Example: Graphing y=-cos(π⋅x)+1.5 | Trigonometry | Algebra 2 | Khan Academy

We're told to graph ( y ) is equal to negative cosine of ( \pi ) times ( x ) plus ( 1.5 ) in the interactive widget, so pause this video and think about how you would do that.

And just to explain how this widget works, if you're trying to do it on Khan Academy, this dot right over here helps to find the midline. You can move that up and down, and then this one right over here is a neighboring extreme point, so either a minimum or a maximum point.

So there's a couple of ways that we can approach this. First of all, let's just think about what cosine of ( \pi x ) looks like, and then we'll think about what the negative does in the plus ( 1.5 ).

So cosine of ( \pi x ), when ( x ) is equal to zero, ( \pi ) times zero is just going to be zero. Cosine of zero is equal to one, and if we're just talking about cosine of ( \pi x ), that's going to be a maximum point when you hit one. Just cosine of ( \pi x ) would oscillate between ( 1 ) and ( -1 ).

And then what would its period be if we're talking about cosine of ( \pi x )? Well, you might remember one way to think about the period is to take ( 2\pi ) and divide it by whatever the coefficient is on the ( x ) right over here. So ( 2\pi ) divided by ( \pi ) would tell us that we have a period of ( 2 ).

And so how do we construct a period of ( 2 ) here? Well, that means that as we start here at ( x = 0 ), we're at ( 1 ). We want to get back to that maximum point by the time ( x ) is equal to ( 2 ).

So let me see how I can do that. If I were to squeeze it a little bit, that looks pretty good. And the reason why I worked on this midline point is I liked having this maximum point at ( 1 ) when ( x ) is equal to ( 0 ) because we said cosine of ( \pi ) times ( 0 ) should be equal to ( 1 ).

So that's why I'm just manipulating this other point in order to set the period right, but this looks right. We're going from this maximum point, we're going all the way down and then back to that maximum point, and it looks like our period is indeed ( 2 ).

So this is what the graph of cosine of ( \pi x ) would look like. Now what about this negative sign? Well, the negative would essentially flip it around, so instead of whenever we're equaling ( 1 ), we should be equal to ( -1 ).

And every time we're equal to ( -1 ), we should be equal to ( 1 ). So what I could do is I could just take that and then bring it down here, and there you have it, I flipped it around. So this is the graph of ( y = -\cos(\pi x) ).

And then last but not least, we have this plus ( 1.5 ), so that's just going to shift everything up by ( 1.5 ). So I'm just going to shift everything up by ( 1.5 ) and shift it up by ( 1.5 ), and there you have it.

That is the graph of ( -\cos(\pi x) + 1.5 ), and you can validate that that's our midline. We're still oscillating one above and one below the negative sign. When cosine of ( \pi ) times zero, that should be ( 1 ), but then you take the negative, we get to ( -1 ). You add ( 1.5 ) to that, you get to positive ( 0.5 ), and so this is all looking quite good.