Trigonometry Examples

$y = 2 \sec (x)$

Step 1

For any $y = \sec (x)$ , vertical asymptotes occur at $x = \frac{π}{2} + n π$ , where $n$ is an integer. Use the basic period for $y = s e c (x)$ , $(- \frac{π}{2}, \frac{3 π}{2})$ , to find the vertical asymptotes for $y = 2 \sec (x)$ . Set the inside of the secant function, $b x + c$ , for $y = a \sec (b x + c) + d$ equal to $- \frac{π}{2}$ to find where the vertical asymptote occurs for $y = 2 \sec (x)$ .

$x = - \frac{π}{2}$

Step 2

Set the inside of the secant function $x$ equal to $\frac{3 π}{2}$ .

$x = \frac{3 π}{2}$

Step 3

The basic period for $y = 2 \sec (x)$ will occur at $(- \frac{π}{2}, \frac{3 π}{2})$ , where $- \frac{π}{2}$ and $\frac{3 π}{2}$ are vertical asymptotes.

$(- \frac{π}{2}, \frac{3 π}{2})$

Step 4

Find the period $\frac{2 π}{| b |}$ to find where the vertical asymptotes exist. Vertical asymptotes occur every half period.

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Step 4.1

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.2

Divide $2 π$ by $1$ .

$2 π$

Step 5

The vertical asymptotes for $y = 2 \sec (x)$ occur at $- \frac{π}{2}$ , $\frac{3 π}{2}$ , and every $π n$ , where $n$ is an integer. This is half of the period.

$π n$

Step 6

There are only vertical asymptotes for secant and cosecant functions.

Vertical Asymptotes: $x = \frac{3 π}{2} + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

Step 7