROMYS program talks about learning things by heart. The difference being if you learn something by heart you've tried so many examples that you learned the pattern, whereas memorizing is just being able to recite something without necessarily understanding. I thought my precalculus students had learned to shift and stretch functions by heart. Many of them probably have, it's certainly a topic they study in Algebra 1 and 2 plus they had to do a big summer project for me that involved 8 parent functions. But, two things happened today that made me wonder if there's a better way than learning "outside the function is vertical, inside is horizontal, multiplying stretches, adding shifts."

The first thing was a comment from a student after I gave back a test (which, by the way, no one studied for! They admit it at least, but apparently the juniors need to read the article on how to study math too.) There was a question asking students to compare y=sec(x) and y=sec(2x)-3. One student said to me "I feel like I just have to memorize all these parts that don't make sense." At which point I thought "Um, yes, but also no" We have done a lot of examples, made connections and they had use of a graphing calculator to answer that question so it wasn't just about memorizing. But it was also concerning that this student hadn't figured the role of each piece out yet since we've been talking about them ad nauseum.

Then I went to a class this evening at the EDC and the awesome Bowen Kerins shared an insight about shifting and stretching functions. First: the whole "inside the function is backwards" doesn't sit well with him or with students (or with me) and, all of these rules are null when you get to something like a circle where there is no "inside the function." Instead, he suggests using a change of variable to get back to the parent function. The strategy of chunking is useful all over the place, and by the time students get to calculus, u substitution will seem obvious. So here's how it works:

    y=(x-5)2 - 7

Make a table for M, N (so much easier since the vertex is at 0,0).

Then, use these linear transformations to get back to x, y.

M=x-5     x=M+5     (Shift right 5)
N=y+7     y=N-7      (Shift down 7)

To find the point you plot, use a fancy tool some call a pointer finger: cover the M and N, plot the x and y. The shifts are right there in the linear transformations you did, and they always work, no inside, outside, intuitive, counterintuitive or solving to find the vertex. It makes sense! Now if only I'd known this before all this work on shifting and stretching.

And, it works for any type of function, let's try a sine wave:

y - 4=sin(πx+π)

Even though the period of this function is 2 and the usual suspects for x values will give ugly y values, the usual suspects work perfectly for M and N.

M=πx+π     x=(M-π)/π     (Shift left π, shrink/change the period by dividing by π)
N=y - 4       y=N+4          (Shift up 4)

My students always struggle with picking the "right" x and y values to get the real shape of the graph, including the maximum and minimum.  With this method there's no need to figure out good x values to choose, they just need to know the parent function well enough and learn that one set of "good x values."

P.S. I left for school at 7 am, ran a club from 2-3 and was in class from 4-8, 45 minutes away. Over twelve hours out of the house and I still blogged! I didn't proofread, but I shared an idea. I will fill in the table and maybe another example tomorrow.

*Edited 11/6 to add tables and trig example


  1. First, congrats on the still blogging!
    As far as the method goes, it looks very similar to the I/O diagrams that I use in my teaching... for instance, let's say we have y=2sin(4x+8)-3. (Um, I'm going to work in degrees because I don't have a pi key.)

    We start with the x, and build up the entire function:
    x -> x4 -> +8 -> function sine -> x2 -> -3 -> =y

    Now simply grab a lot of key points on the original function (in this case sine, but could be a parabola or root or whatever), and follow the arrows to figure out where it maps. If you're tracking an arrow backward, use the inverse operation. So...

    x -> x4 -> +8 -> function sine -> x2 -> -3 -> =y

    -2 <- (0-8)/4 <- (0, 0) -> (0x2)-3 -> -3
    20.5 <- (90-8)/4 <- (90,1) -> (1x2)-3 -> -1
    43 <- (180-8)/4 <-(180,0) -> (0x2)-3 -> -3
    65.5 <- (270-8)/4 <-(270,-1)->(-1x2)-3 ->-5
    88 <- (360-8)/4 <-(360, 0) -> (0x2)-3 -> -3

    Period of 90 comes right out. Nice thing about it is you don't need to worry about factoring out the "4" first, it kind of takes care of itself in the order you're doing the operations. (If you do factor it, divide first, then subtract 2.)

    The main difference I see is that with the M, N you're isolating for the function on one side first, and thus in both cases using inverse operations. Which may make more sense when you have the circle or something, I'd have to think about it.

  2. If there were extra credit available for all you did today, please know that I would definitely give you all the points!!!!!

    Elizabeth (@cheesemonkeysf)