Trigonometry Examples

$y = \frac{1}{5} \cdot \sec (3 π x)$

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 $y = \frac{\sec (3 π x)}{5}$ . 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 $y = \frac{\sec (3 π x)}{5}$ .

$3 π x = - \frac{π}{2}$

Step 1.2

Divide each term in $3 π x = - \frac{π}{2}$ by $3 π$ and simplify.

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Step 1.2.1

Divide each term in $3 π x = - \frac{π}{2}$ by $3 π$ .

$\frac{3 π x}{3 π} = \frac{- \frac{π}{2}}{3 π}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $3$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{3 π x}{3 π} = \frac{- \frac{π}{2}}{3 π}$

Step 1.2.2.1.2

Rewrite the expression.

$\frac{π x}{π} = \frac{- \frac{π}{2}}{3 π}$

Step 1.2.2.2

Cancel the common factor of $π$ .

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Step 1.2.2.2.1

Cancel the common factor.

$\frac{π x}{π} = \frac{- \frac{π}{2}}{3 π}$

Step 1.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{3 π}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{3 π}$

Step 1.2.3.2

Cancel the common factor of $π$ .

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Step 1.2.3.2.1

Move the leading negative in $- \frac{π}{2}$ into the numerator.

$x = \frac{- π}{2} \cdot \frac{1}{3 π}$

Step 1.2.3.2.2

Factor $π$ out of $- π$ .

$x = \frac{π \cdot - 1}{2} \cdot \frac{1}{3 π}$

Step 1.2.3.2.3

Factor $π$ out of $3 π$ .

$x = \frac{π \cdot - 1}{2} \cdot \frac{1}{π \cdot 3}$

Step 1.2.3.2.4

Cancel the common factor.

$x = \frac{π \cdot - 1}{2} \cdot \frac{1}{π \cdot 3}$

Step 1.2.3.2.5

Rewrite the expression.

$x = \frac{- 1}{2} \cdot \frac{1}{3}$

Step 1.2.3.3

Multiply $\frac{- 1}{2}$ by $\frac{1}{3}$ .

$x = \frac{- 1}{2 \cdot 3}$

Step 1.2.3.4

Simplify the expression.

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Step 1.2.3.4.1

Multiply $2$ by $3$ .

$x = \frac{- 1}{6}$

Step 1.2.3.4.2

Move the negative in front of the fraction.

$x = - \frac{1}{6}$

Step 1.3

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

$3 π x = \frac{3 π}{2}$

Step 1.4

Divide each term in $3 π x = \frac{3 π}{2}$ by $3 π$ and simplify.

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Step 1.4.1

Divide each term in $3 π x = \frac{3 π}{2}$ by $3 π$ .

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

Step 1.4.2

Simplify the left side.

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Step 1.4.2.1

Cancel the common factor of $3$ .

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Step 1.4.2.1.1

Cancel the common factor.

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

Step 1.4.2.1.2

Rewrite the expression.

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

Step 1.4.2.2

Cancel the common factor of $π$ .

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Step 1.4.2.2.1

Cancel the common factor.

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

Step 1.4.2.2.2

Divide $x$ by $1$ .

$x = \frac{\frac{3 π}{2}}{3 π}$

Step 1.4.3

Simplify the right side.

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Step 1.4.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 π}{2} \cdot \frac{1}{3 π}$

Step 1.4.3.2

Cancel the common factor of $3 π$ .

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Step 1.4.3.2.1

Cancel the common factor.

$x = \frac{3 π}{2} \cdot \frac{1}{3 π}$

Step 1.4.3.2.2

Rewrite the expression.

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

Step 1.5

The basic period for $y = \frac{\sec (3 π x)}{5}$ will occur at $(- \frac{1}{6}, \frac{1}{2})$ , where $- \frac{1}{6}$ and $\frac{1}{2}$ are vertical asymptotes.

$(- \frac{1}{6}, \frac{1}{2})$

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

$3 π$ is approximately $9.42477796$ which is positive so remove the absolute value

$\frac{2 π}{3 π}$

Step 1.6.2

Cancel the common factor of $π$ .

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Step 1.6.2.1

Cancel the common factor.

$\frac{2 π}{3 π}$

Step 1.6.2.2

Rewrite the expression.

$\frac{2}{3}$

Step 1.7

The vertical asymptotes for $y = \frac{\sec (3 π x)}{5}$ occur at $- \frac{1}{6}$ , $\frac{1}{2}$ , and every $\frac{n}{3}$ , where $n$ is an integer. This is half of the period.

$x = \frac{1}{2} + \frac{n}{3}$

Step 1.8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{1}{2} + \frac{n}{3}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{1}{2} + \frac{n}{3}$ 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.

$a = \frac{1}{5}$

$b = 3 π$

$c = 0$

$d = 0$

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 $\frac{\sec (3 π x)}{5}$ .

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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 $3 π$ in the formula for period.

$\frac{2 π}{| 3 π |}$

Step 4.3

$3 π$ is approximately $9.42477796$ which is positive so remove the absolute value

$\frac{2 π}{3 π}$

Step 4.4

Cancel the common factor of $π$ .

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Step 4.4.1

Cancel the common factor.

$\frac{2 π}{3 π}$

Step 4.4.2

Rewrite the expression.

$\frac{2}{3}$

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: $\frac{0}{3 π}$

Step 5.3

Cancel the common factor of $0$ and $3$ .

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Step 5.3.1

Factor $3$ out of $0$ .

Phase Shift: $\frac{3 (0)}{3 π}$

Step 5.3.2

Cancel the common factors.

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Step 5.3.2.1

Factor $3$ out of $3 π$ .

Phase Shift: $\frac{3 (0)}{3 (π)}$

Step 5.3.2.2

Cancel the common factor.

Phase Shift: $\frac{3 \cdot 0}{3 π}$

Step 5.3.2.3

Rewrite the expression.

Phase Shift: $\frac{0}{π}$

Step 5.4

Divide $0$ by $π$ .

Phase Shift: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{2}{3}$

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: $x = \frac{1}{2} + \frac{n}{3}$ where $n$ is an integer

Amplitude: None

Period: $\frac{2}{3}$

Phase Shift: None

Vertical Shift: None

Step 8