Answer:
[tex]I-y=sec(x)\\III-y=tan(x)[/tex]
Step-by-step explanation:
First, let's understand what the asymptotes are. The asymptotes are straight lines to which the function approaches indefinitely, when at least one of the variables (x or y) tends to infinity. Generally, rational functions have asymptotic behavior. Now, let's check the functions:
[tex]y=sec(x)=\frac{1}{cos(x)}[/tex]
This function will have asymptotes at the points where:
[tex]cos(x)=0[/tex]
Because we can't divide by 0. Now, cos(x)=0 at π/2, 3π/2, 5π/2, 7π/2, and so on... in other terms:
[tex]cos(x)=0\hspace{7}for\hspace{7}x=\frac{\pi}{2} \pm n\pi\hspace{7}n\in N[/tex]
So, this function is a correct answer
[tex]cos(x)[/tex] is is continuous throughout its domain, so it doesn't have asymptotes.
[tex]y=tan(x)=\frac{sin(x)}{cos(x)}[/tex]
Similarly as with sec(x), This function also will have asymptotes at the points where:
[tex]cos(x)=0[/tex]
so, this function is a correct answer as well.
[tex]y=cot(x)=\frac{cos(x)}{sin(x)}[/tex]
This function will have asymptotes at the points where:
[tex]sin(x)=0[/tex]
sin(x)=0 at 0, π, 2π, 3π, and so on... in other terms:
[tex]sin(x)=0\hspace{7}for\hspace{7}x=n \pi\hspace{7}n\in N[/tex]
Therefore, this function isn´t a correct answer.
I attached the graphics where you can see that the answers are correct
