Question

How do you find a vertical asymptote for y = sec(x)?

Answer 1

y=sec(x)

For any y=sec(x), vertical asymptotes occur at x=

π

2

+nπ, where n is an integer. Use the basic period for y=sec(x), (-

π

2

,

3π

2

), to find the vertical asymptotes for y=sec(x). Set the inside of the secant function, bx+c, for y=asec(bx+c)+d equal to -

π

2

to find where the vertical asymptote occurs for y=sec(x).

x=-

π

2

Set the inside of the secant function x equal to

3π

2

.

x=

3π

2

The basic period for y=sec(x) will occur at (-

π

2

,

3π

2

), where -

π

2

Answer 2

The vertical asymptote for y = sec(x) is -

x = ( 2n +1 ) π/2,where n is any integer

now,

y = sec x = 1/cos x

To have a vertical asymptote, the limit (one-sided) has to go either to either  ∞ or  − ∞ which occurs when the denominator is zero.

By solving -

cos x = 0

x = π/2 . 3π/2 , 5π/2

x = π/2 + nπ = ( 2n +1 ) π/2

Therefore, the vertical asymptote will be -

x = ( 2n +1 ) π/2