# Locate the vertical asymptotes,and sketch the graph of this function `y=sec(x/5+pi/5)`

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tiburtius | Certified Educator

Let's first write our function in a different way.

`y=sec(x/5+pi/5)=1/(cos(x/5+pi/5))` because `sec(x)=1/cos x`

Now we can see that vertical asymptotes will be in points where denominator is equal to 0. Thus we need to find solutions of equation

`cos(x/5+pi/5)=0`

Because `cos x` is equal to 0 when `x=pi/2+k pi,` `k in ZZ` we have

`x/5+pi/5=pi/2+k pi`

`x/5=(7 pi)/10+k pi`

`x=(7 pi)/2+5k pi`

So vertical asymptotes will be in points `{x:x=(7pi)/2+5k pi,k in ZZ}`. The graph above shows 4 of such points hence 4 asymptotes which we get for `k=-2,-1,0,1` .