GRAPHING TANGENT FUNCTIONS WITH TRANSFORMATIONS

Problem 1 :

Graph y = 2 tan (x/2)  for -π < x < 3π

Solution :

Step 1 :

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

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

Step 2 :

Midpoint of x =  -𝜋 and x = 𝜋.

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/2, -2) and (𝜋/2, 2).

Step 4 :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

x/2 = k𝜋 + 𝜋/2

x = 2(k𝜋 + 𝜋/2)

x = 2k𝜋 + 𝜋

x = 𝜋(2k + 1)

• When k = -1, x = -𝜋
• When k = 0, x = 𝜋
• When k = 1, x = 3𝜋
• When k = 2, x = 5𝜋

Repeat the same pattern in the interval.

Problem 2 :

Graph two full periods of y = tan (x + π/4)

Solution :

Step 1 :

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

An interval containing one period is (-3𝜋/4, 𝜋/4). Thus two consecutive asymptotes occur at x = -3𝜋/4 and x = 𝜋/4.

Step 2 :

x-intercept :

So, x-intercept is at (-𝜋/4, 0)

Step 3 :

Points on the curve,

So, the required points are (-𝜋/2, -1) and (0, 1).

Step 4 :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

x + 𝜋/4 = k𝜋 + 𝜋/2

x = k𝜋 + 𝜋/2 - 𝜋/4

x = k𝜋 + (2𝜋-𝜋)/4

x = k𝜋 + 𝜋/4

x = 𝜋(k + 1/4)

• When k = -1, x = -3𝜋/4
• When k = 0, x = 𝜋/4
• When k = 1, x = 5𝜋/4
• When k = 2, x = 9𝜋/4

Repeat the same pattern in the interval.

Problem 3 :

Graph two full periods of y = 3 tan (x/4)

Solution :

Step 1 :

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

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

Step 2 :

x-intercept :

So, x-intercept is at (0, 0) at the interval (-2𝜋, 2𝜋). To get the more x-intercepts, we can find the midpoint of any two consecutive asymptotes.

Step 3 :

So, the required points are (-𝜋, -3) and (𝜋, 3).

Step 4 :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

x/4 = k𝜋 + 𝜋/2

x = 4(k𝜋 + 𝜋/2)

x = 4k𝜋 + 4(𝜋/2)

x = 4k𝜋 + 2𝜋

x = 2𝜋(2k + 1)

• When k = -1, x = -2𝜋
• When k = 0, x = 2𝜋
• When k = 1, x = 6𝜋
• When k = 2, x = 10𝜋

At these positions, we have consecutive asymptotes.

Problem 4 :

Graph two full periods of y = (1/2) tan (2x)

Solution :

Step 1 :

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

An interval containing one period is (-𝜋/4, 𝜋/4). Thus two consecutive asymptotes occur at x = -𝜋/4 and x = 𝜋/4.

Step 2 :

x-intercept :

Step 3 :

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

Step 4 :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

2x = k𝜋 + 𝜋/2

x = (1/2)(k𝜋 + 𝜋/2)

x = (1/4)𝜋(2k + 1)

• When k = -1, x = -𝜋/4
• When k = 0, x = 𝜋/4
• When k = 1, x = 3𝜋/4
• When k = 2, x = 5𝜋/4

At these positions, we have consecutive asymptotes.

Problem 5 :

Graph two full periods of y = -2 tan (x/2)

Solution :

Since we have negative sign for 2, we have to reflect the graph across 

Step 1 :

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

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

Step 2 :

x-intercept :

Step 3 :

So, the required points are (-𝜋/2, -2) and (𝜋/2, 2).

Step 4 :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

x/2 = k𝜋 + 𝜋/2

x = 2(k𝜋 + 𝜋/2)

x = 2k𝜋 + 𝜋

x = 𝜋(2k + 1)

• When k = -1, x = -𝜋
• When k = 0, x = 𝜋
• When k = 1, x = 3𝜋
• When k = 2, x = 5𝜋

At these positions, we have consecutive asymptotes.