Graphing square and cube root functions | Algebra 2 | Khan academy

We're told the graph of ( y ) is equal to (\sqrt{x}) is shown below. Fair enough, which of the following is the graph of ( y ) is equal to ( 2\times\sqrt{-x}-1 )? They give us some choices here, and so I encourage you to pause this video and try to figure it out on your own before we work through this together.

All right, now let's work through this together. The way that I'm going to do it is I'm actually going to try to draw what the graph of ( 2\times\sqrt{-x}-1 ) should look like, and then I'll just look at which of the choices is closest to what I drew. The way that I'm going to do that is I'm going to do it step by step.

So we already see what ( y = \sqrt{x} ) looks like. But let's say we just want to build up. So let's say we want to now figure out what is the graph of ( y = \sqrt{-x} )? Instead of an ( x ) under the radical sign, let me put a (-x) under the radical sign. What would that do to it? Well, whatever was happening at a certain value of ( x ) will now happen at the negative of that value of ( x ).

So ( \sqrt{x} ) is not defined for negative numbers; now this one won't be defined for positive numbers. The behavior that you saw at ( x = 2 ) you would now see at ( x = -2 ). The behavior that you saw at ( x = 4 ) you will now see at ( x = -4 ) and so on and so forth. So ( y = \sqrt{-x} ) is going to look like this; you've essentially flipped it over the ( y )-axis.

All right, so we've done this part. Now let's scale that; now let's multiply that by ( 2 ). So what would ( y = 2\times\sqrt{-x} ) look like? Well, it would look like this red curve, but at any given ( x ) value, we're going to get twice as high. So at ( x = -4 ), instead of getting to ( 2 ), we're now going to get to ( 4 ). At ( x = -9 ), instead of getting to ( 3 ), we're now going to get to ( 6 ).

Now, at ( x = 0 ), we're still going to be at zero because ( 2\times 0 = 0 ), so it's going to look like that, something like that. So that's ( y = 2\times\sqrt{-x} ). Then last but not least, what will ( y ) – let me do that in a different color – what will ( y = 2\times\sqrt{-x}-1 ) look like?

Well, whatever ( y ) value we were getting before, we're now just going to shift everything down by ( 1 ). So if we were at ( 6 ) before, we're going to be at ( 5 ) now. If we were at ( 4 ) before, we're now going to be at ( 3 ). If we were at ( 0 ) before, we're now going to be at ( -1 ). And so our curve is going to look something like that.

So let's look for let's see which choices match that. So let me scroll down here, and both ( c ) and ( d ) kind of look right. But notice right at ( 0 ), we wanted to be at ( -1 ); so ( d ) is exactly what we had drawn. At ( -9 ), we're at ( 5 ); at ( -4 ), we're at ( 3 ); and at ( 0 ), we're at ( -1 ), exactly what we had drawn.

Let's do another example. So here this is a similar question. Now they graphed the cube root of ( x ); ( y ) is equal to the cube root of ( x ) and then they say which of the following is the graph of this business? And they give us choices again. So once again, pause this video and try to work it out on your own before we do this together.

All right, now let's work on this together, and I'm going to do the same technique. I'm just going to build it up piece by piece. So this is already ( y = \sqrt[3]{x} ). So now let's build up on that. Let's say we want to now have an ( x + 2) under the radical sign. So let's graph ( y = \sqrt[3]{x + 2} ).

Well, what this does is it shifts the curve to the left, and we've gone over this in multiple videos before. So we are now here and you can even try some values out to verify that at ( x = 0 ) – or actually, let me put it this way – at ( x = -2 ), you're going to take the cube root of ( 0 ), which is right over there. So we hit shifted two to the left, to look something like this.

Now let's build up on that. Let's multiply this times a negative, so ( y = -\sqrt[3]{x + 2} ). What would that look like? Well, if you multiply your whole expression – or the whole graph, or the whole function – by a negative, you're going to flip it over the horizontal axis.

So it is now going to look like this. Whatever ( y ) value you're going to get before for a given ( x ), you're now getting the opposite, the negative of it. So it's going to look like that, something like that. So that is ( y = -\sqrt[3]{x + 2} ).

Then last but not least, we are going to think about – and I'm searching for an appropriate color; I haven't used orange yet – ( y = -\sqrt[3]{x + 2} + 5 ). So all that's going to do is take this last graph and shift it up by ( 5 ). Whatever ( y ) value I'm going to get before, I'm now going to get ( 5 ) higher.

So ( 5 ) higher; let's see I was at ( 0 ) here, so now I'm going to be at ( 5 ) here. So it's going to look something like, something like that. I know I'm not drawing it perfectly, but you get the general idea. Now let's look at the choices, and I think the key point to look at is this point right over here that in our original graph was at ( (0, 0) ); now it is going to be at ( (-2, 5) ).

So let's look for it, and it also should be flipped. So on the left-hand side, we have the top part, and on the right-hand side, we have the part that goes lower. So let's see. So ( a, c, ) and ( b ) all have the left-hand side as the higher part, and then the right-hand side being the lower part. But we wanted this point to be at ( (-2, 5) ).

( a ) doesn't have it there; ( b ) doesn't have it there; ( d ) we already said goes in the wrong direction; it's increasing. So let's see; ( (-2, 5) ) is indeed what we expected. This is pretty close to what we had drawn on our own, so choice ( c ).