f(x) = -2sin[3(x-pi/2)] + 4.


It is obvious that the amplitude is |-2| = 2 and that the period is ;  vertical shift is 4 units up.


For the further analysis, especially for the accurate analysis of the horizontal shift, you MUST write 
the given function with the POSITIVE leading factor (amplitude).


    // If you do not make it (as @MathLover1 NEGLECTS do it at every her attempt), you analysis INEVITABLY will be ERRONEOUS. //


So I write equivalent transformations 

    f(x) =  =  = 

                I want to make the leading coefficient positive. For it, I continue

         =     (I change the leading sign by adding  to the sine argument, and then continue further)

         =  = .


Now, having written the sine function in standard CANONIC form with the positive leading coefficient, 

     I can make the standard analysis for the shift,  and I can safely conclude that the horizontal shift is    units to the right.


Now look into the plot below, which CONFIRMS my analysis.



    


    Plot y =  (the given function, red), 

          and  y = sin(3x)         (the parent function for the shift analysis, green)



Notice that, when we compare the shift, our reference parent function for the comparison is y = sin(3x), shown by green in my plot.

f(x) = -cos(2x -pi/4) -3.


It is obvious that the amplitude is |-1| = 1 and that the period is  = ;  vertical shift is 3 units down.


For the further analysis, especially for the accurate analysis of the horizontal shift, you MUST write 
the given function with the POSITIVE leading factor (amplitude).


    // If you do not make it (as @MathLover1 NEGLECTS do it at every her attempt), you analysis INEVITABLY will be ERRONEOUS. //


So I write equivalent transformations 

    f(x) =  =  = 

                I want to make the leading coefficient positive. For it, I continue

         =     (I change the leading sign by adding  to the cosine argument, and then continue further)

         =  = .


Now, having written the sine function in CANONIC form with the positive leading coefficient, 

     I can make the standard analysis for the shift,  and I can safely conclude that the horizontal shift is    units to the left.


Now look into the plot below, which CONFIRMS my analysis.



    


    Plot y =  (the given function, red), 

          and  y = cos(2x)         (the parent function for the shift analysis, green)



Notice that, when we compare the shift, our reference parent function for the comparison is y = cos(2x), shown by green in my plot.

f(x) = -sec(pi-5x).


In this case, there is NO amplitude. Obviously, the period is ;  there is NO vertical shift.


For the further analysis, especially for the accurate analysis of the horizontal shift, you MUST write 
the given function with the POSITIVE leading factor.


So I write equivalent transformations 

    f(x) =  = 

                I want to make the leading coefficient positive. For it, I change 

                the leading sign by subtracting  to the sec argument, and then continue further

         =  =  = use the fact that sec is an EVEN function = .


Now, having written the sec function in CANONIC form with the positive leading coefficient, 

     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.


Now look into the plot below, which CONFIRMS my analysis.



    


    Plot y =  (the given function, red), 

          and  y = sec(5x) + 0.2         (the parent function for the shift analysis, green)



Notice that, when we compare the shift, our reference parent function for the comparison is y = sec(5x), shown by green in my plot.


To make two plots distinguishable, I added 0.2 to the reference function // othewise, the plot are identical and the difference is not seen.