Calculus Examples

$y = 2 \cot (2 x - 5 π) + 2$

Step 1

Find the asymptotes.

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Step 1.1

For any $y = \cot (x)$ , vertical asymptotes occur at $x = n π$ , where $n$ is an integer. Use the basic period for $y = \cot (x)$ , $(0, π)$ , to find the vertical asymptotes for $y = 2 \cot (2 x - 5 π) + 2$ . Set the inside of the cotangent function, $b x + c$ , for $y = a \cot (b x + c) + d$ equal to $0$ to find where the vertical asymptote occurs for $y = 2 \cot (2 x - 5 π) + 2$ .

$2 x - 5 π = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Add $5 π$ to both sides of the equation.

$2 x = 5 π$

Step 1.2.2

Divide each term in $2 x = 5 π$ by $2$ and simplify.

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Step 1.2.2.1

Divide each term in $2 x = 5 π$ by $2$ .

$\frac{2 x}{2} = \frac{5 π}{2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5 π}{2}$

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5 π}{2}$

Step 1.3

Set the inside of the cotangent function $2 x - 5 π$ equal to $π$ .

$2 x - 5 π = π$

Step 1.4

Solve for $x$ .

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Step 1.4.1

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

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Step 1.4.1.1

Add $5 π$ to both sides of the equation.

$2 x = π + 5 π$

Step 1.4.1.2

Add $π$ and $5 π$ .

$2 x = 6 π$

Step 1.4.2

Divide each term in $2 x = 6 π$ by $2$ and simplify.

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Step 1.4.2.1

Divide each term in $2 x = 6 π$ by $2$ .

$\frac{2 x}{2} = \frac{6 π}{2}$

Step 1.4.2.2

Simplify the left side.

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Step 1.4.2.2.1

Cancel the common factor of $2$ .

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Step 1.4.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{6 π}{2}$

Step 1.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{6 π}{2}$

Step 1.4.2.3

Simplify the right side.

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Step 1.4.2.3.1

Cancel the common factor of $6$ and $2$ .

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Step 1.4.2.3.1.1

Factor $2$ out of $6 π$ .

$x = \frac{2 (3 π)}{2}$

Step 1.4.2.3.1.2

Cancel the common factors.

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Step 1.4.2.3.1.2.1

Factor $2$ out of $2$ .

$x = \frac{2 (3 π)}{2 (1)}$

Step 1.4.2.3.1.2.2

Cancel the common factor.

$x = \frac{2 (3 π)}{2 \cdot 1}$

Step 1.4.2.3.1.2.3

Rewrite the expression.

$x = \frac{3 π}{1}$

Step 1.4.2.3.1.2.4

Divide $3 π$ by $1$ .

$x = 3 π$

Step 1.5

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

$(\frac{5 π}{2}, 3 π)$

Step 1.6

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

Step 1.7

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

$x = \frac{5 π}{2} + \frac{π n}{2}$

Step 1.8

Cotangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{5 π}{2} + \frac{π n}{2}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{5 π}{2} + \frac{π n}{2}$ where $n$ is an integer

Step 2

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

$a = 2$

$b = 2$

$c = 5 π$

$d = 2$

Step 3

Since the graph of the function $c o t$ 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 $2 \cot (2 x - 5 π)$ .

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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 $2$ in the formula for period.

$\frac{π}{| 2 |}$

Step 4.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

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 $2$ in the formula for period.

$\frac{π}{| 2 |}$

Step 4.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

Step 4.3

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

$\frac{π}{2}$

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: $\frac{5 π}{2}$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{π}{2}$

Phase Shift: $\frac{5 π}{2}$ ( $\frac{5 π}{2}$ 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: $x = \frac{5 π}{2} + \frac{π n}{2}$ where $n$ is an integer

Amplitude: None

Period: $\frac{π}{2}$

Phase Shift: $\frac{5 π}{2}$ ( $\frac{5 π}{2}$ to the right)

Vertical Shift: $2$

Step 8