2004-03-13T11:53:00.000+01:00

So recently I came to the decision to teach myself precalc. Precalculus that is. I took it in college, but it didn't quite click. Which was odd for me, because up until that moment, all things mathematick-al and I had always clicked. Maybe, had something to do with the fact: it was my freshman semester, and an incredibly early and dull morning class, taught by a clever, but casually disinterested professor.

Maybe had something to do with my highschool, not filling me in too completely on everything I needed to know math-wise. For instance: I never learned logic. In fact: I never even knew there was a logic. Except of course for: common sense type logic. Like: "it's only logical not to walk off a cliff" logic. Not math-logic, you know? Anyway I started tutoring high-school students, and a bunch of them were like: "can you help us with our logic assignments?" and I was like: "what's logic?"... I suppose that's another thing I should teach myself sometime.

Anyway:

I started my self education by typing "learn precalculus" in a search engine and coming up with this site: Precalculus Problems Website

and I clicked on the first little lesson: Lines.

Here's what I thought of it.

2004-03-12T11:56:00.000+01:00

The Uselessness of the Point Slope Form

So I clicked on the lines thing, and it was lines. Straight line equations, which I figured would be review-slash-easy for me, since I've never had any trouble with them. And there it was: this formula in front of me, I'd never seen in my life.

The Point Slope Form: The line through the point (x1, y1) with slope m has the equation y - y1 = m(x - x1).

I looked it over afew times, and found it completely alien to me. Not only alien, but making no sense whatsoever that I could see. How would y - y1 = m(x - x1) work? It was a strange little sequence to have to memorize. It doesn't role off the tongue at all like y = mx + b, or say... the pythagorean theorum (a squared plus b squared = c squared [for when dealing with the lengths of the sides of a right triangle.]) Anyway, the point is it annoyed me. I didn't get it, and had never heard of it, nor used it before: so it occured to me to wonder how I'd managed to get anywhere with straightline equations without it. I looked it over again, and realized it was a formula to determine slope of a straightline, and then afterwards it's full equation: (AKA: a Slope-Intercept Form, apparently) using two coordinates (x and y,) and the slope. It occured to me that if I had two coordinates, I wouldn't need to remember a formula as oddly formed as the Point Slope, so I simply shrugged it off and moved on to the first problem.

"Find an equation of the line which passes through the point (4, -7) and has slope 3."

I looked at it and quickly multiplied 4 by 3 (12)
and subtracted it from negative 7,
determinining the y intercept to be negative 19,
and the equation to therefore be y = 3x - 19.

So I looked down at the solution to check if my answer was right, expecting to see a simplified version of the math I'd just done. And I found this:

"Sol Using the Point-Slope Form," (( y - y1 = m(x - x1) ))
"we obtain the equation y - (-7) = 3(x - 4)
or, simplifying, y + 7 = 3(x - 4)
or y = 3x - 19."

I studied it for awhile and realized it made sense. Though it was the most complicated way of solving the problem I'd ever seen. Below it, it suggested a simpler method based on steps more similar to the ones I'd taken.

"If instead we use the Slope-Intercept Form," (( good old y = mx + b ))
"we get the equation y = 3x + b.
To determine b, we substitute x = 4
and y = -7
to obtain 7 = 3(4) + b,
so that b = -19.

Now this makes much more sense to me. It feels a little more complicated than the way I think of it in my head, but at least it presents the problem as it is. All you're doing is solving for the y-intercept. There's no need to bring anything as complicated as y - y1 = m(x - x1) into the mix. Still: in other examples throughout the site. The solutions given to the students, are done using the Point-Slope ( y - y1 = m(x - x1) ) Form, without even suggesting the use of Slope-Intercept (y = mx + b) as an alternative. Is this the way we teach people math? It's no wonder so many people think of math, and the most brain dead and useless of the mental disciplines, with worthless and overly complicated equations like point-slope floating around. Teaching people to solve for a y-intercept in an equation using Point-Slope is like teaching children to walk around in semi-circles instead of straight lines to get from one place to another. When I think of the other kids in my class getting taught by this method, it strikes me as no wonder, that I managed to stay ahead, as I was lucky enough to have figured out (through what feels like common sense to me) a simpler method of solving the same problems, that they were all being force-fed complex and unnecessary equations to figure out.

I figured out the math in my head, that I always do, and I guess it turns out to be a variation on Point-Slope that makes much more sense to me.

Shift y – y1 = m(x – x1) around a little and you get this: y = mx + y1 - mx1

so if y = mx + b and y = mx + y1 - mx1

then y1 - mx1 = b

There's your simple equation for solving straight line equations, when you got your slope and your two coordinates.

"Find an equation of the line which passes through the point (4, -7) and has slope 3."

y1 - mx1 = b
-7 - 3(4) = b
-7 -12 = b
-19 = b
y = 3x -19

And it's simple because all your doing is taking your two coordinates, and tracking the line backwards to the y-intercept. You got your slope: 3, and your just taking four hops backwards along the line, and while X-goes back 4, y goes downward from -7 by 3 times as much, because 3's what the slope is. It's easy to explain, and makes alot more sense. This is the way to teach it I think.

Not by Point-Intercept...
that's just wrong.

The J-Formula