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determine amplitude, period, phase shift, vertical shift, asymptotes, domain & range

0 votes

for the function

f(x)=-1/3sin(2x+pi).

asked Mar 4, 2014 in TRIGONOMETRY by angel12 Scholar

2 Answers

0 votes

The function is  image.

The above function is in the form image.

Here A is amplitude, B is stretch along x-axis, d is a constant determines the vertical shift.

The amplitude of the function is image.

1)

The period of a sine function is given by image.

In the given function B is 2.

So, the period is image.

2)

Phase shift of a function is given by image.

In the given function c is image.

Phase shift is image.

3)

In the given function d is zero, that means there is no vertical shift in the function.

Continues....

answered Apr 30, 2014 by Johncena Apprentice

The finction is f(x) = - 1/3 sin(2x + π).

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

a = - 1/3, b = 2, c = - π and d = 0.

Amplitude = | a | = | - 1/3 | = 1/3

Period = 2π/b = 2π/2 = π.

Phase shift = c/b = - π/2.

Vertical shift = d = 0.

 

0 votes

4)

To find the asymptotes of the function, graph the function over a period.

The solutions of the given functions are

image

image.

Taking imageas an interval difference plot the graph.

             x 

                            image
     image

        image

    image

     image

      image

 image

     image

      image

     image

           image

  Now plot these points

pro 9344

Since sine function is a continues sinusoidal function.

So, it has no vertical asymptotes.

And horizontally it is oscillating between image, but it is not converging at either image.

So, it also doesn't have horizontal asypmtotes.

5)

From the graph we can also say the domain and range of the function.

Domain is image.

It is oscillating between image, so the range is image.

answered Apr 30, 2014 by Johncena Apprentice

The left and right endpoints of a one-cycle interval can be determined by solving the equations bx - c = 0 and bx - c = 2π.

Here to find the left and right endpoints of a one-cycle interval can be determined by solving the equations

image.

image.

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