Trigonometry Examples

$f (x) = 4 \sec (4 x)$

Find the asymptotes.

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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 $y = 4 \sec (4 x)$ . Set the inside of the secant function, bx+c, for $y = a \sec (b x + c) + d$ equal to $- \frac{π}{2}$ to find where the vertical asymptote occurs for $y = 4 \sec (4 x)$ .

$4 x = - \frac{π}{2}$

Divide each term by $4$ and simplify.

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Divide each term in $4 x = - \frac{π}{2}$ by $4$ .

$\frac{4 x}{4} = - \frac{π}{2} \cdot \frac{1}{4}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{4 x}{4} = - \frac{π}{2} \cdot \frac{1}{4}$

Divide $x$ by $1$ .

$x = - \frac{π}{2} \cdot \frac{1}{4}$

Simplify the right side of the equation.

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Multiply $- \frac{π}{2}$ by $\frac{1}{4}$ .

$x = - \frac{π}{2} \cdot \frac{1}{4}$

Simplify $- \frac{π}{2} \frac{1}{4}$ .

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Multiply $\frac{1}{4}$ and $\frac{π}{2}$ .

$x = - \frac{π}{4 \cdot 2}$

Multiply $4$ by $2$ .

$x = - \frac{π}{8}$

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

$4 x = \frac{3 π}{2}$

Divide each term by $4$ and simplify.

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Divide each term in $4 x = \frac{3 π}{2}$ by $4$ .

$\frac{4 x}{4} = \frac{3 π}{2} \cdot \frac{1}{4}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{4 x}{4} = \frac{3 π}{2} \cdot \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{3 π}{2} \cdot \frac{1}{4}$

Simplify the right side of the equation.

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Multiply $\frac{3 π}{2}$ by $\frac{1}{4}$ .

$x = \frac{3 π}{2} \cdot \frac{1}{4}$

Simplify $\frac{3 π}{2} \frac{1}{4}$ .

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Multiply $\frac{3 π}{2}$ and $\frac{1}{4}$ .

$x = \frac{3 π}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{3 π}{8}$

The basic period for $y = 4 \sec (4 x)$ will occur at $(- \frac{π}{8}, \frac{3 π}{8})$ , where $- \frac{π}{8}$ and $\frac{3 π}{8}$ are vertical asymptotes.

$(- \frac{π}{8}, \frac{3 π}{8})$

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

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The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$\frac{2 π}{4}$

Reduce the expression by cancelling the common factors.

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Factor $2$ out of $2 π$ .

$\frac{2 (π)}{4}$

Cancel the common factors.

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Factor $2$ out of $4$ .

$\frac{2 π}{2 \cdot 2}$

Cancel the common factor.

$\frac{2 π}{2 \cdot 2}$

Rewrite the expression.

$\frac{π}{2}$

The vertical asymptotes for $y = 4 \sec (4 x)$ occur at $- \frac{π}{8}$ , $\frac{3 π}{8}$ , and every $x = \frac{π n}{4}$ , where $n$ is an integer. This is half of the period.

$x = \frac{π n}{4}$

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

No Horizontal Asymptotes

No Oblique Asymptotes

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

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

$a = 4$

$b = 4$

$c = 0$

$d = 0$

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

Find the period using the formula $\frac{2 π}{| b |}$ .

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

Period: $\frac{2 π}{| b |}$

Replace $b$ with $4$ in the formula for period.

Period: $\frac{2 π}{| 4 |}$

Solve the equation.

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The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

Period: $\frac{2 π}{4}$

Reduce the expression by cancelling the common factors.

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Factor $2$ out of $2 π$ .

Period: $\frac{2 (π)}{4}$

Cancel the common factors.

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Factor $2$ out of $4$ .

Period: $\frac{2 π}{2 \cdot 2}$

Cancel the common factor.

Period: $\frac{2 π}{2 \cdot 2}$

Rewrite the expression.

Period: $\frac{π}{2}$

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

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

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

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

Phase Shift: $\frac{0}{4}$

Divide $0$ by $4$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $0$

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{π}{2}$

Phase Shift: $0$ ( $0$ to the right)

Vertical Shift: $0$

Select a few points to graph.

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Find the point at $x = 0$ .

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Replace the variable $x$ with $0$ in the expression.

$f (0) = 4 \sec (4 (0))$

Simplify the result.

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Multiply $4$ by $0$ .

$f (0) = 4 \sec (0)$

The exact value of $\sec (0)$ is $1$ .

$f (0) = 4 \cdot 1$

Multiply $4$ by $1$ .

$f (0) = 4$

The final answer is $4$ .

$4$

Find the point at $x = \frac{π}{4}$ .

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Replace the variable $x$ with $\frac{π}{4}$ in the expression.

$f (\frac{π}{4}) = 4 \sec (4 (\frac{π}{4}))$

Simplify the result.

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Cancel the common factor of $4$ .

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Write $4$ as a fraction with denominator $1$ .

$f (\frac{π}{4}) = 4 \sec (\frac{4}{1} \cdot \frac{π}{4})$

Factor out the greatest common factor $4$ .

$f (\frac{π}{4}) = 4 \sec (\frac{4 \cdot 1}{1} \cdot \frac{π}{4 \cdot 1})$

Cancel the common factor.

$f (\frac{π}{4}) = 4 \sec (\frac{4 \cdot 1}{1} \cdot \frac{π}{4 \cdot 1})$

Rewrite the expression.

$f (\frac{π}{4}) = 4 \sec (\frac{1}{1} \cdot \frac{π}{1})$

Simplify.

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Multiply $\frac{1}{1}$ and $\frac{π}{1}$ .

$f (\frac{π}{4}) = 4 \sec (\frac{π}{1})$

Divide $π$ by $1$ .

$f (\frac{π}{4}) = 4 \sec (π)$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$f (\frac{π}{4}) = 4 (- \sec (0))$

The exact value of $\sec (0)$ is $1$ .

$f (\frac{π}{4}) = 4 (- 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$f (\frac{π}{4}) = 4 \cdot - 1$

Multiply $4$ by $- 1$ .

$f (\frac{π}{4}) = - 4$

The final answer is $- 4$ .

$- 4$

Find the point at $x = \frac{π}{2}$ .

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Replace the variable $x$ with $\frac{π}{2}$ in the expression.

$f (\frac{π}{2}) = 4 \sec (4 (\frac{π}{2}))$

Simplify the result.

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Cancel the common factor of $2$ .

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Write $4$ as a fraction with denominator $1$ .

$f (\frac{π}{2}) = 4 \sec (\frac{4}{1} \cdot \frac{π}{2})$

Factor out the greatest common factor $2$ .

$f (\frac{π}{2}) = 4 \sec (\frac{2 \cdot 2}{1} \cdot \frac{π}{2 \cdot 1})$

Cancel the common factor.

$f (\frac{π}{2}) = 4 \sec (\frac{2 \cdot 2}{1} \cdot \frac{π}{2 \cdot 1})$

Rewrite the expression.

$f (\frac{π}{2}) = 4 \sec (\frac{2}{1} \cdot \frac{π}{1})$

Simplify.

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

$f (\frac{π}{2}) = 4 \sec (\frac{2 π}{1})$

Divide $2 π$ by $1$ .

$f (\frac{π}{2}) = 4 \sec (2 π)$

$2 π$ is a full rotation so replace with $0$ .

$f (\frac{π}{2}) = 4 \sec (0)$

The exact value of $\sec (0)$ is $1$ .

$f (\frac{π}{2}) = 4 \cdot 1$

Multiply $4$ by $1$ .

$f (\frac{π}{2}) = 4$

The final answer is $4$ .

$4$

Find the point at $x = \frac{3 π}{4}$ .

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Replace the variable $x$ with $\frac{3 π}{4}$ in the expression.

$f (\frac{3 π}{4}) = 4 \sec (4 (\frac{3 π}{4}))$

Simplify the result.

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Cancel the common factor of $4$ .

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Write $4$ as a fraction with denominator $1$ .

$f (\frac{3 π}{4}) = 4 \sec (\frac{4}{1} \cdot \frac{3 π}{4})$

Factor out the greatest common factor $4$ .

$f (\frac{3 π}{4}) = 4 \sec (\frac{4 \cdot 1}{1} \cdot \frac{3 π}{4 \cdot 1})$

Cancel the common factor.

$f (\frac{3 π}{4}) = 4 \sec (\frac{4 \cdot 1}{1} \cdot \frac{3 π}{4 \cdot 1})$

Rewrite the expression.

$f (\frac{3 π}{4}) = 4 \sec (\frac{1}{1} \cdot \frac{3 π}{1})$

Simplify.

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Multiply $\frac{1}{1}$ and $\frac{3 π}{1}$ .

$f (\frac{3 π}{4}) = 4 \sec (\frac{3 π}{1})$

Divide $3 π$ by $1$ .

$f (\frac{3 π}{4}) = 4 \sec (3 π)$

Remove full rotations of $2 π$ until the angle is between $0$ and $2 π$ .

$f (\frac{3 π}{4}) = 4 \sec (π)$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$f (\frac{3 π}{4}) = 4 (- \sec (0))$

The exact value of $\sec (0)$ is $1$ .

$f (\frac{3 π}{4}) = 4 (- 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{4}) = 4 \cdot - 1$

Multiply $4$ by $- 1$ .

$f (\frac{3 π}{4}) = - 4$

The final answer is $- 4$ .

$- 4$

Find the point at $x = π$ .

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Replace the variable $x$ with $π$ in the expression.

$f (π) = 4 \sec (4 (π))$

Simplify the result.

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Multiply $4$ by $π$ .

$f (π) = 4 \sec (4 π)$

$4 π$ is a full rotation so replace with $0$ .

$f (π) = 4 \sec (0)$

The exact value of $\sec (0)$ is $1$ .

$f (π) = 4 \cdot 1$

Multiply $4$ by $1$ .

$f (π) = 4$

The final answer is $4$ .

$4$

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 4 \\ \frac{π}{4} & - 4 \\ \frac{π}{2} & 4 \\ \frac{3 π}{4} & - 4 \\ π & 4 \end{array}$

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

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

Amplitude: None

Period: $\frac{π}{2}$

Phase Shift: $0$ ( $0$ to the right)

Vertical Shift: $0$

$\begin{array}{c} x & f (x) \\ 0 & 4 \\ \frac{π}{4} & - 4 \\ \frac{π}{2} & 4 \\ \frac{3 π}{4} & - 4 \\ π & 4 \end{array}$

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