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Sal is given a drawing with four graphs and four formulas of square-root functions. He uses transformations to match each formula with its appropriate graph. Created by Sal Khan.
Video transcript
Match each function with its graph. And we have graph D, A, B, and C. And let's just start with the graph of B because, actually, this one looks the closest to the square root of x, which would look something like that. But it's clearly shifted. And it's flipped over the horizontal axis. The fact that it's flipped over the square horizontal axis, that means that we're taking the negative square root. So it's going to be one of these two cases right over here. We're shifting it down by 2. So we should have a negative 2, and both of these cases have a negative 2 here. And we're shifting it relative to the square root of x. We're shifting it to the right by 1. So if we're shifting to the right by 1, what we see under the radical should be x minus 1. And both of these have an x minus 1 on it. So which of these is B? And which of these is C? And I encourage you to think about that. Pause the video if you'd like. Well, the difference between these two is this one has a scaling factor of 2 here while this one doesn't. And of course, this is going to turn into more and more negative values. This is going to be getting negative faster. And you see that graph C here gets negative faster. And you could even try some points. In either case, when x is equal to 1, in either case, what we have under the radical becomes 0. x minus 1 is 0. x minus 1 is 0 when x is equal to 1. And in either case, our y value's negative 2. But then, you see, as x increases, this one right over here gets negative twice as fast. You see that right over here. This one right over here, y has gone from negative 2 to negative 4 here. For graph C, y has gone from negative 2 to negative 6. So it's gone down by 4. This has only gone down by 2. So it's pretty clear that graph B corresponds to the one that doesn't have the 2 out front. So this is graph B right over here. And graph C corresponds to the one that has a negative 2 out front. So let me try that. The negative 2 out front. So now we just have to think about these two equations and match them to these two graphs. And so both of these, they haven't been flipped around the horizontal axis. They've been flipped around the vertical axis. And that's why whatever we have under the radical we've essentially taken the negative of it. And actually, we could figure out which ones these are just by looking at how much they've been shifted in the y direction relative to the square root of x-- which would look something like that. I know I can't draw on this one right over here. Well, this one has been shifted up by 2. D has been shifted up by 2. This one over here, g of x has been shifted up by 2. And you could also see that it's been shifted to the left by 1. If you're shifting to the left by 1, normally, under the radical, you would have x plus 1. But then, we flipped it over around the vertical axis. And so that takes the negative of that. That's why it's negative x minus 1. You could view this as the negative of x plus 1. So either way, that is D. Throw that under the D. And just deductive reasoning tells us A. This is A, and it makes sense. We see it has shifted up by 1. So it's plus 1. And we could view this as the negative of x plus 4. So it's been shifted 4 to the left. And we see that is definitely the case relative to the square root of x. We got it right.