Intercept TheoremDate: 06/02/2003 at 23:10:10 From: Travis Subject: My own Theorem A few teachers helped me prove my own theorem. My teachers have told me they have never seen a theorem like it before. My theorem states that by dividing (B) by the opposite of (M) or (-M) you get the x-intercept. Date: 06/03/2003 at 08:56:13 From: Doctor Peterson Subject: Re: My own Theorem Hi, Travis. This is not new, but is certainly a useful fact. You can prove it easily: y = mx + b 0 = mx + b for x-intercept -b = mx subtracting b from both sides -b/m = x dividing both sides by m So it doesn't save a lot of work, but if you are working on a lot of these problems, it can add up. I myself probably derive this fact the way I just showed every time I solve one of these; that is, rather than memorizing it, I am aware in the back of my mind that there is such a trick, go through the bit of algebra in my head quickly to remind myself what is divided by what, and then apply it to each problem. That's a good way to keep your mind uncluttered but be able to save time on repetitive work. I'm a little surprised your teachers haven't seen it; but that may be because it is not generally taught as a fact by itself. Since you and I solve these problems on our own (you as a student, I in helping students without having an answer book!), we are more likely than a teacher to do them enough to want and discover a shortcut like this. Keep looking for easier ways to do things, because that's how a lot of useful math is discovered. If after every problem that's at all hard, you look back and ask yourself, "How could I have done this more quickly?" - or, "How can I do the next one with less work?" - you will be using your mind at its best. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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