How do you graph #y + 1 = 3 cos 4 (x-2)#?

Write the equation as #y=3cos(4x-8)+1#. Now you're in the generic form

#Acos(omega x + phi)+k#.

A function in this form has four important informations:

1. #A# is the amplitude, which is the maximum value reached by the function. Of course, the standard amplitude is #1#, since #cos(x)# ranges between #-1# and #1#. And in fact, a function with amplitude #A# ranges from #-A# to #A#.

2. #omega# affects the period, because it changes the "speed" with which the function grows. Look at this example: if we have the standard function #cos(x)#, if you want to go from #cos(0)# to #cos(2\pi)#, the variable #x# must from from #0# to #2\pi#. Now try #cos(2x)#: in this case, if #x# runs from #0# to #pi#, your function ranges from #cos(0)# to #cos(2\pi)#. So, we needed "half" the #x# travel to cover a whole period. In general, the formula states that the period #T# is #T=(2\pi)/\omega#.

3. #\phi# is a phase shift, and again look at this example: with the standard function #cos(x)#, you have #cos(0)# for #x=0#, of course. Now we try #cos(x-1)#. To have #cos(0)#, we must input #x=1#. So, the same value has been shifted ahead of #1# unit. In general, if #phi# is positive, it shifts the function backwards (which means to the left on the #x#-axis) of #phi# units, and if #phi# is negative, the shift is to the right.

4. Finally, the #+1# at the end is a vertical shift. Think of it like this: when you have #y=cos(x)#, it means that you are associating with every #x# the #y# value "#cos(x)#". Now, you change to #y=cos(x)+1#. This means that now you associate to the same old #x# the new value #cos(x)+1#, which is one more than the old value. So, if you add one unit on the #y# axis, you shift upwards. Of course, if #k# is negative, the shift is downwards.

So, in the end, you start from the standard cosine function. Then, you do all the transformations:

1. Change the amplitude

2. Change the period

3. Horizontal shift

4. Vertical shift