Precalculus Examples

$y = 3 \sec (2 x)$

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 $y = 3 \sec (2 x)$ . 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 $y = 3 \sec (2 x)$ .

$2 x = - \frac{π}{2}$

Step 2

Divide each term in $2 x = - \frac{π}{2}$ by $2$ and simplify.

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Step 2.1

Divide each term in $2 x = - \frac{π}{2}$ by $2$ .

$\frac{2 x}{2} = \frac{- \frac{π}{2}}{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $2$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- \frac{π}{2}}{2}$

Step 2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{2}$

Step 2.3.2

Multiply $- \frac{π}{2} \cdot \frac{1}{2}$ .

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Step 2.3.2.1

Multiply $\frac{1}{2}$ by $\frac{π}{2}$ .

$x = - \frac{π}{2 \cdot 2}$

Step 2.3.2.2

Multiply $2$ by $2$ .

$x = - \frac{π}{4}$

Step 3

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

$2 x = \frac{3 π}{2}$

Step 4

Divide each term in $2 x = \frac{3 π}{2}$ by $2$ and simplify.

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Step 4.1

Divide each term in $2 x = \frac{3 π}{2}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{3 π}{2}}{2}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 π}{2} \cdot \frac{1}{2}$

Step 4.3.2

Multiply $\frac{3 π}{2} \cdot \frac{1}{2}$ .

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Step 4.3.2.1

Multiply $\frac{3 π}{2}$ by $\frac{1}{2}$ .

$x = \frac{3 π}{2 \cdot 2}$

Step 4.3.2.2

Multiply $2$ by $2$ .

$x = \frac{3 π}{4}$

Step 5

The basic period for $y = 3 \sec (2 x)$ will occur at $(- \frac{π}{4}, \frac{3 π}{4})$ , where $- \frac{π}{4}$ and $\frac{3 π}{4}$ are vertical asymptotes.

$(- \frac{π}{4}, \frac{3 π}{4})$

Step 6

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 6.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Step 6.2

Cancel the common factor of $2$ .

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Step 6.2.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 6.2.2

Divide $π$ by $1$ .

$π$

Step 7

The vertical asymptotes for $y = 3 \sec (2 x)$ occur at $- \frac{π}{4}$ , $\frac{3 π}{4}$ , and every $x = \frac{π n}{2}$ , where $n$ is an integer. This is half of the period.

$x = \frac{π n}{2}$

Step 8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π n}{2}$ where $n$ is an integer

Step 9