Trigonometry Examples

$f (x) = 3 \tan (4 x)$

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 $y = 3 \tan (4 x)$ . Set the inside of the tangent function, bx+c, for $y = a \tan (b x + c) + d$ equal to $- \frac{π}{2}$ to find where the vertical asymptote occurs for $y = 3 \tan (4 x)$ .

$4 x = - \frac{π}{2}$

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 $1$ to get $x$ .

$x = - \frac{π}{2} \cdot \frac{1}{4}$

Simplify the right side of the equation.

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

$x = - \frac{π}{2} \cdot \frac{1}{4}$

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

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

$x = - \frac{π}{4 \cdot 2}$

Multiply $4$ by $2$ to get $8$ .

$x = - \frac{π}{8}$

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

$4 x = \frac{π}{2}$

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 $1$ to get $x$ .

$x = \frac{π}{2} \cdot \frac{1}{4}$

Simplify the right side of the equation.

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

$x = \frac{π}{2} \cdot \frac{1}{4}$

Simplify $\frac{π}{2} \frac{1}{4}$ .

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

$x = \frac{π}{2 \cdot 4}$

Multiply $2$ by $4$ to get $8$ .

$x = \frac{π}{8}$

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

$(- \frac{π}{8}, \frac{π}{8})$

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

$\frac{π}{4}$

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

$x = \frac{π n}{4}$

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

$a = 3$

$b = 4$

$c = 0$

$d = 0$

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

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

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

Period: $\frac{π}{| b |}$

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

Period: $\frac{π}{| 4 |}$

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

Period: $\frac{π}{4}$

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 $4$ to get $0$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $0$

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{π}{4}$

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.

$f (0) = 3 \tan (4 (0))$

Simplify the result.

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Multiply $4$ by $0$ to get $0$ .

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

The exact value of $\tan (0)$ is $0$ .

$f (0) = 3 \cdot 0$

Multiply $3$ by $0$ to get $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

Find the point at $x = \frac{π}{16}$ .

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Replace the variable $x$ with $\frac{π}{16}$ in the expression.

$f (\frac{π}{16}) = 3 \tan (4 (\frac{π}{16}))$

Simplify the result.

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Cancel the common factor of $4$ .

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

$f (\frac{π}{16}) = 3 \tan (\frac{4}{1} \cdot \frac{π}{16})$

Factor out the greatest common factor $4$ .

$f (\frac{π}{16}) = 3 \tan (\frac{4 \cdot 1}{1} \cdot \frac{π}{4 \cdot 4})$

Cancel the common factor.

$f (\frac{π}{16}) = 3 \tan (\frac{4 \cdot 1}{1} \cdot \frac{π}{4 \cdot 4})$

Rewrite the expression.

$f (\frac{π}{16}) = 3 \tan (\frac{1}{1} \cdot \frac{π}{4})$

Multiply $\frac{1}{1}$ and $\frac{π}{4}$ to get $\frac{π}{4}$ .

$f (\frac{π}{16}) = 3 \tan (\frac{π}{4})$

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$f (\frac{π}{16}) = 3 \cdot 1$

Multiply $3$ by $1$ to get $3$ .

$f (\frac{π}{16}) = 3$

The final answer is $3$ .

$3$

Find the point at $x = \frac{3 π}{16}$ .

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Replace the variable $x$ with $\frac{3 π}{16}$ in the expression.

$f (\frac{3 π}{16}) = 3 \tan (4 (\frac{3 π}{16}))$

Simplify the result.

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Cancel the common factor of $4$ .

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

$f (\frac{3 π}{16}) = 3 \tan (\frac{4}{1} \cdot \frac{3 π}{16})$

Factor out the greatest common factor $4$ .

$f (\frac{3 π}{16}) = 3 \tan (\frac{4 \cdot 1}{1} \cdot \frac{3 π}{4 \cdot 4})$

Cancel the common factor.

$f (\frac{3 π}{16}) = 3 \tan (\frac{4 \cdot 1}{1} \cdot \frac{3 π}{4 \cdot 4})$

Rewrite the expression.

$f (\frac{3 π}{16}) = 3 \tan (\frac{1}{1} \cdot \frac{3 π}{4})$

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

$f (\frac{3 π}{16}) = 3 \tan (\frac{3 π}{4})$

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

$f (\frac{3 π}{16}) = 3 (- \tan (\frac{π}{4}))$

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$f (\frac{3 π}{16}) = 3 (- 1 \cdot 1)$

Multiply $- 1$ by $1$ to get $- 1$ .

$f (\frac{3 π}{16}) = 3 \cdot - 1$

Multiply $3$ by $- 1$ to get $- 3$ .

$f (\frac{3 π}{16}) = - 3$

The final answer is $- 3$ .

$- 3$

Find the point at $x = \frac{π}{4}$ .

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

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

Simplify the result.

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Cancel the common factor of $4$ .

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

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

Factor out the greatest common factor $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

Divide $π$ by $1$ to get $π$ .

$f (\frac{π}{4}) = 3 \tan (π)$

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

$f (\frac{π}{4}) = 3 (- \tan (0))$

The exact value of $\tan (0)$ is $0$ .

$f (\frac{π}{4}) = 3 (- 0)$

Multiply $- 1$ by $0$ to get $0$ .

$f (\frac{π}{4}) = 3 \cdot 0$

Multiply $3$ by $0$ to get $0$ .

$f (\frac{π}{4}) = 0$

The final answer is $0$ .

$0$

Find the point at $x = \frac{5 π}{16}$ .

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Replace the variable $x$ with $\frac{5 π}{16}$ in the expression.

$f (\frac{5 π}{16}) = 3 \tan (4 (\frac{5 π}{16}))$

Simplify the result.

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Cancel the common factor of $4$ .

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

$f (\frac{5 π}{16}) = 3 \tan (\frac{4}{1} \cdot \frac{5 π}{16})$

Factor out the greatest common factor $4$ .

$f (\frac{5 π}{16}) = 3 \tan (\frac{4 \cdot 1}{1} \cdot \frac{5 π}{4 \cdot 4})$

Cancel the common factor.

$f (\frac{5 π}{16}) = 3 \tan (\frac{4 \cdot 1}{1} \cdot \frac{5 π}{4 \cdot 4})$

Rewrite the expression.

$f (\frac{5 π}{16}) = 3 \tan (\frac{1}{1} \cdot \frac{5 π}{4})$

Multiply $\frac{1}{1}$ and $\frac{5 π}{4}$ to get $\frac{5 π}{4}$ .

$f (\frac{5 π}{16}) = 3 \tan (\frac{5 π}{4})$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (\frac{5 π}{16}) = 3 \tan (\frac{π}{4})$

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$f (\frac{5 π}{16}) = 3 \cdot 1$

Multiply $3$ by $1$ to get $3$ .

$f (\frac{5 π}{16}) = 3$

The final answer is $3$ .

$3$

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 0 \\ \frac{π}{16} & 3 \\ \frac{3 π}{16} & - 3 \\ \frac{π}{4} & 0 \\ \frac{5 π}{16} & 3 \end{array}$

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: $\frac{π}{4}$

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

Vertical Shift: $0$

$\begin{array}{c} x & f (x) \\ 0 & 0 \\ \frac{π}{16} & 3 \\ \frac{3 π}{16} & - 3 \\ \frac{π}{4} & 0 \\ \frac{5 π}{16} & 3 \end{array}$

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Trigonometry Examples

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