SOLUTION: Find the first TWO non-negative asymptotes and the first negative asymptote of the graph of y in radians y=2sec(x−π/2)+1 . smallest non-negative asymptote: x=

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Question 592892: Find the first TWO non-negative asymptotes and the first negative asymptote of the graph of y in radians
y=2sec(x−π/2)+1 .
smallest non-negative asymptote: x=
second non-negative asymptote: x=
first negative asymptote: x=

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
We will first find a general solution for all the asymptotes. Then we will use that to find the specific ones the problem asks you to find.

The (vertical) asymptotes will occur for x values that make sec%28x+-+pi%2F2%29 undefined. Since sec is the reciprocal of cos, sec is undefined when cos is zero. So we are interested in the solution to:
cos%28x-pi%2F2%29+=+0
You might be able to figure this out in your head. If not, we can make it easier to solve by using the cos(A-B) formula:
cos(A-B) = cos(A)*cos(B) + sin(A)sin(B)
With A = x and B = pi%2F2,
cos%28x-pi%2F2%29+=+0
becomes
cos%28x%29%2Acos%28pi%2F2%29+%2B+sin%28x%29%2Asin%28pi%2F2%29+=+0
The cos and sin of pi%2F2 are known. Substituting these values in we get:
cos(x)*0 + sin(x)*1 = 0
which simplifies to:
sin(x) = 0

This is easily solved. Since sin(x) = 0 at 0 and at pi, the general solution is
x = 0+%2B+2pi%2An (where "n" is any integer)
or
x = pi+%2B+2pi%2An (where "n" is any integer)

From the general solution above, we can now find the desired specific solutions. Just play around with different integer values for "n" until you find the two smallest non-negative asymptotes and the first negative asymptote. (Reread the start of this solution to remind yourself why these solutions to sin(x)=0 turn out to be asymptotes for your original equation.)

(You'll find that using n=0 in the first equation and n=0 in the second equation will give you the two smallest non-negative asymptotes and using n = -1 in the second equation will give you the first negative asymptote.)