Example: Graphing y=3⋅sin(½⋅x)-2 | Trigonometry | Algebra 2 | Khan Academy

So we're asked to graph ( y ) is equal to three times sine of one half ( x ) minus two in the interactive widget. And this is the interactive widget that you would find on Khan Academy. It first bears mentioning how this widget works. So this point right over here helps you define the midline, the thing that you can imagine your sine or cosine function oscillates around.

Then, you also define a neighboring extreme point, either a maximum or a minimum point, to graph your function. So let's think about how we would do this. Like always, I encourage you to pause this video and think about how you would do it yourself.

But the first way I like to think about it is: What would a regular, just, if I if this just said ( y ) is equal to sine of ( x )? How would I graph that? Well, sine of 0 is 0, sine of ( \frac{\pi}{2} ) is 1, and then sine of ( \pi ) is 0 again. So this is what just regular sine of ( x ) would look like.

But let's think about how this is different. Well, first of all, it's not just sine of ( x ); it's sine of one half ( x ). So what would be the graph of just sine of one half ( x )? Well, one way to think about it (there's actually two ways to think about it) is a coefficient right over here on your ( x ) term that tells you how lar— how fast the thing that's being inputted into sine is growing. And now it's going to grow half as fast.

So one way to think about it is your period is now going to be twice as long. Instead of getting to this next maximum point at ( \frac{\pi}{2} ), you're going to get there at ( \pi ). You could test that: when ( x ) is equal to ( \pi ), this will be one half ( \pi ); sine of one half ( \pi ) is indeed equal to 1.

Another way to think about it is you might be familiar with the formula (although I always like you to think about where these formulas come from) that to figure out the period of a sine or cosine function, you take ( 2\pi ) and you divide it by whatever this coefficient is. So ( 2\pi ) divided by one half is going to be ( 4\pi ).

You can see the period here: we go up, down, and back to where we were over ( 4\pi ). That makes sense, because if you just had a 1 coefficient here, your period would be ( 2\pi ). ( 2\pi ) radians makes one circle around the unit circle, which is one way to think about it.

So right here we have the graph of sine of one half ( x ). Now, what if we wanted to instead think about three times the graph of sine of one half ( x ), or three sine one half ( x )? Well then our amplitude is just going to be three times as much.

So instead of our maximum point going from— instead of our maximum point being at 1, it will now be at 3. Or another way to think about it is we're going three above the midline and three below the midline. So this right over here is the graph of three sine of one half ( x ).

Now we have one thing left to do, and this is this minus two. So this minus two is just going to shift everything down by two. We just have to shift everything down. So let me shift this one down by two, and let me shift this one down by two.

And so there you have it. Notice our period is still ( 4\pi ); our amplitude, how much we oscillate above or below the midline, is still 3, and now we have this minus two. Another way to think about it: when ( x ) is equal to zero, this whole first term is going to be zero, and ( y ) should be equal to negative two, and we're done.