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 intercepts and asymptotes of tangent functions. Here’s a harder example; y equals 6 tangent of ½ x plus pi over 2.
Now remember the tangent of theta equals zero when theta is an integer multiple of pi. So if I said this inside stuff equal to theta then I can just substitute 1/2x plus pi over 2 equals nPi. The first thing I want to do is subtract this pi over 2 from both sides and I get 1/2x equals negative pi over 2 plus nPi. Then I multiply through by 2 and I get x equals, now remember to distribute the 2, 2 times negative pi over 2 is negative pi and 2 times nPi is 2nPi. So the zeros will be negative pi, pi, 3pi and so on. And the x intercepts will be negative pi zero, pi zero, 3pi zero and so on. These are the odd multiples of pi.
What about the asymptotes? Tangent theta is undefined when theta is pi over 2 plus nPi. Again we substitute 1/2x plus pi over 2 for theta and that equals pi over 2 plus nPi. That’s nice, the pi over 2's cancel and we’re left with 1/2 x equals nPi and then we just multiply by 2. X equals 2nPi. These are the even multiples of pi so the asymptotes are going to be x equals well for example 0, 2pi, 4pi and so on. The asymptotes will be x equals the even multiples of pi, the intercepts will be x equals the odd multiples of pi.
That’s a harder example but even when you have both the horizontal stretch or shrink and a shift to the left or right, setting theta equal to the inside stuff always works.