HOW TO GRAPH COTANGENT FUNCTIONS WITH TRANSFORMATIONS

Problem 1 :

Graph y = 3 cot 2x

Solution :

• A = 3, B = 2
• period = 𝜋/|B|
• period for the function y = 3 cot 2x is (0, 𝜋/2)

Step 1 :

Find two consecutive asymptotes, we do this by finding an interval containing one period.

An interval containing one period is (0, 𝜋/2). Thus two consecutive asymptotes occur at x = 0 and x = 𝜋/2.

Step 2 :

Midpoint of x =  0 and x = 𝜋/2 is an x-intercept of the function.

An x-intercept is 0 and the graph passes through (𝜋/4, 0).

Step 3 :

To find the points on the graph which is 1/4 and 3/4 of the way between two consecutive asymptotes, we follow

So, the required points on the curve are (𝜋/8, 3) and (3𝜋/8, -3).

Step 4 :

Vertical asymptote of y = cot x is k𝜋, where k is integer. Here

2x = k𝜋

x = k𝜋/2

• When k = -1, x = -𝜋/2
• When k = 0, x = 0
• When k = 1, x = 𝜋/2
• When k = 2, x = 𝜋

Repeat the same pattern in the interval.

Problem 2 :

Graph y = (1/2) cot 2x

Solution :

• A = 1/2, B = 2
• period = 𝜋/|B|
• period for the function y = (1/2) cot 2x is (0, 𝜋/2)

Step 1 :

Find two consecutive asymptotes, we do this by finding an interval containing one period.

An interval containing one period is (0, 𝜋/2). Thus two consecutive asymptotes occur at x = 0 and x = 𝜋/2.

Step 2 :

Midpoint of x =  0 and x = 𝜋/2 is an x-intercept of the function.

An x-intercept is 0 and the graph passes through (𝜋/4, 0).

Step 3 :

To find the points on the graph which is 1/4 and 3/4 of the way between two consecutive asymptotes, we follow

So, the required points on the curve are (𝜋/8, 0.5) and (3𝜋/8, -0.5).

Step 4 :

Vertical asymptote of y = cot x is k𝜋, where k is integer. Here

2x = k𝜋

x = k𝜋/2

• When k = -1, x = -𝜋/2
• When k = 0, x = 0
• When k = 1, x = 𝜋/2
• When k = 2, x = 𝜋

Repeat the same pattern in the interval.

Problem 3 :

Graph y = -3 cot (𝜋/2)x

Solution :

• A = 3, B = 1/2
• period = 𝜋/|B|
• period for the function y =  -3 cot (𝜋/2)x is (0, 2)

Step 1 :

Find two consecutive asymptotes, we do this by finding an interval containing one period.

An interval containing one period is (0, 2). Thus two consecutive asymptotes occur at x = 0 and x = 2.

Since there is a reflection, we can draw the graph of y = 3 cot (𝜋/2)x, then using the concept of reflection we can get a new graph.

Step 2 :

Midpoint of x =  0 and x = 2 is an x-intercept of the function.

An x-intercept is 0 and the graph passes through (1, 0).

Step 3 :

To find the points on the graph which is 1/4 and 3/4 of the way between two consecutive asymptotes, we follow

So, the required points on the curve are (0.5, 3) and (1.5, -3).

Using reflection,

(0.5, -3) and (1.5, 3)

Step 4 :

Vertical asymptote of y = cot x is k𝜋, where k is integer. Here

(𝜋/2)x = k𝜋

x = k𝜋/(𝜋/2)

x = 2k

• When k = -1, x = -2
• When k = 0, x = 0
• When k = 1, x = 2
• When k = 2, x = 4

Repeat the same pattern in the interval.

After reflection :

Problem 4 :

Graph y = 3 cot (x + 𝜋/2)

Solution :

• A = 3, B = 1
• period = 𝜋/|B| == > 𝜋
• period for the function y = 3 cot (x + 𝜋/2) is (0, 𝜋)

Step 1 :

Find two consecutive asymptotes, we do this by finding an interval containing one period.

An interval containing one period is (-𝜋/2, 𝜋/2). Thus two consecutive asymptotes occur at x = -𝜋/2 and x = 𝜋/2.

Step 2 :

Midpoint of x = -𝜋/2 and x = 𝜋/2 is an x-intercept of the function.

An x-intercept is 0 and the graph passes through (0, 0).

Step 3 :

To find the points on the graph which is 1/4 and 3/4 of the way between two consecutive asymptotes, we follow

So, the required points on the curve are (-𝜋/4, 3) and (𝜋/4, -3).

Step 4 :

Vertical asymptote of y = cot x is k𝜋, where k is integer. Here

x+𝜋/2 = k𝜋

x = k𝜋 - (𝜋/2)

x = (2k𝜋 - 𝜋)/2

x = (𝜋/2)(2k-1)

• When k = 0, x = -𝜋/2
  
• When k = 1, x = 𝜋/2
• When k = 2, x = 3𝜋/2

Repeat the same pattern in the interval.