Trigonometry Examples

$y = \tan (x - π) - 2$

Step 1

Find the asymptotes.

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Step 1.1

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 $y = \tan (x - π) - 2$ . 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 $y = \tan (x - π) - 2$ .

$x - π = - \frac{π}{2}$

Step 1.2

Move all terms not containing $x$ to the right side of the equation.

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Step 1.2.1

Add $π$ to both sides of the equation.

$x = - \frac{π}{2} + π$

Step 1.2.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = - \frac{π}{2} + π \cdot \frac{2}{2}$

Step 1.2.3

Combine $π$ and $\frac{2}{2}$ .

$x = - \frac{π}{2} + \frac{π \cdot 2}{2}$

Step 1.2.4

Combine the numerators over the common denominator.

$x = \frac{- π + π \cdot 2}{2}$

Step 1.2.5

Simplify the numerator.

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Step 1.2.5.1

Move $2$ to the left of $π$ .

$x = \frac{- π + 2 \cdot π}{2}$

Step 1.2.5.2

Add $- π$ and $2 π$ .

$x = \frac{π}{2}$

Step 1.3

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

$x - π = \frac{π}{2}$

Step 1.4

Move all terms not containing $x$ to the right side of the equation.

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Step 1.4.1

Add $π$ to both sides of the equation.

$x = \frac{π}{2} + π$

Step 1.4.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{π}{2} + π \cdot \frac{2}{2}$

Step 1.4.3

Combine $π$ and $\frac{2}{2}$ .

$x = \frac{π}{2} + \frac{π \cdot 2}{2}$

Step 1.4.4

Combine the numerators over the common denominator.

$x = \frac{π + π \cdot 2}{2}$

Step 1.4.5

Simplify the numerator.

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Step 1.4.5.1

Move $2$ to the left of $π$ .

$x = \frac{π + 2 \cdot π}{2}$

Step 1.4.5.2

Add $π$ and $2 π$ .

$x = \frac{3 π}{2}$

Step 1.5

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

$(\frac{π}{2}, \frac{3 π}{2})$

Step 1.6

Find the period $\frac{π}{| b |}$ to find where the vertical asymptotes exist.

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Step 1.6.1

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 1.6.2

Divide $π$ by $1$ .

$π$

Step 1.7

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

$x = \frac{π}{2} + π n$

Step 1.8

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

No Horizontal Asymptotes

No Oblique Asymptotes

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

$a = 1$

$b = 1$

$c = π$

$d = - 2$

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

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Step 4.1

Find the period of $\tan (x - π)$ .

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Step 4.1.1

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

$\frac{π}{| b |}$

Step 4.1.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 4.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 4.1.4

Divide $π$ by $1$ .

$π$

Step 4.2

Find the period of $- 2$ .

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Step 4.2.1

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

$\frac{π}{| b |}$

Step 4.2.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 4.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 4.2.4

Divide $π$ by $1$ .

$π$

Step 4.3

The period of addition/subtraction of trig functions is the maximum of the individual periods.

$π$

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: $\frac{π}{1}$

Step 5.3

Divide $π$ by $1$ .

Phase Shift: $π$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: $- 2$

Step 7

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

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

Amplitude: None

Period: $π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: $- 2$

Step 8