Precalculus Examples

$f (x) = 6 \sec (\frac{1}{2} x) - 4$

Step 1

Combine $\frac{1}{2}$ and $x$ .

$y = 6 \sec (\frac{x}{2}) - 4$

Step 2

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 $y = 6 \sec (\frac{x}{2}) - 4$ . 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 $y = 6 \sec (\frac{x}{2}) - 4$ .

$\frac{x}{2} = - \frac{π}{2}$

Step 3

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = - π$

Step 4

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

$\frac{x}{2} = \frac{3 π}{2}$

Step 5

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = 3 π$

Step 6

The basic period for $y = 6 \sec (\frac{x}{2}) - 4$ will occur at $(- π, 3 π)$ , where $- π$ and $3 π$ are vertical asymptotes.

$(- π, 3 π)$

Step 7

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 7.1

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{2}}$

Step 7.2

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 2$

Step 7.3

Multiply $2$ by $2$ .

$4 π$

Step 8

The vertical asymptotes for $y = 6 \sec (\frac{x}{2}) - 4$ occur at $- π$ , $3 π$ , and every $2 π n$ , where $n$ is an integer. This is half of the period.

$x = 3 π + 2 π n$

Step 9

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 3 π + 2 π n$ where $n$ is an integer

Step 10