Trigonometry Examples

$y = \tan (x + 2) - π$

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

$x + 2 = - \frac{π}{2}$

Step 1.2

Subtract $2$ from both sides of the equation.

$x = - \frac{π}{2} - 2$

Step 1.3

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

$x + 2 = \frac{π}{2}$

Step 1.4

Subtract $2$ from both sides of the equation.

$x = \frac{π}{2} - 2$

Step 1.5

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

$(- \frac{π}{2} - 2, \frac{π}{2} - 2)$

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

$x = - \frac{π}{2} - 2 + π n$

Step 1.8

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

No Horizontal Asymptotes

No Oblique Asymptotes

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

$a = 1$

$b = 1$

$c = - 2$

$d = - π$

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 $\tan (x + 2)$ .

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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 $- π$ .

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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: $\frac{- 2}{1}$

Step 5.3

Divide $- 2$ by $1$ .

Phase Shift: $- 2$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $π$

Phase Shift: $- 2$ ( $2$ to the left)

Vertical Shift: $- π$

Step 7

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

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

Amplitude: None

Period: $π$

Phase Shift: $- 2$ ( $2$ to the left)

Vertical Shift: $- π$

Step 8