Trigonometry Examples

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

Step 1

Find the asymptotes.

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For any $y = \tan (x)$ , vertical asymptotes occur at $x = \frac{π}{2} + n π$ , where $n$ is an integer. Use the basic period for $y = \tan (x)$ , $(- \frac{π}{2}, \frac{π}{2})$ , to find the vertical asymptotes for $y = \frac{\tan (3 x)}{2}$ . Set the inside of the tangent function, $b x + c$ , for $y = a \tan (b x + c) + d$ equal to $- \frac{π}{2}$ to find where the vertical asymptote occurs for $y = \frac{\tan (3 x)}{2}$ .

$3 x = - \frac{π}{2}$

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

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

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

Simplify the left side.

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Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{3}$

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{3}$

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

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

$x = - \frac{π}{3 \cdot 2}$

Multiply $3$ by $2$ .

$x = - \frac{π}{6}$

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

$3 x = \frac{π}{2}$

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

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

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

Simplify the left side.

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Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{3}$

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{3}$

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

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

$x = \frac{π}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{π}{6}$

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

$(- \frac{π}{6}, \frac{π}{6})$

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

$\frac{π}{3}$

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

$x = \frac{π}{6} + \frac{π n}{3}$

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{6} + \frac{π n}{3}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{6} + \frac{π n}{3}$ where $n$ is an integer

Step 2

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

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

$b = 3$

$c = 0$

$d = 0$

Step 3

Since the graph of the function $t a n$ does not have a maximum or minimum value, there can be no value for the amplitude.

Amplitude: None

Step 4

Find the period of $\frac{\tan (3 x)}{2}$ .

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

$\frac{π}{| b |}$

Replace $b$ with $3$ in the formula for period.

$\frac{π}{| 3 |}$

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

$\frac{π}{3}$

Step 5

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: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{π}{3}$

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: $x = \frac{π}{6} + \frac{π n}{3}$ where $n$ is an integer

Amplitude: None

Period: $\frac{π}{3}$

Phase Shift: None

Vertical Shift: None

Step 8