Graph y=sec(3x) - MathMaster

$y = \sec (3 x)$

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 $y = \sec (3 x)$ . 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 $y = \sec (3 x)$ .

$3 x = - \frac{π}{2}$

Divide each term by $3$ and simplify.

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

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

Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

$x = - \frac{π}{2} \cdot \frac{1}{3}$

Multiply $- \frac{π}{2} \cdot \frac{1}{3}$ .

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

$x = - \frac{π}{3 \cdot 2}$

Multiply $3$ by $2$ .

$x = - \frac{π}{6}$

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

$3 x = \frac{3 π}{2}$

Divide each term by $3$ and simplify.

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

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

Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

$x = \frac{3 π}{2} \cdot \frac{1}{3}$

Cancel the common factor of $3$ .

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Factor $3$ out of $3 π$ .

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

Cancel the common factor.

$x = \frac{3 π}{2} \cdot \frac{1}{3}$

Rewrite the expression.

$x = \frac{π}{2}$

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

$(- \frac{π}{6}, \frac{π}{2})$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

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

$x = \frac{π n}{3}$

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π n}{3}$ 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.

$a = 1$

$b = 3$

$c = 0$

$d = 0$

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 $3$ in the formula for period.

Period: $\frac{2 π}{| 3 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

Period: $\frac{2 π}{3}$

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: $\frac{0}{3}$

Divide $0$ by $3$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $0$

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{2 π}{3}$

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.

$f (0) = \sec (3 (0))$

Simplify the result.

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Multiply $3$ by $0$ .

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

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

$f (0) = 1$

The final answer is $1$ .

$1$

Find the point at $x = \frac{π}{3}$ .

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

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

Simplify the result.

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Cancel the common factor of $3$ .

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Cancel the common factor.

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

Rewrite the expression.

$f (\frac{π}{3}) = \sec (π)$

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.

$f (\frac{π}{3}) = - \sec (0)$

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

$f (\frac{π}{3}) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f (\frac{π}{3}) = - 1$

The final answer is $- 1$ .

$- 1$

Find the point at $x = \frac{2 π}{3}$ .

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Replace the variable $x$ with $\frac{2 π}{3}$ in the expression.

$f (\frac{2 π}{3}) = \sec (3 (\frac{2 π}{3}))$

Simplify the result.

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Cancel the common factor of $3$ .

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Cancel the common factor.

$f (\frac{2 π}{3}) = \sec (3 (\frac{2 π}{3}))$

Rewrite the expression.

$f (\frac{2 π}{3}) = \sec (2 π)$

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

$f (\frac{2 π}{3}) = \sec (0)$

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

$f (\frac{2 π}{3}) = 1$

The final answer is $1$ .

$1$

Find the point at $x = π$ .

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

$f (π) = \sec (3 (π))$

Simplify the result.

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

$f (π) = \sec (π)$

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.

$f (π) = - \sec (0)$

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

$f (π) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f (π) = - 1$

The final answer is $- 1$ .

$- 1$

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

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

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

Simplify the result.

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Cancel the common factor of $3$ .

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Cancel the common factor.

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

Rewrite the expression.

$f (\frac{4 π}{3}) = \sec (4 π)$

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

$f (\frac{4 π}{3}) = \sec (0)$

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

$f (\frac{4 π}{3}) = 1$

The final answer is $1$ .

$1$

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 1 \\ \frac{π}{3} & - 1 \\ \frac{2 π}{3} & 1 \\ π & - 1 \\ \frac{4 π}{3} & 1 \end{array}$

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

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

Amplitude: None

Period: $\frac{2 π}{3}$

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

Vertical Shift: $0$

$\begin{array}{c} x & f (x) \\ 0 & 1 \\ \frac{π}{3} & - 1 \\ \frac{2 π}{3} & 1 \\ π & - 1 \\ \frac{4 π}{3} & 1 \end{array}$

Graph y=sec(3x)

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