y= -2 tan ( 1/2 x + 5(3.14)/3)

Question

how do you do this step by step, draw it on a graph?

Answer 1

The function is y = - 2 tan (1/2 x + 5π/3).

Compare the equation y = - 2 tan (1/2 x + 5π/3) with y = a tan(bx - c).

a = - 2, b = 1/2, c = - 5π/3 and Period = π/b = π/(1/2) = 2π.

Two consecutive vertical asymptotes can be found by solving the equations bx - c = - π/2 and bx - c = π/2.

1/2 x + 5π/3 = - π/2         and       1/2 x + 5π/3 = π/2.

1/2 x = - π/2 - 5π/3          and       1/2 x = π/2 - 5π/3.

1/2 x = (- 3π - 10π)/6      and       1/2 x = (- 3π + 10π)/6.

1/2 x = - 13π/6                 and       1/2 x = - 7π/6

x = - 13π/3                       and        x = - 7π/3.

The that two consecutive vertical asymptotes occur at x = - 13π/3 and x = - 7π/3.

The interval [- 13π/3, - 7π/3] corresponds to one cycle of the graph. Dividing this interval into four equal parts produces the key points.

one fourth of part is [- 7π/3 + 13π/3]/4 = (6π/3)(1/4) = π/2.

The x-coordinates of the five key points are

x = - 13π/3.

x = - 13π/3 + π/2 = (- 26π + 3π)/6 = - 23π/6.

x = - 23π/6 + π/2 = (- 23π + 3π)/6 = - 20π/6 = - 10/3π.

x = - 10π/3 + π/2 = (- 20π + 3π)/6 = - 17π/6.

x = - 17π/6 + π/2 = (- 17π + 3π)/6 = - 14π/6 = - 7π/3.

Between these two asymptotes, plot a few points, including the -intercept, as shown in the table.

x

First ploting the asymptotes.

The midpoint between two consecutive vertical asymptotes is an x - intercept of the graph. The period of the function
y = a tan(bx - c) is the distance between two consecutive vertical asymptotes. The amplitude of a tangent function is not defined.

After plotting the asymptotes and the x - intercept, plot a few additional points between the two asymptotes and sketch one cycle. Finally, sketch one or two additional cycles to the left and right.

Plot these five points and fill in the graph of the tangent function as shown in Figure.

 

Answer 2

The function is .

The first term of zero will not shift the graph of vertically.

The factor of 2 will double the y - coordinates.To determine the horizontal shift and period, We check one cycle.

One cycle :  

               

                

                  .

A cycle will begin at , so the graph is units to the left.

To find the period, we compute the difference between end points for the cycle :

Period

          .

Dividing the period by 4 gives us , so we mark the x - axis at intervals of .

Beginning with as follows(we only need to find the three middle points).

A cycle will begin and end with x - intercepts at and (on the shifted axis).

There will be a vertical asymptote at .

Graph the function :

From the graph, we summarize the characteristics for the tangent .The derivative of these formulaes is similar, except that a period of  π is used.