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## Algebra 2

# Example: Graphing y=3⋅sin(½⋅x)-2

CCSS.Math: ,

Sal graphs y=3⋅sin(½⋅x)-2 by thinking about the graph of y=sin(x) and analyzing how the graph (including the midline, amplitude, and period) changes as we perform function transformations to get from y=sin(x) to y=3⋅sin(½⋅x)-2. Created by Sal Khan.

## Video transcript

- [Instructor] So we're asked to graph y is equal to three times sine of 1/2x minus 2 in the interactive widget. And this is the interactive widget that you would find on Khan Academy. And it first bears mentioning
how this widget works. So this point right over here, it helps you define the midline, the thing that you could imagine your sine or cosine
function oscillates around, and 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, and 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 this just said y is equal to sine of x,
how would I graph that? Well, sine of 0 is 0. Sine of pi over 2 is 1. And then sine of pi is 0, again. And 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 1/2x. So what would be the graph
of just sine of 1/2x? 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 fast the thing that's being
inputted into sine is growing. And now it's going to grow half as fast. And so one way to think about it is your period is now going
to be twice as long. So one way to think about it is instead of getting to this next
maximum point at pi over 2, you're going to get there at pi. And you could test that. If you at, when x is equal to pi, this will be 1/2 pi, sine of 1/2 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 1/2
is going to be 4 pi. And you could see the period here, we go up, down, and back
to where we were over 4 pi. And that makes sense, because if you just had
a 1 coefficient here, your period would be 2 pi, 2 pi radians. You make one circle around the unit circle is one way to think about it. So right here we have the
graph of sine of 1/2x. Now what if we wanted to, instead, think about 3 times the
graph of sine of 1/2x, or 3 sine 1/2x? Well then our amplitude's just going to be three times as much. And 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 3 above the midline
and 3 below the midline. So this right over here is
the graph of 3 sine of 1/2x. Now we have one thing left to do, and this is this minus 2. So this minus 2 is just going
to shift everything down by 2. So we just have to shift everything down. So let me shift this one down by 2 and let me shift it this one down by 2. 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 2. Another way to think about
it, when x is equal to 0, this whole first term is going to be 0, and y should be equal to negative 2? And we're done.