Answer:
Explanation:
If we refer to the unit circle, we can see that the first period of the original tangent function is undefined at plus or minus pi/2 (because tan(x) = opposite/adjacent and opposite =1 and adjacent = 0, we also know that regardless of multiplicative x transformations, tan(0)=0 because 0/1=0), this is precisely where there are vertical asymptotes. With the transformation of multiplying all x value inputs by 2, our new function's asymptotes will have to be at x= plus or minus pi/4 to compensate for the (tan2x), but we can also use values in the second and third quadrant like 3pi/4 and 5pi/4. The mid points shouldn't prove too much an issue, since the midpoint between two points is half the distance from each, we can easily find the absolute value. From 0 to the asymptote at pi/4, the total distance is pi/4, but half of that, or the midpoint would be pi/8. If we want to we could effectively say that all of the mid points from the x intercepts to the asymptotes for the function tan(2x) are every (pi/4)(2z-1) plus or minus pi/8, where z is any integer.