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

Question

for the function

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

Answer 1

The function is  .

The above function is in the form .

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

The amplitude of the function is .

1)

The period of a sine function is given by .

In the given function B is 2.

So, the period is .

2)

Phase shift of a function is given by .

In the given function c is .

Phase shift is .

3)

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

Continues....

Comments

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.

 

Answer 2

4)

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

The solutions of the given functions are

.

Taking as an interval difference plot the graph.

x

  Now plot these points

Since sine function is a continues sinusoidal function.

So, it has no vertical asymptotes.

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

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 .

It is oscillating between , so the range is .

Comments

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

.

.