Question

For what values of x, if any, does `f(x) = sec((-15pi)/8+2x)` have vertical asymptotes?

Answer 1

`x=(pi(19+8k))/16,kinZZ`

Explanation:

`f(x)=sec((-15pi)/8+2x)`

Rewriting using the definition of secant:

`f(x)=1/cos((-15pi)/8+2x)`

This function will have vertical asymptotes when its denominator equals zero, or when:

`cos((-15pi)/8+2x)=0`

Note that the function `cos(theta)=0` when `theta=-pi/2,pi/2,(3pi)/2`, which can be generalized as saying that `cos(theta)=0` for `theta=pi/2+kpi,kinZZ`. Note that `kinZZ` is the mathematical way of writing that `k` is an integer.

Thus, we see that there is a vertical asymptote in `f(x)` when:

`(-15pi)/8+2x=pi/2+kpi" "" "" "" ",kinZZ`

Adding `(15pi)/8` to both sides:

`2x=pi/2+(15pi)/8+kpi" "" "" "" ",kinZZ`

`2x=(19pi)/8+kpi" "" "" "" ",kinZZ`

`x=(19pi)/16+(kpi)/2" "" "" "" ",kinZZ`

`x=(pi(19+8k))/16" "" "" "" ",kinZZ`

So, there are an infinite number of times when there are vertical asymptotes. We can identify those close to the origin by trying `k=-1`, which gives an asymptote at `x=(11pi)/16`, or any other value of `k` for a different asymptote.