let's take a look at writing an equation for a graph when the graph given is either a side or a cosine function so looking at this graph we can see here at the horizontal units are each one PI and the vertical units are all one so to get started here let's try to identify our amplitude is now the easiest way to do that I think would be to just kind of eyeball where your midline would be so we can see here that the line that would divide this function into two halves would be the line y equals negative two from the midline we can see that we have to go down three units to get to the minimum or up three units to get to our maximum I the way you want to think about it and that is going to be our amplitude so in our amplitude is equal to three which would imply that our leading coefficient a wouldn't either need to be positive or negative three depending on the situation the other way to think about determining the amplitude would it be just go the entire distance from maximum to the minimum which in this case would be six and then just cut that in half you'll get the same answer either way next up let's take a look at calculating the period the easiest way again to do that would be to go from peak to peak so if we measure from this minimum to this mineral on the right we can just measure the distance it would take to go from one minimum to the other you can see here it would have taken for horizontal units to get from one minimum to the other so that is four pi the period would be four pi since we know that the period is equal to two PI over B we can kind of set up this equation here 2 pi over b is equal to 4 pi we can solve for B and see that B should be equal to 1/2 so we've got an A we've got a B let's take a look at finding the vertical shift see that from our old mid line here of the x-axis everything has been shifted down 2 units so if we know our vertical shift is down 2 units then we know that our K is going to be equal to negative 2 so again we've got our a we've got our B and we've got our K the only variable that might be left for us to find is the H for the phase shift and we may not even need that so let's take a look at our graph here taking a look at when x equals 0 we can see that our graph currently takes on a maximum value the function that also take out a maximum value when x is 0 is cosine so thinking about this as a graph that is a cosine function you can see that we wouldn't need to worry about a phase shift because everything lines up where it's supposed to be x is 0 our f of X is the maximum value that we're going to see and everything else kind of follows suit from there go to intercept minimum intercept defer to continue out it would go out to a maximum so no phase shift needed here to make everything line up last thing we need to do now is just kind of plug in the variables that we found so we know that if RB is 1/2 we'll have 1/2 X which is the same thing as x over 2 our phase shift again we don't have one so we don't have to worry about doing a plus or minus zero here for our H just close up our parenthesis vertical shift down 2 gives us a minus 2 on the end and the last names worry about is our amplitude now our a we said is if you're positive or negative 3 since cosine normally has a maximum when x is 0 which is what we see here on our graph tall so we know there hasn't been a reflection if there was our our first point lives in a minimum so since there's been no reflection we're going to take the positive a value for this particular graph and all said and done this is our final answer our f of X is equal to 3 times the cosine of X over 2 minus 2 you