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

##### 1 Answer

#### 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

Thus, we see that there is a vertical asymptote in

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

Adding

#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