Trigonometry Examples

$y = \sec (x - \frac{π}{4})$

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

$x - \frac{π}{4} = - \frac{π}{2}$

Step 1.2

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

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Step 1.2.1

Add $\frac{π}{4}$ to both sides of the equation.

$x = - \frac{π}{2} + \frac{π}{4}$

Step 1.2.2

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

$x = - \frac{π}{2} \cdot \frac{2}{2} + \frac{π}{4}$

Step 1.2.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.2.3.1

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

$x = - \frac{π \cdot 2}{2 \cdot 2} + \frac{π}{4}$

Step 1.2.3.2

Multiply $2$ by $2$ .

$x = - \frac{π \cdot 2}{4} + \frac{π}{4}$

Step 1.2.4

Combine the numerators over the common denominator.

$x = \frac{- π \cdot 2 + π}{4}$

Step 1.2.5

Simplify the numerator.

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Step 1.2.5.1

Multiply $2$ by $- 1$ .

$x = \frac{- 2 π + π}{4}$

Step 1.2.5.2

Add $- 2 π$ and $π$ .

$x = \frac{- π}{4}$

Step 1.2.6

Move the negative in front of the fraction.

$x = - \frac{π}{4}$

Step 1.3

Set the inside of the secant function $x - \frac{π}{4}$ equal to $\frac{3 π}{2}$ .

$x - \frac{π}{4} = \frac{3 π}{2}$

Step 1.4

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

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Step 1.4.1

Add $\frac{π}{4}$ to both sides of the equation.

$x = \frac{3 π}{2} + \frac{π}{4}$

Step 1.4.2

To write $\frac{3 π}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{3 π}{2} \cdot \frac{2}{2} + \frac{π}{4}$

Step 1.4.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.4.3.1

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

$x = \frac{3 π \cdot 2}{2 \cdot 2} + \frac{π}{4}$

Step 1.4.3.2

Multiply $2$ by $2$ .

$x = \frac{3 π \cdot 2}{4} + \frac{π}{4}$

Step 1.4.4

Combine the numerators over the common denominator.

$x = \frac{3 π \cdot 2 + π}{4}$

Step 1.4.5

Simplify the numerator.

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Step 1.4.5.1

Multiply $2$ by $3$ .

$x = \frac{6 π + π}{4}$

Step 1.4.5.2

Add $6 π$ and $π$ .

$x = \frac{7 π}{4}$

Step 1.5

The basic period for $y = \sec (x - \frac{π}{4})$ will occur at $(- \frac{π}{4}, \frac{7 π}{4})$ , where $- \frac{π}{4}$ and $\frac{7 π}{4}$ are vertical asymptotes.

$(- \frac{π}{4}, \frac{7 π}{4})$

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 $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.6.2

Divide $2 π$ by $1$ .

$2 π$

Step 1.7

The vertical asymptotes for $y = \sec (x - \frac{π}{4})$ occur at $- \frac{π}{4}$ , $\frac{7 π}{4}$ , and every $x = - \frac{π}{4} + π n$ , where $n$ is an integer. This is half of the period.

$x = - \frac{π}{4} + π n$

Step 1.8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - \frac{π}{4} + π n$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - \frac{π}{4} + π n$ 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.

$a = 1$

$b = 1$

$c = \frac{π}{4}$

$d = 0$

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 $\sec (x - \frac{π}{4})$ .

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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 $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.4

Divide $2 π$ by $1$ .

$2 π$

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: $\frac{\frac{π}{4}}{1}$

Step 5.3

Divide $\frac{π}{4}$ by $1$ .

Phase Shift: $\frac{π}{4}$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $2 π$

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: $x = - \frac{π}{4} + π n$ where $n$ is an integer

Amplitude: None

Period: $2 π$

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

Vertical Shift: None

Step 8