Algebra 2 (Eureka Math/EngageNY)
- Sinusoidal function from graph
- Construct sinusoidal functions
- Trig word problem: modeling daily temperature
- Trig word problem: modeling annual temperature
- Modeling with sinusoidal functions
- Trig word problem: length of day (phase shift)
- Modeling with sinusoidal functions: phase shift
Sal finds the equation of a sinusoidal function from its graph where the minimum point (-2,-5) and the maximum point (2,1) are highlighted. Created by Sal Khan.
Write the equation of the function f of x graphed below. And so we have this clearly periodic function. So immediately you might say, well, this is either going to be a sine function or a 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 1. At the minimum points, it's a value of y is equal to negative 5. So halfway between those, 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 is y is equal to negative 2. So it's clearly shifted down. 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, it 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 immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether 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 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 the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking 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 to 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 could this constant actually be? And to think about that, let's look at the period of this function. Let's see. If we went from this point-- where we 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. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let us 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 2pi. 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 1/k'th as long. So now your period is going to be 2 pi over k. Notice, 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. It's going to take you less distance for the whole argument to get to the same point on the unit circle. So let's think about it this way-- so if we wanted to say 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 1/8. Multiply both sides by 2pi. And we get k is equal to-- let's see. This is 1. This is 4. k is equal to pi/4. And we are done. And you can verify that by trying out some of these points right over here. This function is equal to 3 sine of pi over 4x minus 2.