Trigonometry Examples

$y = \frac{1}{4} \cdot \cot (\frac{1}{5} x) + 3$

Step 1

Simplify each term.

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Step 1.1

Combine $\frac{1}{5}$ and $x$ .

$y = \frac{1}{4} \cdot \cot (\frac{x}{5}) + 3$

Step 1.2

Combine $\frac{1}{4}$ and $\cot (\frac{x}{5})$ .

$y = \frac{\cot (\frac{x}{5})}{4} + 3$

Step 2

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 $y = \frac{\cot (\frac{x}{5})}{4} + 3$ . 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 $y = \frac{\cot (\frac{x}{5})}{4} + 3$ .

$\frac{x}{5} = 0$

Step 3

Set the numerator equal to zero.

$x = 0$

Step 4

Set the inside of the cotangent function $\frac{x}{5}$ equal to $π$ .

$\frac{x}{5} = π$

Step 5

Solve for $x$ .

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Step 5.1

Multiply both sides of the equation by $5$ .

$5 \frac{x}{5} = 5 π$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $5$ .

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Step 5.2.1.1

Cancel the common factor.

$5 \frac{x}{5} = 5 π$

Step 5.2.1.2

Rewrite the expression.

$x = 5 π$

Step 6

The basic period for $y = \frac{\cot (\frac{x}{5})}{4} + 3$ will occur at $(0, 5 π)$ , where $0$ and $5 π$ are vertical asymptotes.

$(0, 5 π)$

Step 7

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

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Step 7.1

$\frac{1}{5}$ is approximately $0.2$ which is positive so remove the absolute value

$\frac{π}{\frac{1}{5}}$

Step 7.2

Multiply the numerator by the reciprocal of the denominator.

$π \cdot 5$

Step 7.3

Move $5$ to the left of $π$ .

$5 π$

Step 8

The vertical asymptotes for $y = \frac{\cot (\frac{x}{5})}{4} + 3$ occur at $0$ , $5 π$ , and every $5 π n$ , where $n$ is an integer.

$x = 5 π n$

Step 9

Cotangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 5 π n$ where $n$ is an integer

Step 10