Graph of Sine, a*sin(bx+c), Function

Graphing and sketching sine functions of the form

f (x) = a*sin (bx + c):

step by step tutorial.

The properties such as domain, range and intercepts of the graphs of these functions are also looked at in detail.

Once you finish the present tutorial, you may want to go through a self test on trigonometric graphs. Free graph paper is available.



Review

We first start with the graph of the basic sine function

f (x) = sin (x)

The domain of function f is the set of all real numbers. The range of f is the interval [-1,1].

-1 <= sin (x) <= 1        (<= means less than or equal)

Also function f is periodic with period equal to 2p.

The graph of f over one period can be sketched by first finding points that give important information such as x intercepts, y intercept, maxima and minima.

Let us make a table of values for function f over the interval one period: [0 , 2p].

x 0 p/2 p 3p/2 2p
f (x) 0 1 0 -1 0

The choice of the values of x in the table correspond  to x  and y intercepts, maxima  and minima points. These are useful points to graph the sine function over one period: [0 , 2p]. To graph f, we first graph the points in the table then join these points. Of course you may add extra points if you wish.  But the five points used are key points.  Another important point to note is that the 5 key points divide the period into 4 equal parts. See figures below.

key points of sine function

 

graph of sin(x)

To have a complete picture of why the graph of the sin(x) changes with x as shown above, you may want to go through an interactive tutorial on the trigonometric unit circle trigonometric unit circle.

Graphing f (x) = a*sin(b x + c)

We first need to understand how do the parameters a, b and c affect the graph of f (x)=a*sin(bx+c) when compared to the graph of sin(x) ?

You may want to go through an interactive tutorial on sine functions.

The domain of f is the set of all real numbers. The range of expression bx + c is the set of all real numbers. Therefore the range of sin(bx+c) is [-1,1]. Hence

-1 <= sin(bx+c) <= 1

Multiply both sides by a. If a > 0

-a <= a*sin(bx+c) <= a

If a < 0 (change symbols of inequality)

-a >= a*sin(bx+c) >= a

or       a <= a*sin(bx+c) <= - a

 

We can say that parameter a affect the range of f which can be written as [-|a| , |a|].

| a | is called the amplitude.

Period of f

Let us first assume that c = 0 and f (x) = a*sin (bx). For f to complete one cycle (period), expression bx needs to vary from 0 to 2p.

0 <= bx <= 2p.

Divide all terms in the inequality by b.

If b > 0

0 <= x <= 2p/b.

Period = 2p/b - 0 = 2p/b.

If b < 0 (change symbols of inequalities)

0 >= x >= 2p/b.

Which is equivalent to

2p/b <= x <=0.

Period = 0 - 2p/b = - 2p/b

Using the absolute value notation we can write

the period of f  = 2p/| b |.

Phase Shift

We now consider the whole argument bx + c. For f to complete one cycle (period), expression bx+c needs to vary from 0 to 2p.

0 <= bx + c<= 2p.

Assume b > 0 and solve for x

-c <= bx <= 2p-c.

-c / b <= x <= 2p/b - c/b.

Period of f =  2p/b - c/b - (-c/b) = 2p/| b |. c does not affect the period.

Let us now compare the cycle [0 , 2p/b] when c = 0 with the cycle [-c/b , 2p/b - c/b]. This indicates that there is a shift of -c/b. -c/b is called the phase shift. If -c/b < 0, the shift will be to the left. If -c/b >0, the shift will be to the right.

phase shift -c/b < 0

phase shift -c/b > 0


Example 1: f is a function given by

f (x) = 2sin(3x - p/2)

 

a - Find the domain of f and range of f.

b - Find the period and the phase shift of the graph of f.

c - Sketch the graph of function f over one period.

Answer to Example 1

a - The domain of f is the set of all real numbers . The range is given by the interval [-2 , 2].

b - Period = 2p/|b| = 2p /3

Phase shift = - c / b = - (- p/2) / 3 =  p/6

c - To sketch the graph of f over one period, we need to find the 5 key points first. Let 3x - p/2 vary from 0 to 2p in order to have a complete period then find the values of f (x). See table below.

3x-p/2 0 p/2 p 3p/2 2p
f (x) 0 2 0 -2 0

We now need to find the corresponding values of x. The first row in the table above gives

0 <= 3x - p/2 <= 2p

Solve for x

p/2 <= 3x <= 2p+p/2

p/2 <= 3x <= 5p/2

p/6 <= x <= 5p/6

We now complete the table with the x values. Once the x values  p/6 and 5p/6 describing a whole period are found, the other 3 points are found as follows:

The middle value  =( p/6 + 5p/6) / 2 = 3p/6

Value at the first quarter = (p/6 + 3p/6) / 2 = 2p/6

Value at the third quarter = (3p/6 + 5p/6) / 2  = 4p/6

The fractions have not been reduced, this will make it easy to scale the x axis in units of p/6 and graph the points.

3x-p/2 0 p/2 p 3p/2 2p
f (x) 0 2 0 -2 0
x p/6 2p/6 3p/6 4p/6 5p/6

We now put the points given (x , f (x)) and join them by a smooth curve.

graph of f(x) = 2sin(3x-pi/2)

 

Matched Problem: f is a function given by

f (x) = (1/2)sin(4x + p/2)

 

a - Find the domain of f and range of f.

b - Find the period and the phase shift of the graph of f.

c - Sketch the graph of function f over one period.

More references and links to graphing, graphs of functions and sine functions.



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Updated: 2 April 2013

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