Question 1203784
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Sketch two periods of the graph for the following function.
f(x) = 5 sec(3x)
Identify the stretching factor and period.
Identify the asymptotes in the displayed domain of the graph you selected above. 
(Enter your answers as a comma-separated list of equations.)
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        Teaching by  @MathLover1 in her post is  INCORRECT  and  WRONG.

        Many of her numbers are wrong; her argumentation is defective; her explanations are incorrect.


        @Theo uses inappropriate terminology/terms as  " amplitude ",  " frequency ",  that are used 

        for harmonic sinusoidal functions  ONLY,  so if you will use it,  you may find yourself

        in unpleasant situation,  when everybody around will look at you with squared eyes.


        In addition,  @Theo in his post makes a  RUDE  MISTAKE,  stating that the given function 

        has vertical asymptotes at  x= 0 degrees and  x= 120 degrees.  The asymptotes are different - see my post.


        I came to bring correct answers to posed questions and to make this job in a way as it should be done.

        And using  ( it is  IMPORTANT ! )  right words,  connected in right order into right sequence 

        for readers to use my post as a universal mantra.



<pre>
(1)  To sketch the plot, go to web-site www.desmos.com/calculator

     Find there free of charge plotting tool for common use.

     Print the equation of the function and get the plot in the next instance.



(2)  In this problem, there are 2 (two) stretching factors: one vertical and one horizontal, 
     and both deserves to be discussed.

     Vertical stretching factor is 5.  It is the coefficient (multiplier) before the function "sec".

     In this problem, this coefficient is greater than 1, so it really produces vertical stretching 
     with coefficient 5.


     Horizontal stretching coefficient is 1/3, and because it is less than 1, in this problem 
     it is better to speak about horizontal compression with coefficient 3.

     The period is  {{{2pi/3}}}.


(3)  Since  5sec(3x) = {{{5/cos(3x)}}}, vertical asymptotes are where cos(3x) is zero.

     cos(3x) is zero where 3x = {{{pi/2 + k*pi}}},  or  at  x= {{{pi/6 + k*(pi/3)}}},  k= 0, +/-1, +/-2, . . . 

     So, vertical asymptotes are at x= {{{pi/6 + k*(pi/3)}}},  k= 0, +/-1, +/-2, . . . 

     First two vertical asymptotes in positive domain are at x= {{{pi/6 + pi/3}}} = {{{3pi/6}}} = {{{pi/2}}}  and   x= {{{pi/6 + (2pi)/3}}} = {{{5pi/3}}}.
</pre>

Solved and answered in correct form, &nbsp;as it should be done,
with explanations where necessary.