Trigonometry Examples

$y = 2 \cot (3 x) + 4$

Step 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 $y = 2 \cot (3 x) + 4$ . 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 $y = 2 \cot (3 x) + 4$ .

$3 x = 0$

Step 2

Divide each term in $3 x = 0$ by $3$ and simplify.

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Step 2.1

Divide each term in $3 x = 0$ by $3$ .

$\frac{3 x}{3} = \frac{0}{3}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $3$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{0}{3}$

Step 2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{3}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Divide $0$ by $3$ .

$x = 0$

Step 3

Set the inside of the cotangent function $3 x$ equal to $π$ .

$3 x = π$

Step 4

Divide each term in $3 x = π$ by $3$ and simplify.

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Step 4.1

Divide each term in $3 x = π$ by $3$ .

$\frac{3 x}{3} = \frac{π}{3}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $3$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{π}{3}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{3}$

Step 5

The basic period for $y = 2 \cot (3 x) + 4$ will occur at $(0, \frac{π}{3})$ , where $0$ and $\frac{π}{3}$ are vertical asymptotes.

$(0, \frac{π}{3})$

Step 6

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

$\frac{π}{3}$

Step 7

The vertical asymptotes for $y = 2 \cot (3 x) + 4$ occur at $0$ , $\frac{π}{3}$ , and every $\frac{π n}{3}$ , where $n$ is an integer.

$x = \frac{π n}{3}$

Step 8

Cotangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

Step 9