How do you graph #y + 1 = 3 cos 4 (x2)#?
1 Answer
Write the equation as
A function in this form has four important informations:

#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# . 
#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# . 
#\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(x1)# . 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. 
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: