Trigonometry Examples

$f (x) = 5 \tan (3 x - 2) + 1$

Step 1

Find the asymptotes.

Tap for more steps...

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 $y = 5 \tan (3 x - 2) + 1$ . 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 $y = 5 \tan (3 x - 2) + 1$ .

$3 x - 2 = - \frac{π}{2}$

Step 1.2

Solve for $x$ .

Tap for more steps...

Step 1.2.1

Add $2$ to both sides of the equation.

$3 x = - \frac{π}{2} + 2$

Step 1.2.2

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

Tap for more steps...

Step 1.2.2.1

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

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

Step 1.2.2.2

Simplify the left side.

Tap for more steps...

Step 1.2.2.2.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 1.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.2.3

Simplify the right side.

Tap for more steps...

Step 1.2.2.3.1

Simplify each term.

Tap for more steps...

Step 1.2.2.3.1.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.2.2.3.1.2

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

Tap for more steps...

Step 1.2.2.3.1.2.1

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

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

Step 1.2.2.3.1.2.2

Multiply $3$ by $2$ .

$x = - \frac{π}{6} + \frac{2}{3}$

Step 1.3

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

$3 x - 2 = \frac{π}{2}$

Step 1.4

Solve for $x$ .

Tap for more steps...

Step 1.4.1

Add $2$ to both sides of the equation.

$3 x = \frac{π}{2} + 2$

Step 1.4.2

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

Tap for more steps...

Step 1.4.2.1

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

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

Step 1.4.2.2

Simplify the left side.

Tap for more steps...

Step 1.4.2.2.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 1.4.2.2.1.1

Cancel the common factor.

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

Step 1.4.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.4.2.3

Simplify the right side.

Tap for more steps...

Step 1.4.2.3.1

Simplify each term.

Tap for more steps...

Step 1.4.2.3.1.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.4.2.3.1.2

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

Tap for more steps...

Step 1.4.2.3.1.2.1

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

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

Step 1.4.2.3.1.2.2

Multiply $2$ by $3$ .

$x = \frac{π}{6} + \frac{2}{3}$

Step 1.5

The basic period for $y = 5 \tan (3 x - 2) + 1$ will occur at $(- \frac{π}{6} + \frac{2}{3}, \frac{π}{6} + \frac{2}{3})$ , where $- \frac{π}{6} + \frac{2}{3}$ and $\frac{π}{6} + \frac{2}{3}$ are vertical asymptotes.

$(- \frac{π}{6} + \frac{2}{3}, \frac{π}{6} + \frac{2}{3})$

Step 1.6

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

$\frac{π}{3}$

Step 1.7

The vertical asymptotes for $y = 5 \tan (3 x - 2) + 1$ occur at $- \frac{π}{6} + \frac{2}{3}$ , $\frac{π}{6} + \frac{2}{3}$ , and every $x = - \frac{π}{6} + \frac{2}{3} + \frac{π n}{3}$ , where $n$ is an integer.

$x = - \frac{π}{6} + \frac{2}{3} + \frac{π n}{3}$

Step 1.8

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - \frac{π}{6} + \frac{2}{3} + \frac{π n}{3}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - \frac{π}{6} + \frac{2}{3} + \frac{π n}{3}$ 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.

$a = 5$

$b = 3$

$c = 2$

$d = 1$

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 $\frac{π}{| b |}$ .

Tap for more steps...

Step 4.1

Find the period of $5 \tan (3 x - 2)$ .

Tap for more steps...

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 $3$ in the formula for period.

$\frac{π}{| 3 |}$

Step 4.1.3

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

$\frac{π}{3}$

Step 4.2

Find the period of $1$ .

Tap for more steps...

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 $3$ in the formula for period.

$\frac{π}{| 3 |}$

Step 4.2.3

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

$\frac{π}{3}$

Step 4.3

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

$\frac{π}{3}$

Step 5

Find the phase shift using the formula $\frac{c}{b}$ .

Tap for more steps...

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: $\frac{2}{3}$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{π}{3}$

Phase Shift: $\frac{2}{3}$ ( $\frac{2}{3}$ to the right)

Vertical Shift: $1$

Step 7

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

Vertical Asymptotes: $x = - \frac{π}{6} + \frac{2}{3} + \frac{π n}{3}$ where $n$ is an integer

Amplitude: None

Period: $\frac{π}{3}$

Phase Shift: $\frac{2}{3}$ ( $\frac{2}{3}$ to the right)

Vertical Shift: $1$

Step 8