PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
We’re finding the intercepts and asymptotes of a tangent function. Let’s do this one; y equals 1/3 tangent of pi over 2x.
First thing I want to do is make a substitution, theta equals pi over 2x, because remember tangent theta equals zero when theta is an integer multiple of pi. Theta equals nPi. So then I can substitute back. Pi over 2 x equals nPi and in order to solve for x I multiply both sides by 2 over pi, pi's cancel, over here and I get x equals 2n.
Tangent is going to be zero when x is for example, zero, 2, 4, 6 and so on and the intercepts are going to be (0,0), (2,0), (4,0) and so on. Even numbers comma zero.
What about the asymptotes? Tangent theta is undefined when theta is pi over 2 plus nPi. Again we go back to theta equals pi over 2x. I substitute pi over 2x equals pi over 2 plus nPi and I just multiply through by 2 over pi again, same trick. Just be careful because you’ve go to distribute this 2 over pi over both of these terms. Pi over 2 times 2 over pi is 1 and nPi times 2 over pi is 2n. So the asymptotes are going to be x equals, and of course they’ll be many of them, 1, 3, 5, 7, and so on. Also x equals -1, -3, -5, -7 and so on. All the odd numbers, those will be the asymptotes.
Remember this trick, set theta equal to zero and remember that tangent theta equals zero when theta equals n pi and tangent’s undefined when theta is pi over 2 plus n pi.