Conversation between me and a learning-disabled student. The quotes aren't exact, but I think I've correctly captured the overall pace, tone, and spirit of the exchange.
ME: OK, so you remember the slope-intercept form for graphing a line?
LD STUDENT: It's "y equals mx plus b."
ME: Perfect. And you've worked with finding a line's equation by using two points, right?
LD: Yup. (makes a weird, bubble-popping sound when pronouncing the final "p")
ME: So let's review that. Here are two points. Point A will be... (writing while talking) at (5,5), and Point B will be at... let's see... (-1,-3). Can you find the slope from that?
LD: Uh...
ME: The slope, in general, is what over what?
LD: Umm... 2 over 3?
ME: No, it's rise over...
LD: Oh, yeah-- rise over run.
ME: Right. So in this case, with the two points I gave you, can you calculate the slope?
LD: Uh...
ME: We talked about this before. You're basically making a fraction, but you have to subtract something from something, and something else from something else.
LD: Huh?
ME: The top of the fraction is going to be what minus what?
LD: Uh... this and this? (points to "5" and "-1," the x-coordinates)
ME: No, those are the x-coordinates. Is the top of the fraction going to be the difference of y's or the difference of x's?
LD: Difference of y's?
ME: Right. Slope is rise over run, and the y-coordinates are all about the up/down while the x-coordinates are all about side-to-side. So where are the y-coordinates?
LD: Here? (points to "5" and "-3," the proper y-coordinates)
ME: Good. So those are the y's. And which ones are the x's?
LD: These? (points to "5" and "-1," the proper x-coordinates)
ME: Great! So can you make this fraction?
LD: Is it like this? (writes "(5 - (-3))/(5 - (-1))")
ME: Perfect! So what's next?
LD: (writes "8/4")
ME: Hm. Something's wrong here.
LD: (writes "-8/4"; he's obviously just guessing)
ME: OK, you did the subtraction correctly on the top of the fraction: minus a negative is like adding a positive, so 5 minus negative 3 is like 5 plus 3, or 8. That part's fine. So--
LD: Oh! (redoes the subtraction in the denominator and gets the proper result: 6)
ME: So it's 8 over 6, right?
LD: Yup. (weird popping sound)
ME: Are we done yet? Can this fraction be reduced?
LD: (wordlessly reduces 8/6 to 4/3)
ME: You got it! So now we have our slope. What do we do next?
LD: (stares at me uncomprehendingly)
ME: What's the whole point of this exercise?
LD: (still staring, brow furrowed in earnest concentration)
ME: We're trying to find the equation of this line by using two points, right?
LD: (nods, not really following)
ME: So now we've got the slope, right?
LD: (nods)
ME: So where do I plug that into the slope-intercept form? Which part is the slope?
LD: b?
ME: No, it's m. Remember, m is the slope. So now we can plug m into the original equation, like so-- (jotting down "y=mx+b," then replacing m with 4/3). But we're not done, right? We still have to find b. So how do we do that?
LD: I don't know.
ME: Sure, you do! We've already been over this three or four times, you've done it at school, and you went over it with my coworker the other day.
LD: Uh...
ME: You've already been given two points, so all you need to do is choose one of those points, then plug the x and y values into the equation we're making. So choose a point. Point A or Point B?
LD: (looking distracted) A.
ME: OK. So Point A is (5,5). How do I plug these values into the equation?
LD: b equals 5?
ME: No. Remember that each point is in the form (x,y), so--
LD: Subtract?
ME: No. You--
LD: Multiply?
ME: No; you need to plug the numbers into the equation. Do you understand what I mean by "plug the numbers in"?
LD: (blinks rapidly, uncomprehendingly)
ME: (sighing) OK, so we've got this equation: y = (4/3)x + b. We've chosen (5,5) to plug into the equation, so I write a "5" instead of the y and a "5" instead of the x. What's the equation?
LD: y = mx + b?
ME: No-- what's our new equation after I plug in all the numbers?
LD: Oh. (starts writing... eventually writes "5 = (4/3)(5) + b")
ME: Good! So where do we go from here?
LD: (stares again)
ME: We're trying to solve for b, right?
LD: (nods and blinks rapidly)
ME: So we've got to get b all by itself. How do we do that? What's the first step?
LD: (clearly no clue, and this is pre-algebra stuff)
ME: Let's multiply 4/3 by 5 first. What's that going to be?
LD: (calculates on paper) 20 over 15?
ME: Where'd that extra 5 come from?
LD: I don't know.
ME: Do you remember how to multiply fractions? Top to top, bottom to bottom?
LD: Yeah, so I did 4 times 5 and 3 times 5.
ME: But there's no second 5.
LD: (sullen silence)
ME: The whole number 5 can be rewritten as a fraction, right? What's that fraction?
LD: 5 over 5.
ME: No, it's 5 over 1.
LD: Oh, riiiiight...
ME: OK, so if we multiply 4/3 by 5/1, we get...
LD: Oh, wait-- that's 20 over 3.
ME: Right! So if we're trying to get b by itself, what happens next?
LD: (blank stare)
ME: Don't we need to subtract...?
LD: OK. (does nothing)
ME: (sighing again) So we subtract 20/3 from both sides (writing out the subtraction), which means the left side of the equation now says "5 minus 20/3," and on the right side, all that's left is...?
LD: Zero?
ME: No...
LD: One?
ME: No...
LD: I don't know.
ME: It's b. All that's left is b.
LD: (sounding skeptical) O...kay.
ME: So b equals...?
LD: 5 minus 20/3?
ME: But we need to evaluate that expression. We can't leave it that way. So...
LD: It's minus fifteen thirds?
ME: Huh? No, that's not it. You've got to convert your whole number into a fraction-- something over 3.
LD: (blinking while staring at the paper we've been working on)
ME: (writing "5/1 = [ ]/3") 5 over 1 equals something over 3.
LD: 3?
ME: No. How did I go from 1 to 3 for the denominator? What times one is 3?
LD: 3?
ME: Right.
LD: (irritated) But that's what I said.
ME: No, you were saying "three" for the wrong thing before. So if 1 times 3 is 3, then 5 times 3 is...?
LD: 15.
ME: Good! So what's the fraction?
LD: 15 over 3?
ME: Right. So now we can subtract. 15/3 minus 20/3 equals...?
LD: Negative 5?
ME: Negative 5 over 3.
LD: OK.
ME: So where do we go from here?
LD: (that famously blank stare)
ME: We've just solved for b, right? So what do we do?
LD: (still staring)
ME: We've got our slope and our y-intercept. Can't we plug those into the equation?
LD: I guess.
ME: So give it a try.
LD: (poised, looking ready to write something on the page... eventually writes nothing)
ME: What's our slope?
LD: Negative 5 over 3?
ME: No, that's our y-intercept. (pointing at the slope written on the page) What's the slope?
LD: Oh, yeah-- four thirds.
ME: And what's the y-intercept?
LD: Negative 5 over 3?
ME: And what's the slope-intercept form?
LD: y equals four-thirds x plus b?
ME: No, no-- the general form?
LD: (stares, unsure what to say next, fearful of being wrong again)
ME: It's y = mx + b.
LD: Oh. Right.
ME: So now let's plug in our slope and our intercept...
LD: y equals four-thirds x plus five thirds?
ME: Almost there, but remember that the y-intercept is negative.
LD: y equals four-thirds x minus five thirds?
ME: Good!
LD: (frustrated) But I said four thirds a second ago and you said I was wrong!
ME: I was asking you for the general equation before; now we're talking about the equation for this particular line. But at long last, we've got our equation, so congratulations. So, one more time-- what's the slope of this line?
LD: (long pause) Five thirds?
ME: (sighing)
_
Damn you, Kevin! You promised me you wouldn't post this on your blog! I'll never ask your help on math again!
ReplyDeleteBy the way, the word verification is "aurce"! That's a British word. A quant like you probably doesn't know what it means, but it's what you are for telling on me!
Jeffery Hodges
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Sorry, Jeff, but the world has a right to know about the many crosses I bear.
ReplyDelete