Trigonometry Examples

$y = \sec (2 x - \frac{π}{2})$

Step 1

Find the asymptotes.

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

$2 x - \frac{π}{2} = - \frac{π}{2}$

Step 1.2

Solve for $x$ .

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Step 1.2.1

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

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Step 1.2.1.1

Add $\frac{π}{2}$ to both sides of the equation.

$2 x = - \frac{π}{2} + \frac{π}{2}$

Step 1.2.1.2

Combine the numerators over the common denominator.

$2 x = \frac{- π + π}{2}$

Step 1.2.1.3

Add $- π$ and $π$ .

$2 x = \frac{0}{2}$

Step 1.2.1.4

Divide $0$ by $2$ .

$2 x = 0$

Step 1.2.2

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 1.2.2.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Divide $0$ by $2$ .

$x = 0$

Step 1.3

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

$2 x - \frac{π}{2} = \frac{3 π}{2}$

Step 1.4

Solve for $x$ .

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Step 1.4.1

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

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Step 1.4.1.1

Add $\frac{π}{2}$ to both sides of the equation.

$2 x = \frac{3 π}{2} + \frac{π}{2}$

Step 1.4.1.2

Combine the numerators over the common denominator.

$2 x = \frac{3 π + π}{2}$

Step 1.4.1.3

Add $3 π$ and $π$ .

$2 x = \frac{4 π}{2}$

Step 1.4.1.4

Cancel the common factor of $4$ and $2$ .

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Step 1.4.1.4.1

Factor $2$ out of $4 π$ .

$2 x = \frac{2 (2 π)}{2}$

Step 1.4.1.4.2

Cancel the common factors.

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Step 1.4.1.4.2.1

Factor $2$ out of $2$ .

$2 x = \frac{2 (2 π)}{2 (1)}$

Step 1.4.1.4.2.2

Cancel the common factor.

$2 x = \frac{2 (2 π)}{2 \cdot 1}$

Step 1.4.1.4.2.3

Rewrite the expression.

$2 x = \frac{2 π}{1}$

Step 1.4.1.4.2.4

Divide $2 π$ by $1$ .

$2 x = 2 π$

Step 1.4.2

Divide each term in $2 x = 2 π$ by $2$ and simplify.

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Step 1.4.2.1

Divide each term in $2 x = 2 π$ by $2$ .

$\frac{2 x}{2} = \frac{2 π}{2}$

Step 1.4.2.2

Simplify the left side.

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Step 1.4.2.2.1

Cancel the common factor of $2$ .

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Step 1.4.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{2 π}{2}$

Step 1.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2 π}{2}$

Step 1.4.2.3

Simplify the right side.

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Step 1.4.2.3.1

Cancel the common factor of $2$ .

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Step 1.4.2.3.1.1

Cancel the common factor.

$x = \frac{2 π}{2}$

Step 1.4.2.3.1.2

Divide $π$ by $1$ .

$x = π$

Step 1.5

The basic period for $y = \sec (2 x - \frac{π}{2})$ will occur at $(0, π)$ , where $0$ and $π$ are vertical asymptotes.

$(0, π)$

Step 1.6

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 1.6.1

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

$\frac{2 π}{2}$

Step 1.6.2

Cancel the common factor of $2$ .

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Step 1.6.2.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 1.6.2.2

Divide $π$ by $1$ .

$π$

Step 1.7

The vertical asymptotes for $y = \sec (2 x - \frac{π}{2})$ occur at $0$ , $π$ , and every $x = 0 + \frac{π n}{2}$ , where $n$ is an integer. This is half of the period.

$x = 0 + \frac{π n}{2}$

Step 1.8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 0 + \frac{π n}{2}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 0 + \frac{π n}{2}$ 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.

$a = 1$

$b = 2$

$c = \frac{π}{2}$

$d = 0$

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 $\sec (2 x - \frac{π}{2})$ .

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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 $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 4.3

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

$\frac{2 π}{2}$

Step 4.4

Cancel the common factor of $2$ .

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Step 4.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 4.4.2

Divide $π$ by $1$ .

$π$

Step 5

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

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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: $\frac{\frac{π}{2}}{2}$

Step 5.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $\frac{π}{2} \cdot \frac{1}{2}$

Step 5.4

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

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Step 5.4.1

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

Phase Shift: $\frac{π}{2 \cdot 2}$

Step 5.4.2

Multiply $2$ by $2$ .

Phase Shift: $\frac{π}{4}$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $π$

Phase Shift: $\frac{π}{4}$ ( $\frac{π}{4}$ to the right)

Vertical Shift: None

Step 7

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

Vertical Asymptotes: $x = 0 + \frac{π n}{2}$ where $n$ is an integer

Amplitude: None

Period: $π$

Phase Shift: $\frac{π}{4}$ ( $\frac{π}{4}$ to the right)

Vertical Shift: None

Step 8