Trigonometry Examples

$y = 3 \sec (\frac{1}{2} x)$

Step 1

Find the asymptotes.

Tap for more steps...

Step 1.1

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 $y = 3 \sec (\frac{x}{2})$ . 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 $y = 3 \sec (\frac{x}{2})$ .

$\frac{x}{2} = - \frac{π}{2}$

Step 1.2

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

$x = - π$

Step 1.3

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

$\frac{x}{2} = \frac{3 π}{2}$

Step 1.4

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

$x = 3 π$

Step 1.5

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

$(- π, 3 π)$

Step 1.6

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

Tap for more steps...

Step 1.6.1

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

$\frac{2 π}{\frac{1}{2}}$

Step 1.6.2

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 2$

Step 1.6.3

Multiply $2$ by $2$ .

$4 π$

Step 1.7

The vertical asymptotes for $y = 3 \sec (\frac{x}{2})$ 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 1.8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

No Horizontal Asymptotes

No Oblique Asymptotes

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

Step 2

Use the form $a \sec (b x - c) + d$ to find the variables used to find the amplitude, period, phase shift, and vertical shift.

$a = 3$

$b = \frac{1}{2}$

$c = 0$

$d = 0$

Step 3

Since the graph of the function $s e c$ does not have a maximum or minimum value, there can be no value for the amplitude.

Amplitude: None

Step 4

Find the period of $3 \sec (\frac{x}{2})$ .

Tap for more steps...

Step 4.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2

Replace $b$ with $\frac{1}{2}$ in the formula for period.

$\frac{2 π}{| \frac{1}{2} |}$

Step 4.3

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

$\frac{2 π}{\frac{1}{2}}$

Step 4.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 2$

Step 4.5

Multiply $2$ by $2$ .

$4 π$

Step 5

Find the phase shift using the formula $\frac{c}{b}$ .

Tap for more steps...

Step 5.1

The phase shift of the function can be calculated from $\frac{c}{b}$ .

Phase Shift: $\frac{c}{b}$

Step 5.2

Replace the values of $c$ and $b$ in the equation for phase shift.

Phase Shift: $\frac{0}{\frac{1}{2}}$

Step 5.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $0 \cdot 2$

Step 5.4

Multiply $0$ by $2$ .

Phase Shift: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $4 π$

Phase Shift: None

Vertical Shift: None

Step 7

The trig function can be graphed using the amplitude, period, phase shift, vertical shift, and the points.

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

Amplitude: None

Period: $4 π$

Phase Shift: None

Vertical Shift: None

Step 8