If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:5:22

Video transcript

right the equation of the function f of X graphed below so we have this clearly periodic function so immediately you might say well this is either going to be a sine function or cosine function but its midline and its amplitude are not just the plain vanilla sine or cosine function and we can see that right over here the midline is halfway between the maximum point and the minimum point the maximum point right over here it hits a value of y equals one at the minimum points it hits a value of y is equal to negative five so halfway between those want the average of 1 and negative 5 1 plus negative 5 is negative 4 divided by 2 is negative 2 so this right over here is the midline so this thing is clearly so this is y is equal to negative 2 this is y is equal to negative 2 so it's clearly clearly shifted down so it's we're going to have so I'll actually I'll talk in a second about what type of an expression it might be but now also let's think about its amplitude its amplitude that's how far it might get away from the midline we see here it went 3 above the midline going from negative 2 to 1 and went 3 above the midline at the maximum point and it can also go 3 below the midline at the minimum point so this thing clearly has an amplitude of 3 so amplitude amplitude of 3 so immediately we can say well look this is going to have a form something like f of X is equal to the amplitude 3 3 it we haven't figured out yet whether we this is going to be a cosine function or a sine function so I'll write cosine first cosine maybe some coefficient times X plus the midline the midline we already figured out was minus 2 or negative 2 so it could take that form or it could take f of X is equal to is equal to 3 times it could be sine of X or sine of some coefficient times X sine of KX minus 2 plus the midline so minus 2 so how do we figure out which of these are well let's just think about the behavior of this function when X is equal to 0 when X is equal to 0 if this is KX then the input into this into the cosine is going to be 0 cosine of 0 is 1 whether you're talk about degrees or radians cosine of 0 is 1 while sine of 0 so if x is 0 K times 0 is going to be 0 sine of 0 is 0 so what's this thing doing when X is equal to 0 well when X is equal to 0 we are at the midline if we're at the midline that means that all of this stuff right over here evaluated to 0 so since when x equals 0 all of this stuff evaluated is 0 we can rule out the cosine function when x equals 0 here this stuff doesn't evaluate to 0 so we can rule out this one right over there and so we are left with this and we just really need to figure out what is what is this what could be what could this constant actually be and to think about that let's look at the period of this of this function so to go from to go from and we could we could let's see if we went from this point where we intersect the midline we go this point to intersect the midline and we have a positive slope the next point that we do that is right over here so our period is 8 our period is 8 so what coefficient could we have here to make a period make the period of this thing be equal to 8 well let's just remind ourselves what the period of sine of X is so the period of sine of X so I'll write period right over here is 2 pi 2 pi you increase your angle by 2 pi radians or decrease it you're back at the same point on the unit circle so what would be the period of sine of KX well now your X your input is increasing K times faster so you're going to get to the same point K times faster so your period is going to be is going to be 1 K as as long so now your period is going to be 2 pi over K notice you're increasing your argument as x increases your argument into the sine function is increasing K times as fast you're multiplying it by K so your period is going to be short is going to take you less distance to get for the whole argument for to get to the same point on the unit circle so let's think about it this way now so if we wanted to say two pi over K is equal to 8 so if 2 PI over K is equal to 8 well what is our K well we could take the reciprocal of both sides we get K over 2 pi is equal to 1 over 8 multiply both sides by 2 pi and we get we get K is equal to let's see this is 1 this is 4 K is equal to PI over 4 and we are done and you can verify that by trying out some of these points right over here that this function is equal to 3 sine of PI over 4 X minus 2