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document.write( "for each of the following, give the amplitude, the period, and describe any vertical or horizontal shifts
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document.write( "away from the origin(i.e. from the standard parent graph)
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document.write( "2a f(x) = -2sin[3(x-pi/2)] + 4
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document.write( "b. f(x) = -cos(2x -pi/4) -3
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document.write( "c. f(x) = 2tan(x/3) -1
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document.write( "d. f(x) = -sec(pi- 5x)
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document.write( "~~~~~~~~~~~~~~~~~\r
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document.write( "
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document.write( "
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document.write( "\n" );
document.write( " The analysis given by @MathLover1, has ERRORS, that are extremely dangerous for a beginning student.\r
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document.write( "
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document.write( "\n" );
document.write( " Therefore, I came to FIX her errors and to bring a CORRECT ANALYSIS with detailed explanations.\r
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document.write( "
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document.write( "
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document.write( "\n" );
document.write( "Case (a) \r
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document.write( "
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document.write( "\n" );
document.write( "\r\n" );
document.write( "f(x) = -2sin[3(x-pi/2)] + 4.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "It is obvious that the amplitude is |-2| = 2 and that the period is 
; vertical shift is 4 units up.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "For the further analysis, especially for the accurate analysis of the horizontal shift, you MUST write \r\n" );
document.write( "the given function with the POSITIVE leading factor (amplitude).\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " // If you do not make it (as @MathLover1 NEGLECTS do it at every her attempt), you analysis INEVITABLY will be ERRONEOUS. //\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So I write equivalent transformations \r\n" );
document.write( "\r\n" );
document.write( " f(x) = 
= 
= \r\n" );
document.write( "\r\n" );
document.write( " I want to make the leading coefficient positive. For it, I continue\r\n" );
document.write( "\r\n" );
document.write( " = 
(I change the leading sign by adding 
to the sine argument, and then continue further)\r\n" );
document.write( "\r\n" );
document.write( " = 
= 
.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now, having written the sine function in standard CANONIC form with the positive leading coefficient, \r\n" );
document.write( "\r\n" );
document.write( " I can make the standard analysis for the shift, and I can safely conclude that the horizontal shift is 
units to the right.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now look into the plot below, which CONFIRMS my analysis.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " 
\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " Plot y = (the given function, red), \r\n" );
document.write( "\r\n" );
document.write( " and y = sin(3x) (the parent function for the shift analysis, green)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Notice that, when we compare the shift, our reference parent function for the comparison is y = sin(3x), shown by green in my plot.\r\n" );
document.write( "
\r
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document.write( "
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document.write( "\n" );
document.write( "Case (b) \r
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document.write( "
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document.write( "\n" );
document.write( "\r\n" );
document.write( "f(x) = -cos(2x -pi/4) -3.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "It is obvious that the amplitude is |-1| = 1 and that the period is 
= ; vertical shift is 3 units down.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "For the further analysis, especially for the accurate analysis of the horizontal shift, you MUST write \r\n" );
document.write( "the given function with the POSITIVE leading factor (amplitude).\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " // If you do not make it (as @MathLover1 NEGLECTS do it at every her attempt), you analysis INEVITABLY will be ERRONEOUS. //\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So I write equivalent transformations \r\n" );
document.write( "\r\n" );
document.write( " f(x) = 
= = \r\n" );
document.write( "\r\n" );
document.write( " I want to make the leading coefficient positive. For it, I continue\r\n" );
document.write( "\r\n" );
document.write( " = 
(I change the leading sign by adding to the cosine argument, and then continue further)\r\n" );
document.write( "\r\n" );
document.write( " = 
= 
.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now, having written the sine function in CANONIC form with the positive leading coefficient, \r\n" );
document.write( "\r\n" );
document.write( " I can make the standard analysis for the shift, and I can safely conclude that the horizontal shift is 
units to the left.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now look into the plot below, which CONFIRMS my analysis.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " 
\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " Plot y = 
(the given function, red), \r\n" );
document.write( "\r\n" );
document.write( " and y = cos(2x) (the parent function for the shift analysis, green)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Notice that, when we compare the shift, our reference parent function for the comparison is y = cos(2x), shown by green in my plot.\r\n" );
document.write( "
\r
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document.write( "
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document.write( "\n" );
document.write( "Case (d)\r
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document.write( "
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document.write( "\n" );
document.write( "\r\n" );
document.write( "f(x) = -sec(pi-5x).\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "In this case, there is NO amplitude. Obviously, the period is 
; there is NO vertical shift.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "For the further analysis, especially for the accurate analysis of the horizontal shift, you MUST write \r\n" );
document.write( "the given function with the POSITIVE leading factor.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So I write equivalent transformations \r\n" );
document.write( "\r\n" );
document.write( " f(x) = 
= \r\n" );
document.write( "\r\n" );
document.write( " I want to make the leading coefficient positive. For it, I change \r\n" );
document.write( "\r\n" );
document.write( " the leading sign by subtracting to the sec argument, and then continue further\r\n" );
document.write( "\r\n" );
document.write( " = 
= 
= use the fact that sec is an EVEN function = 
.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now, having written the sec function in CANONIC form with the positive leading coefficient, \r\n" );
document.write( "\r\n" );
document.write( " I can make the standard analysis for the shift, and I can safely conclude that the horizontal shift is 0 (zero, ZERO) in this case.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now look into the plot below, which CONFIRMS my analysis.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " 
\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " Plot y = (the given function, red), \r\n" );
document.write( "\r\n" );
document.write( " and y = sec(5x) + 0.2 (the parent function for the shift analysis, green)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Notice that, when we compare the shift, our reference parent function for the comparison is y = sec(5x), shown by green in my plot.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "To make two plots distinguishable, I added 0.2 to the reference function // othewise, the plot are identical and the difference is not seen.\r\n" );
document.write( "
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document.write( "==============\r
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document.write( "
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document.write( "\n" );
document.write( "At this point, my solution is COMPLETED.\r
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document.write( "
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document.write( "
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document.write( "\n" );
document.write( "I hope that the reader sees now all errors of the solution by @MathLover1.\r
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document.write( "
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document.write( "\n" );
document.write( "She really made all possible and typical errors in her analysis.\r
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document.write( "
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document.write( "\n" );
document.write( "I really glad that I had the opportunity to fix all these errors and to teach the reader TO AVOID them.\r
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document.write( "
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document.write( "
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document.write( " \n" );
document.write( "![\"\"](\"/images/stars/stars-13.gif\")
