Precalculus Examples

$y = \sec (x - π)$

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 $y = \sec (x - π)$ . 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 $y = \sec (x - π)$ .

$x - π = - \frac{π}{2}$

Step 2

Move all terms not containing $x$ to the right side of the equation.

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Step 2.1

Add $π$ to both sides of the equation.

$x = - \frac{π}{2} + π$

Step 2.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.3

Combine $π$ and $\frac{2}{2}$ .

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

Step 2.4

Combine the numerators over the common denominator.

$x = \frac{- π + π \cdot 2}{2}$

Step 2.5

Simplify the numerator.

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Step 2.5.1

Move $2$ to the left of $π$ .

$x = \frac{- π + 2 \cdot π}{2}$

Step 2.5.2

Add $- π$ and $2 π$ .

$x = \frac{π}{2}$

Step 3

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

$x - π = \frac{3 π}{2}$

Step 4

Move all terms not containing $x$ to the right side of the equation.

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Step 4.1

Add $π$ to both sides of the equation.

$x = \frac{3 π}{2} + π$

Step 4.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 4.3

Combine $π$ and $\frac{2}{2}$ .

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

Step 4.4

Combine the numerators over the common denominator.

$x = \frac{3 π + π \cdot 2}{2}$

Step 4.5

Simplify the numerator.

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Step 4.5.1

Move $2$ to the left of $π$ .

$x = \frac{3 π + 2 \cdot π}{2}$

Step 4.5.2

Add $3 π$ and $2 π$ .

$x = \frac{5 π}{2}$

Step 5

The basic period for $y = \sec (x - π)$ will occur at $(\frac{π}{2}, \frac{5 π}{2})$ , where $\frac{π}{2}$ and $\frac{5 π}{2}$ are vertical asymptotes.

$(\frac{π}{2}, \frac{5 π}{2})$

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 $1$ is $1$ .

$\frac{2 π}{1}$

Step 6.2

Divide $2 π$ by $1$ .

$2 π$

Step 7

The vertical asymptotes for $y = \sec (x - π)$ occur at $\frac{π}{2}$ , $\frac{5 π}{2}$ , and every $x = \frac{π}{2} + π n$ , where $n$ is an integer. This is half of the period.

$x = \frac{π}{2} + π n$

Step 8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

Step 9