<PRE><B>Question 41472</B>
&lt;pre&gt;&lt;font size = 4&gt;&lt;b&gt;
graph y=3sinx and y=sin 3x on the same axes. 
label the graph of each function.

I know the amp. of first function is 3 and
period is 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt; and for second function amp. is 1 
and period is 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3 i just dont know how to graph

=======================================================

Draw the y-axis long enough to include both amplitudes 
as well as the opposite signs of the amplitudes:

 3|-
  |
 2|-
  |
 1|- 
  |
 0|-----------------------------------------------------------------------
  |
-1|-
  |
-2|-
  |
-3|- 

Divide the period of each by 4. 

Dividing the first function&#39;s period by 4: 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;÷4 is &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2
Multiply this by 0, 1, 2, 3, and 4

0&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, 1&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, 3&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, 4&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2

or, reducing:

0, &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;, 3&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;

Positive sine functions go x-intercept, maximum, x-intercept, minimum,
x-intercept. 

Therefore it will have an x intercept at 0, a maximum at &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, 
an x-intercept at &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;, a minimum at 3&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2 and an x-intercept at 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;

Dividing the second function&#39;s period by 4: (2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3)÷4 is &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6

Multiply this by 0, 1, 2, 3, and 4

0&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 1&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 3&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 4&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6

or, reducing:

0, &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;, &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3, 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3

Positive sine functions go x-intercept, maximum, x-intercept, minimum,
x-intercept. 

Therefore it will have an x intercept at 0, a maximum at &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 
an x-intercept at &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;, a minimum at &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3 and an x-intercept at 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3

The LCD of periods &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 and &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2 is 6, so mark off the x-axis in units of 
&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, that is

1&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 3&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 4&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 5&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, ... , 12&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6

 3|-
  |
 2|-
  |
 1|- 
  |
-0|----·----·----·----·----·----·----·----·----·----·----·----·------   
  |  &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6  2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 3&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 4&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 5&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 6&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 7&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 8&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 9&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 10&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 11&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 12&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6
-1|-
  |
-2|-
  |
-3|-

I stopped at 12&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 because that equals 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;, the larger period.

Reducing the ones of those that will reduce:

&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3, &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3, 5&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;, 7&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 4&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3, 3&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2, 5&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3, 11&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6, 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;.

 3|-
  |
 2|-
  |
 1|-
  | 
-0|----·----·----·----·----·----·----·----·----·----·----·----·------   
  |   &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6  &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3  &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3 5&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6   &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;  7&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 4&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3 3&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/2 5&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/3 11&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;/6 2&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;
-1|-
  |
-2|-
  |
-3|-

Note that those values on the x-axis are about

.5, 1, 1.6, 2.1, 2.6, 3.14, 3.7, 4.2, 4.7, 5.2, 5.8, 6.28 

So the two graphs are

{{{ graph( 300, 100, 0, 6.28, -3, 3, sin(3*x)*sqrt(2.1-x)/sqrt(2.1-x), 3*sin(x)) }}}

But you should extend the graph of y = sin(3x) to the
length of the period of the graph of y = 3sin(x) like
this:

{{{ graph( 300, 100, 0, 6.28, -3, 3, sin(3*x), 3*sin(x)) }}}

You should also leave the x-axis markings in terms of &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;,
rather than their numerical values as are on the graphs above. 

Edwin
AnlytcPhil@aol.com&lt;/pre&gt;</PRE>